Euler's work VIII

Nice Euler sum involving the Apéry's constant $\zeta(3)$ and the Catalan's constant $\beta(2)$

Today we show the proof of this nice Euler sum proposed by @Ali39342137 $$\sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = \frac{\pi^2 \ln\left(2\right)}{8}-\frac{\pi}{2}\beta(2) + \frac{21}{32}\zeta(3)$$ The proof will rely on some properties of the digamma function and the use of Fourier series.

Proof:

Recall the following integral for the difference of digamma functions (proof in Appendix 1): $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$ Let $n \in \mathbb{N}$. If we put $\mu -1 = n$ $$\int_{0}^{1} \frac{x^{n}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{n}{2}+1\right)-\psi\left(\frac{n}{2}+\frac{1}{2}\right)\right]$$ Hence $$\int_{0}^{1} \frac{x^{n}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{n}{2}+1\right)+\gamma-\left(\psi\left(\frac{n-1}{2}+1\right)+\gamma\right)\right] = \frac{1}{2}\left[H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right]$$ $$\Longrightarrow H_{\frac{n}{2}}-H_{\frac{n-1}{2}} = 2\int_{0}^{1} \frac{x^{n}}{1+x} dx$$ Therefore $$\Longrightarrow \frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = 2\int_{0}^{1} \frac{1}{1+x} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n dx$$ Hence $$\begin{align*} S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = &\sum_{n=0}^{\infty} 2\int_{0}^{1} \frac{1}{1+x} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n dx\\ = &2\int_{0}^{1} \frac{1}{1+x} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^n\right) dx \quad \textrm{ (Fubini-Tonelli)} \\ = &4\int_{0}^{1} \frac{1}{1+w^2} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}w^{2n+1}\right) dw \quad \left(w^2 \mapsto x\right) \end{align*}$$ We proved previously (proof in Appendix 2): $$\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}w^{2n+1} = \frac{1}{2}\arctan(w)\ln(1+w^2)$$ Hence $$S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = 2\int_{0}^{1} \frac{\arctan(w)\ln(1+w^2)}{1+w^2} dw$$ We just have to find the integral in the right hand side $$\begin{align*} S = 2\int_{0}^{1} \frac{\arctan(w)\ln(1+w^2)}{1+w^2} dw =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(1+\tan^2(\theta)\right)d\theta \quad \left(\theta = \arctan(w)\right) \\ =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\frac{\cos^2(\theta)}{\cos^2(\theta)}+\frac{\sin^2(\theta)}{\cos^2(\theta)}\right)d\theta \\ =& 2\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\frac{1}{\cos^2(\theta)}\right)d\theta \\ =& -4\int_{0}^{\frac{\pi}{4}} \theta \ln\left(\cos(\theta)\right)d\theta \\ =& -\int_{0}^{\frac{\pi}{2}} t \ln\left(\cos\left(\frac{t}{2}\right)\right)dt \quad \left( t \mapsto 2\theta\right) \\ =& -\int_{0}^{\frac{\pi}{2}} t\ln\left(\frac{2}{2}\cos\left(\frac{t}{2}\right)\right)dt \\ =& \ln(2)\int_{0}^{\frac{\pi}{2}} t dt - \int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt \\ =& \frac{\pi^2 \ln\left(2\right)}{8} -\underbrace{\int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt}_{I} \\ \end{align*}$$ We can expand the integrand in $I$ with the following Fourier series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kx)}{k} = \ln\left(2\cos\left(\frac{x}{2}\right)\right) \quad -\pi\lt x\lt\pi$$ Therefore $$\begin{align*} I = \int_{0}^{\frac{\pi}{2}} t \ln\left(2\cos\left(\frac{t}{2}\right)\right)dt =& \int_{0}^{\frac{\pi}{2}} t \sum_{k=1}^{\infty} \frac{(-1)^{k-1}\cos(kt)}{k}dt\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \int_{0}^{\frac{\pi}{2}} t\cos(kt)dt \quad \textrm{ (Fubini-Tonelli)}\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3} \int_{0}^{\frac{k\pi}{2}} s\cos(s)ds \quad (s \mapsto kt) \\ \stackrel{IBP}{= }&\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\left[ s\sin(s)\Big|_{0}^{\frac{k\pi}{2}} - \int_{0}^{\frac{k\pi}{2}}\sin(s) ds\right] \\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\left[ \left(\frac{k\pi}{2}\right)\sin\left(\frac{k\pi}{2}\right) + \cos\left(\frac{k\pi}{2}\right)-1\right] \\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin\left(\frac{k\pi}{2}\right)}{k^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}\cos\left(\frac{k\pi}{2}\right)}{k^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3} \\ \end{align*}$$ Note that $$\left(\sin\left(\frac{k\pi}{2}\right)\right)_{k} = \left(1,0,-1,0,1,0,-1,...\right)$$ $$\left(\cos\left(\frac{k\pi}{2}\right)\right)_{k} = \left(0,-1,0,1,0,-1,0,...\right)$$ Therefore $$\begin{align*} I =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}\sin\left(\frac{k\pi}{2}\right)}{k^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}\cos\left(\frac{k\pi}{2}\right)}{k^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2} + \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k)^3} - \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}\\ =& \frac{\pi}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2} - \frac{7}{8}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^3}\\ =& \frac{\pi}{2}\beta(2) - \frac{7}{8}\eta(3) \quad \textrm{ (eta and beta dirichlet functions)}\\ \end{align*}$$ Finally, using the identity: $$\eta(v) = (1-2^{1-v})\zeta(v)$$ $$I = \frac{\pi}{2}\beta(2) - \frac{21}{32}\zeta(3)$$ Therefore $$\boxed{S = \sum_{n=0}^{\infty}\frac{(-1)^{n-1}H_{2n}}{2n+1}\left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right) = \frac{\pi^2 \ln\left(2\right)}{8}-\frac{\pi}{2}\beta(2) + \frac{21}{32}\zeta(3)}$$

Appendix 1

We show the proof of this nice integral representation $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$

Proof

$$\begin{align*} \int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx =& \int_{0}^{1} \sum_{j=0}^{\infty} (-1)^jx^jx^{\mu-1} dx\\ =& \sum_{j=0}^{\infty} (-1)^j\int_{0}^{1} x^{j+\mu-1} dx\\ =& \sum_{j=0}^{\infty} \frac{(-1)^j}{j+\mu} \\ =& \sum_{j=0}^{\infty}\left[ \frac{1}{2j+\mu} - \frac{1}{2j+1+\mu}\right] \\ =& \frac{1}{2}\sum_{j=0}^{\infty}\left[ \frac{1}{j+\frac{\mu}{2}} - \frac{1}{j+\frac{1+\mu}{2}}\right] \\ =& \frac{1}{4}\sum_{j=0}^{\infty}\frac{1}{\left(j+\frac{\mu}{2}\right)\left(j+\frac{1+\mu}{2}\right)} \end{align*}$$ The difference between two digamma functions may be expressed as $$\psi(x)-\psi(y) = (x-y)\sum_{j=0}^{\infty} \frac{1}{(j+x)(j+y)}$$ If we put $\displaystyle x = \frac{\mu}{2}, y = \frac{\mu+1}{2}$ $$\int_{0}^{1} \frac{x^{\mu-1}}{1+x} dx = \frac{1}{4}\sum_{j=0}^{\infty}\frac{1}{\left(j+\frac{\mu}{2}\right)\left(j+\frac{1+\mu}{2}\right)} = \frac{1}{2}\left[\psi\left(\frac{\mu+1}{2}\right)-\psi\left(\frac{\mu}{2}\right)\right]$$

Appendix 2

$$\sum_{n=0}^{\infty} \frac{(-1)^{n-1}H_{2n}}{2n+1}x^{2n+1} = \frac{1}{2}\arctan(x)\ln(1+x^2)$$

Proof

One of the special cases of the Gauss hypergometric is the following: $${}_{2}F_{2} \left(a,\frac{1}{2}+a;\frac{3}{2};z^{2}\right)=\frac{1}{(2-4a)z}\left((1+z)^{1-2a}-(1-z)^{1-2a}\right)$$ If we put $\displaystyle a = \frac{t}{2} , z = ix$: $${}_{2}F_{2} \left(\frac{t}{2},\frac{1}{2}+\frac{t}{2};\frac{3}{2};-x\right)=\frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)$$ From the definition of $_{2}F_{1}$: $$\begin{align*} {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2}, -x\right) =& \sum_{n=0}^{\infty} \frac{\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{\left(\frac{3}{2}\right)_{n}} \frac{(-1)^nx^n}{n!} \\ =& \sum_{n=0}^{\infty} \frac{4^n\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{4^n(1)_{n}\left(\frac{3}{2}\right)_{n}} (-1)^nx^n \\ =& \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n \end{align*}$$ Therefore $$\sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n = \frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)$$ Differentiating with respect to $t$: $$\frac{d}{dt}\Big|_{t=1} \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^nx^n = \sum_{n=0}^{\infty} \frac{(1)_{2n}\left[\psi(2n+1)+\gamma\right]}{(2)_{2n}} (-1)^n= \sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n}$$ Hence $$\begin{align*} \sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n} =& \frac{d}{dt}\Big|_{t=1} \left[\frac{1}{2(1-t)ix}\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)\right]\\ =& \lim_{t \to 1} \left[\underbrace{-\frac{i\left((1+ix)^{1-t}-(1-ix)^{1-t}\right)}{2(1-t)^2x}}_{=0} -\frac{i\left((1-ix)^{1-t}\ln(1-ix)-(1+ix)^{1-t}\ln(1+ix)\right)}{2(1-t)x}\right]\\ =& \lim_{t \to 1} -\frac{i\left((1-ix)^{1-t}\ln(1-ix)-(1+ix)^{1-t}\ln(1+ix)\right)}{2(1-t)x}\\ =& \frac{i\left(\ln^2(1+ix)-\ln^2(1-ix)\right)}{4x} \quad \textrm{ (L'Hôpital's rule) } \\ =& \frac{i\left(\ln(1+ix)-\ln(1-ix)\right)\left(\ln(1+ix)+\ln(1-ix)\right)}{4x}\\ =& \frac{i\left(\ln(1+ix)-\ln(1-ix)\right)\left(\ln(1+ix)+\ln(1-ix)\right)}{4x}\\ =& \frac{1}{2x}\left[-\frac{i}{2}\ln\left(\frac{1-ix}{1+ix}\right)\ln(1+x^2)\right]\\ =& \frac{1}{2x}\arctan(x)\ln(1+x^2) \end{align*}$$ Therefore we can conclude $$\sum_{n=0}^{\infty} \frac{(-1)^{n}H_{2n}}{2n+1}x^{2n+1} = \frac{1}{2}\arctan(x)\ln(1+x^2)$$

Integral of the day XVIII

Integral involving a discrete Laplace transform

Today we evaluate the following integral posted by @integralsbot $$\int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = \frac{\pi}{4e^{\alpha\beta}\sinh(\beta) \cosh^{1-\alpha}(\beta)} \quad \alpha >-1, \beta>0$$ The trick to evaluate this integral is the use of a discrete Laplace transform.

Proof.

This is the discrete Laplace transform that we need (Proof in the appendix): $$1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 } \tag{1}$$ Therefore $$\begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = & \frac{1}{2}\int_{0}^{\pi} \frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\cos (2\beta) - \cos (w)} dw \quad (w \mapsto 2x) \\ =& \frac{1}{4}\int_{-\pi}^{\pi} \frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\cos (2\beta) - \cos( w)} dw \quad \textrm{ (the integrand is even)}\\ =& \frac{1}{4}\int_{-\pi}^{\pi} \frac{\sinh (2\beta)}{\cos (2\beta) - \cos( w)}\frac{\cos \left( \frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\sinh (2\beta )} dw \\ =& \frac{1}{4}\int_{-\pi}^{\pi}\left(1+2\sum_{k=1}^{\infty} e^{-k2\beta}\cos (kw) \right)\frac{\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right)}{\sinh (2\beta) } dw \quad \textrm{ from (1) } \\ =& \frac{1}{4\sinh (2\beta)}\underbrace{\int_{-\pi}^{\pi}\cos \left(\alpha \frac{w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw}_{J} + \frac{1}{2\sinh (2\beta)}\sum_{k=1}^{\infty} e^{-k 2\beta}\underbrace{\int_{-\pi}^{\pi} \cos(kw) \cos \left(\alpha \frac{w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw}_{K} \tag{2} \\ \end{align*}$$ Lets start with $J$. Using the fact that $\Re(e^{ix}) = \cos(x)$: $$J = \int_{-\pi}^{\pi}\cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw = \Re\left(\int_{-\pi}^{\pi} e^{i\frac{\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw \right)$$ If we make the substitution $$e^{iw} = z$$ $$dw = \frac{dz}{zi}$$ we transform the integral in a contour integral round the unit complex circle: $$\begin{align*} J = \Re\left(\int_{-\pi}^{\pi} e^{i\frac{\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw \right) =& \Re\left(\frac{1}{i} \oint_{|z|=1} \frac{z^{\frac{\alpha}{2}}(z+1)^{\alpha} }{2^{\alpha}z^{\frac{\alpha}{2}}z} dz\right)\\ =& \Re\left(\frac{1}{2^{\alpha}i} \oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz\right) \end{align*}$$ The function $\displaystyle f(z) = \frac{(z+1)^{\alpha} }{z}$ has two singularities at $z=0$ and $z=-1$. However, since $\alpha >-1$, $z=-1$ is a removable singularity and its residue is zero. Therefore we can use the binomial theorem to find the residue at $z=0$: $$(1+z)^{\alpha} = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j} \quad |z|\lt 1$$ $$f(z) = \frac{(z+1)^{\alpha} }{z} = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j-1}$$ The residue is the coefficient of $z^{-1}$ which means that $j=0$. By the Residue theorem: $$\oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz = 2\pi i \operatorname{Res}(f,0) = 2\pi i \binom{\alpha}{0} = 2\pi i$$ Hence $$J = \Re\left(\frac{1}{2^{\alpha}i} \oint_{|z|=1} \frac{(z+1)^{\alpha} }{z} dz\right) = \Re\left( \frac{\pi}{2^{\alpha-1}} \right) = \frac{\pi}{2^{\alpha-1}} \tag{3}$$ Now for $K$: $$\begin{align*} K = \int_{-\pi}^{\pi} \cos(kw) \cos \left(\frac{\alpha w}{2}\right) \cos^{\alpha} \left(\frac{w}{2}\right) dw = \Re\left( \int_{-\pi}^{\pi} \cos(kw) e^{\frac{i\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw\right) \end{align*}$$ Making the same substitution as before: $$e^{iw} = z$$ $$dw = \frac{dz}{zi}$$ $$\begin{align*} K = \Re\left( \int_{-\pi}^{\pi} \cos(kw) e^{\frac{i\alpha w}{2}} \cos^{\alpha} \left(\frac{w}{2}\right) dw\right) = \Re\left( \frac{1}{2^{\alpha+1}i}\oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz\right) \end{align*}$$ The function $$g(z)= \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}}$$ has two singularities at $z=-1$ and $z=0$ but again the singularity at $z=-1$ is a removable singularity and its residue is zero. To find the residue at $z=0$ we will use the binomial expansion of $(z+1)^{\alpha}$. $$g(z)= \frac{(z^{2k}+1)}{z^{k+1}}\sum_{j=0}^{\infty} \binom{\alpha}{j} z^j = \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j+k-1} + \sum_{j=0}^{\infty} \binom{\alpha}{j} z^{j-k-1} \quad |z|\lt 1$$ Therefore the residue is the coefficient of $z^{-1}$. Note that the residue in the first sum is zero while the second sum attains the residue at $j=k$ $$\oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz = 2\pi i \operatorname{Res}(g,0) = 2\pi i \binom{\alpha}{k}$$ Therefore $$K = \Re\left( \frac{1}{4i}\oint_{|z|=1} \frac{(z^{2k}+1)(z+1)^{\alpha}}{z^{k+1}} dz\right) = \Re\left( \frac{\pi}{2^{\alpha}}\binom{\alpha}{k} \right) = \frac{\pi}{2^{\alpha}}\binom{\alpha}{k} \tag{4}$$ Hence $$\begin{align*} I = \int_{0}^{\frac{\pi}{2}} \frac{\cos \alpha x \cos^{\alpha} x}{\cos 2\beta - \cos 2x} dx =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta) } + \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}\sum_{k=1}^{\infty}\binom{\alpha}{k} e^{-k 2\beta} \quad \textrm{ from (2),(3),(4)} \\ =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}\sum_{k=0}^{\infty}\binom{\alpha}{k} e^{-k 2\beta} \quad \textrm{ (re-indexing) } \\ =& \frac{\pi}{2^{\alpha+1}\sinh (2\beta)}(1+e^{-2\beta})^{\alpha} \quad \textrm{ (binomial theorem) } \\ =& \frac{\pi}{2 e^{\alpha\beta}\sinh (2\beta)}\left(\frac{e^{\beta}+e^{-\beta}}{2}\right)^{\alpha} \\ =& \frac{\pi \cos^{\alpha}(\beta)}{2e^{\alpha\beta}\sinh (2\beta)} \\ =& \frac{\pi}{4 e^{\alpha\beta}\sinh(\beta)\cosh^{1-\alpha}(\beta)} \quad \left(\sinh(2\beta) = 2\sinh(\beta)\cosh(\beta)\right)\\ \end{align*}$$ Therefore, we can conclude $$\boxed{ \int_{0}^{\frac{\pi}{2}} \frac{\cos (\alpha x) \cos^{\alpha} (x)}{\cos (2\beta) - \cos (2x)} dx = \frac{\pi}{4e^{\alpha\beta}\sinh(\beta) \cosh^{1-\alpha}(\beta)} \quad \alpha >-1, \beta>0 }$$ Appendix.

Proposition. $$1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 }$$ Proof: $$\begin{align*} 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{-kt}e^{ikx} \right)\\ =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{k(ix-t)} \right)\\ =& 1+2\Re\left(\frac{1}{e^{t-ix}-1} \right)\\ =& 1+2\Re\left(\frac{e^{-t}}{e^{-ix}-e^{-t}} \right)\\ =& 1+2e^{-t}\Re\left(\frac{1}{\cos x-i\sin x- e^{-t}}\right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{(\cos x-e^{-t})^2 + \sin^2 x} \right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{e^{-2t}-2e^{t}\cos x + 1} \right)\\ =& 1+\Re\left(\frac{\cos x -e^{-t}+ i\sin x}{\cosh t-\cos x} \right)\\ =& \frac{\cosh t- \cos x}{\cosh t- \cos x}+\frac{\cos x-e^{-t}}{(\cosh t- \cos x )} \\ =& \frac{\sinh t}{\cosh t - \cos x} \end{align*}$$

Integral of the day XVII

Integral involving the complete Gamma function $\Gamma$

Today we evaluate this integral proposed by @BenriBot_ccbs: $$\int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x(1+x^2})}dx = -\sqrt{\pi}\Gamma^2\left(\frac{5}{4}\right)$$ Update: I added another solution which involves the complete Beta function and seems to be far easier.

Proof

Solution 1 (New). Using the complete beta function $B(x,y)$: $$\begin{align*} I = \int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x(1+x^2})}dx = & \sqrt{2}\int_{0}^{1} \frac{\ln(w)}{\sqrt{1-w^4}}dw \quad \left(w \mapsto \frac{1-x}{1+x} \right) \\ =& \sqrt{2}\int_{0}^{1} \frac{\left(\frac{d}{dt}\Big|_{t=0+} w^t \right)}{\sqrt{1-w^4}} dw \\ =&\frac{d}{dt}\Big|_{t=0+} \sqrt{2}\int_{0}^{1} \frac{ w^t}{\sqrt{1-w^4}} dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sqrt{2}\int_{0}^{1} w^t(1-w^4)^{-\frac{1}{2}} dw\\ =& \frac{d}{dt}\Big|_{t=0+} \frac{\sqrt{2}}{4} \int_{0}^{1} s^{\frac{t-3}{4}}(1-s)^{-\frac{1}{2}} ds\\ =& \frac{d}{dt}\Big|_{t=0+} \frac{\sqrt{2}}{4} B\left(\frac{t+1}{4}, \frac{1}{2} \right)\\ \end{align*}$$ Recall the derivative of the complete beta function: $$\frac{\partial}{\partial x} \mathrm{B}(x, y) = \mathrm{B}(x, y) \big(\psi(x) - \psi(x + y)\big)$$ Hence $$\begin{align*} I = \frac{d}{dt}\Big|_{t=0+} \frac{\sqrt{2}}{4} B\left(\frac{t+1}{4}, \frac{1}{2} \right) = & \lim_{t \to 0+} \frac{\sqrt{2}}{16} B\left(\frac{t+1}{4}, \frac{1}{2} \right)\left[ \psi\left(\frac{t+1}{4}\right)-\psi\left(\frac{t+3}{4} \right)\right]\\ =& \frac{\sqrt{2}}{16} B\left(\frac{1}{4}, \frac{1}{2} \right)\left[ \psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4} \right)\right]\\ =& \frac{\sqrt{2}}{16} \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}\left[ \psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4} \right)\right] \end{align*}$$ From the reflection formula $$\psi(1-x) - \psi(x) = \pi\cot(\pi x)$$ we have $$I= \frac{\sqrt{2}}{16} \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}\left[ \psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4} \right)\right] = -\frac{\sqrt{2}\pi^{\frac{3}{4}}}{16} \frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}$$ which is bascially the same as $-\sqrt{\pi}\Gamma^2\left(\frac{5}{4}\right)$.

Solution 2. Using ${}_{3}F_{2}$: $$\begin{align*} I = \int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x(1+x^2})}dx = & \sqrt{2}\int_{0}^{1} \frac{\ln(w)}{\sqrt{1-w^4}}dw \quad \left(w \mapsto \frac{1-x}{1+x} \right) \\ =& \sqrt{2}\int_{0}^{1} \frac{\left(\frac{d}{dt}\Big|_{t=0+} w^t \right)}{\sqrt{1-w^4}} dw \\ =&\frac{d}{dt}\Big|_{t=0+} \sqrt{2}\int_{0}^{1} \frac{ w^t}{\sqrt{1-w^4}} dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sqrt{2}\int_{0}^{1} \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-1)^n w^{4n} w^{t} dw\\ =& \frac{d}{dt}\Big|_{t=0+} \sqrt{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-1)^n \int_{0}^{1} w^{4n+t} dw\\ =& \frac{d}{dt}\Big|_{t=0+} \sqrt{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \frac{(-1)^n}{4n+t+1} \\ =& \sqrt{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \left[\frac{d}{dt}\Big|_{t=0+} \frac{(-1)^n}{4n+t+1} \right]\\ =& -\sqrt{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \frac{(-1)^n}{(4n+1)^2}\\ \end{align*}$$ Using the formula that relate the binomial coefficient to the Pochhammer polynomial (rising factorial): $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ and the reflection formula $$(-x)_{n} = (-1)^n(x-n+1)_{n}$$ Therefore $$\begin{align*} I = -\sqrt{2}\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \frac{(-1)^n}{(4n+1)^2} =& -\sqrt{2}\sum_{n=0}^{\infty} \frac{\left(-\frac{1}{2}-n+1\right)_{n}}{n!} \frac{(-1)^n}{(4n+1)^2} \quad \textrm{ from (1)}\\ =& -\sqrt{2}\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{n!} \frac{1}{(4n+1)^2} \quad \textrm{ from (2)}\\ \end{align*}$$ Recall the recursion formula for the argument of the rising factorial: $$n+x = \frac{x(x+1)_{n}}{(x)_{n}} \tag{3}$$ Hence $$\begin{align*} I = -\sqrt{2}\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{n!} \frac{1}{(4n+1)^2} = & -\frac{\sqrt{2}}{16}\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{n!} \frac{1}{(n+\frac{1}{4})^2}\\ =& -\sqrt{2}\sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}\left(\frac{1}{4}\right)_{n}\left(\frac{1}{4}\right)_{n}}{\left(\frac{5}{4}\right)_{n}\left(\frac{5}{4}\right)_{n}}\frac{1}{n!} \quad \textrm{ from (3)} \\ =& -\sqrt{2} {}_{3}F_{2} \left[{\frac{1}{2}, \frac{1}{4}, \frac{1}{4} \atop \frac{5}{4}, \frac{5}{4} }; 1 \right] \end{align*}$$ This is the Generalized hypergeometric function ${}_{3}F_{2}(a,b,c;d,e;z)$. This function satisfies the Dixon's well-poised sum: $${{}_{3}F_{2}}\left[{a,b,c\atop a-b+1,a-c+1};1\right]=\frac{\Gamma\left(\frac{1 }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left( \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1 \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)} \quad \Re(a-2b-2c)>-2$$ Putting $\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c=\frac{1}{4}$: $$\begin{align*} I = -\sqrt{2}{}_{3}F_{2} \left[{\frac{1}{2}, \frac{1}{4}, \frac{1}{4} \atop \frac{5}{4}, \frac{5}{4} }; 1 \right]=&-\sqrt{2} \frac{\Gamma\left(\frac{1 }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left( \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1 \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)} \\ =& -\sqrt{2} \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma(1)\Gamma(1)\Gamma(1)}\\ =& -\sqrt{2} \frac{\Gamma^3\left(\frac{5}{4}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{2}\right) }\\ = &-\sqrt{\pi}\Gamma^2\left(\frac{5}{4}\right) \end{align*}$$ Therefore, we can conclude $$\boxed{ \int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x(1+x^2})}dx = -\sqrt{\pi}\Gamma^2\left(\frac{5}{4}\right)}$$ which is basically the same result as before

Infinite products

Product related to a series expansion of $\cot(\pi x)$

Today we show the proof of this nice result proposed by @replicamethod $$\prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \frac{2\pi}{3\sqrt{3}}$$ The proof will rely on the expansion in zeta numbers of the $\cot(\pi x)$ function.

Proof

Consider the infinite product $$A = \prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \prod_{k=1}^{\infty} \frac{k^2}{k^2-\frac{1}{9}} = \prod_{k=1}^{\infty} \frac{1}{1-\frac{1}{9k^2}} = \frac{1}{\prod_{k=1}^{\infty}1-\frac{1}{9k^2}}$$ Taking logarithm in both sides: $$\ln(A) = -\sum_{k=1}^{\infty} \ln\left(1-\frac{1}{9k^2}\right) = \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n9^nk^{2n}} = \underbrace{\sum_{n=1}^{\infty}\frac{1}{n9^n}\sum_{k=1}^{\infty}}_{\textrm{Rearrangement}}\frac{1}{k^{2n}} = \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n9^n}$$ Now using the following expansion (proof in the Appendix): $$\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n} x^{2n} = \ln(\pi x) + \ln(\csc \pi x)$$ If we put $\displaystyle x = \frac{1}{3}$ $$\sum_{n=1}^{\infty} \frac{\zeta(2n)}{n9^n} = \ln\left(\frac{\pi}{3} \right) + \ln\left(\frac{2}{\sqrt{3}}\right)$$ Hence $$\ln(A) = \ln\left(\frac{\pi}{3} \right) + \ln\left(\frac{2}{\sqrt{3}}\right) = \ln\left(\frac{2\pi}{3\sqrt{3}}\right)$$ $$\Longrightarrow A = \frac{2\pi}{3\sqrt{3}}$$ Hence $$\boxed{ \prod_{k=1}^{\infty} \frac{(3k)^2}{(3k-1)(3k+1)} = \frac{2\pi}{3\sqrt{3}}}$$

Appendix

Proposition: $$\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k}= \ln\left(\frac{\pi x}{\sin(\pi x)}\right) \quad 0\lt x\lt 1$$ Proof:

Recall the expansion in partial fractions of $\cot(x\pi)$ $$\cot(x \pi) = \frac{2}{\pi}\left[\frac{1}{2 x}+ \sum_{n=1}^{\infty} \frac{x}{(x^2-n^2)}\right] \quad x\neq \pm 1, \pm 2, \pm 3,..$$ From these we can obtain the exapnsion in zeta numbers. Suppose that $|x|\lt1$, using the first equation we have $$\begin{align*} \cot(x \pi) =& \frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{(n^2-x^2)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2\left(1-\left(\frac{x}{n}\right)^2\right)} \\ = &\frac{1}{\pi x}- \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{x}{n^2} \sum_{j=0}^{\infty} \left(\frac{x}{n}\right)^{2j}\\ =&\frac{1}{\pi x}- \frac{2}{\pi}\underbrace{\sum_{j=0}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^{2j+2}} x^{2j+1}}_{\textrm{Rearrangement}} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{j=0}^{\infty}\zeta(2j+2) x^{2j+1} \\ =&\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad (j = k-1)\\ \end{align*}$$ Therefore $$\boxed{ \cot(\pi x) =\frac{1}{\pi x}- \frac{2}{\pi}\sum_{k=1}^{\infty}\zeta(2k) x^{2k-1} \quad |x|\lt1 }$$ Integrating the expansion of $\pi \cot(\pi x)$: $$\int \pi \cot(\pi x) dx =\int \frac{1}{ x}dx- 2\sum_{k=1}^{\infty}\zeta(2k) \int x^{2k-1}dx + C$$ Given that $$\int \pi \cot(\pi x) dx = \ln(\sin(\pi x))+ C$$ Hence $$\ln(\sin(\pi x)) = \ln(x) -\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + C$$ $$\ln\left(\frac{\sin(\pi x)}{x}\right) = -\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + C$$ Taking the limit as $x \to 0$: $$\ln\left( \lim_{x \to 0 } \frac{\sin(\pi x)}{x}\right) = \underbrace{-\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} }_{=0}+ C$$ Given that $\displaystyle \lim_{x \to 0} \frac{\sin(\pi x) }{x} =\pi \Longrightarrow C = \ln(\pi)$ $$\therefore \ln(\sin(\pi x)) = \ln(x) - \sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k} + \ln(\pi)\Longrightarrow \ln\left(\frac{\pi x}{\sin(\pi x)}\right) = \sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k}$$ $$\boxed{\sum_{k=1}^{\infty}\zeta(2k)\frac{x^{2k}}{k}= \ln\left(\frac{\pi x}{\sin(\pi x)}\right) }$$

Integral of the day XV

Monstrous integrals posted by @infseriesbot

Today we show the proof of this triplet of very weird integrals posted by @infseriesbot here, here and here:

Let $|z|\lt1$ , then $$\int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx = \pi$$ $$\int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx = \pi$$ $$\int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx = \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right]$$ These results are consequence of the residue theorem and some properties of the hypergeometric functions. We previously solved similar integrals here and here using similar techniques.

Proof:

1. First integral

Let $$I = \int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx$$ First, note that the function $$f(x) = \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}$$ is an even function. Then, $$I = \frac{1}{2}\int_{-\pi}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx$$ Using the fact that $\Re (e^{i\varphi}) = \cos(\varphi)$: $$I = \Re\left( \frac{1}{2}\int_{-\pi}^{\pi} e^{\left(ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}} dx\right)$$ Making the following substitution: $$e^{ix} = w$$ $$\cos(x) = \frac{w+w^{-1}}{2}$$ $$\sin(x) = \frac{w-w^{-1}}{2i}$$ $$dx = \frac{dw}{wi}$$ Using the logarithmic definition of $\arctan$ and after a lengthy but easy calculation we can obtain: $$\begin{align*}\arctan\left(\frac{z\sin x}{1+z\cos x}\right) =& \arctan\left(\frac{i(z-w^2z)}{w^2z+2w+z}\right)\\ =& -\frac{i}{2}\ln\left(\frac{w^2z+w}{w+z}\right) \tag{1} \end{align*}$$ $$e^{\left(ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)} = e^{\frac{a}{2}\ln\left(\frac{w^2z+w}{w+z}\right)} = \left(\frac{w^2z+w}{w+z}\right)^{\frac{a}{2}}$$ Also note that $$(1+2z\cos x + z^2) = \frac{(w+z)(wz+1)}{w} \tag{2}$$ Then, our integral became a contour integral round the unit complex circle: $$I = \Re \left(\frac{1}{2}\oint_{|w|=1} \left(\frac{w^2z+w}{w+z}\right)^{\frac{a}{2}}\frac{w^{\frac{a}{2}}}{{(w+z)^{\frac{a}{2}}}(wz+1)^{\frac{a}{2}}wi} dw\right) = \Re \left(\frac{1}{2i}\oint_{|w|=1} \frac{w^{a-1}}{(w+z)^a} dw\right)$$ If $|w|=1>|z|$, we can expand the denominator with the binomial theorem: $$\Re \left(\frac{1}{2i}\oint_{|w|=1} \frac{w^{a-1}}{(w+z)^a} dw \right)= \Re\left( \frac{1}{2i}\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{-a}{j} w^{-j-1}z^{j} \right) dw$$ The function: $$g(w) = \sum_{j=0}^{\infty}\binom{-a}{j} w^{-j-1}z^{j}$$ Has a pole at $w=0$, then the residue is the coefficient of $w^{-1}$: $$-j-1 = -1 \Longrightarrow j=0$$ $$\operatorname{Res}\left(g(w),0\right) = 1$$ By the residue theorem: $$\Re\left( \frac{1}{2i}\oint_{|w|=1} \sum_{j=0}^{\infty} w^{-j-1}z^{j} dw\right) = \Re \left(\pi \operatorname{Res}\left(g(w),0\right)\right) = \Re (\pi) = \pi$$ Therefore $$\boxed{ \int_{0}^{\pi} \cos\left(a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}} dx = \pi}$$

2. Second integral:

$$J = \int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx$$ The integrand is also even, then $$J = \frac{1}{2}\int_{-\pi}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx$$ and using the fact that $\Re (e^{i\varphi}) = \cos(\varphi)$ $$J = \Re\left(\frac{1}{2}\int_{-\pi}^{\pi} e^{\left(ibx -ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx\right)$$ We can expand $\displaystyle \left(2\cos \frac{x}{2}\right)^{2b}$ with the binomial theorem $$\left(2\cos \frac{x}{2}\right)^{2b} = (e^{\frac{ix}{2}}+ e^{\frac{-ix}{2}})^{2b} = \sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}$$ Hence $$J = \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2}\int_{-\pi}^{\pi} e^{\left(-ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)}(1+2z\cos x+z^2)^{-\frac{a}{2}}e^{ix(2b-j)}dx\right)$$ If we make the substitution: $$e^{ix} = w$$ $$dx = \frac{dw}{wi}$$ and using (1) and (2). We have: $$e^{\left(-ia\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)} = \left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}$$ and $$(1+2z\cos x + z^2) = \frac{(w+z)(wz+1)}{w}$$ Hence, $$\begin{align*} J =& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}\frac{w^{\frac{a}{2}}}{(w+z)^{\frac{a}{2}}(wz+1)^{\frac{a}{2}}w} w^{(2b-j)}dw\right)\\ =& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \frac{w^{2b-j-1}}{(wz+1)^a}dw\right)\\ \end{align*}$$ If $|z|\lt1$ $$\frac{1}{(wz+1)^a} = \sum_{k=0}^{\infty} \binom{-a}{k}w^kz^k$$ Hence, $$\begin{align*}J=& \Re\left(\sum_{j=0}^{\infty} \binom{2b}{j} \frac{1}{2i}\oint_{|w|=1} \sum_{k=0}^{\infty} \binom{-a}{k}w^{2b-j+k-1}z^k dw\right)\\ =& \Re\left(\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \binom{2b}{j} \binom{-a}{k}\frac{z^k }{2i}\oint_{|w|=1} w^{2b-j+k-1}dw\right) \end{align*}$$ By the residue theorem $$\begin{align*} J =& \Re\left(\sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \binom{2b}{j} \binom{-a}{k}\frac{z^k }{2i}\oint_{|w|=1} w^{2b-j+k-1}dw\right) \\ =&\Re\left(\pi \sum_{k=0}^{\infty} \binom{2b}{2b+k} \binom{-a}{k}z^k\right) \end{align*}$$ Note that $$\binom{2b}{2b+k} = 0 \Longleftrightarrow k>0 \wedge b>0$$ Therefore $$J = \Re\left(\pi \sum_{k=0}^{\infty} \binom{2b}{2b+k} \binom{-a}{k}z^k\right) = \pi \binom{-a}{0}\binom{2b}{2b} = \pi$$ Hence, we can conclude $$\boxed{\int_{0}^{\pi} \cos\left(bx -a\arctan\left(\frac{z\sin x}{1+z\cos x}\right)\right)(1+2z\cos x+z^2)^{-\frac{a}{2}}\left(2\cos \frac{x}{2}\right)^{2b} dx = \pi }$$

3. Third integral

Finally, we evaluate the last monstrous integral of this triplet using the results obtained in the previous integrals. Additionally, we need to make the assumption that $\displaystyle \tfrac{c-1+b}{2} \in \mathbb{N}$ $$K = \int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx$$ Do the substitution $\displaystyle t = 2x$. $$K = \frac{1}{2} \int_{0}^{\pi} \cos\left(\frac{bt}{2}-a\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt$$ Again, the integrand is an even function of $t$: $$K = \frac{1}{4} \int_{-\pi}^{\pi} \cos\left(\frac{bt}{2}-a\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt$$ Using the fact that $\displaystyle \Re(e^{i\varphi}) = \cos(\varphi)$ $$K = \Re \left(\frac{1}{4} \int_{-\pi}^{\pi} e^{\left(i\frac{bt}{2}-ia\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)}(1+2z\cos t +z^2)^{-\frac{a}{2}}\cos^{c-1}\left(\frac{t}{2}\right) dt\right)$$ Expanding $\displaystyle \cos^{c-1}\left(\frac{t}{2}\right)$ with the binomial theorem: $$K = \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}} \int_{-\pi}^{\pi} e^{\left(-ia\arctan\left(\frac{z\sin t}{1+z\cos t}\right)\right)}(1+2z\cos t +z^2)^{-\frac{a}{2}}e^{it\left(\frac{b+c-1}{2}-j\right)} dt\right)$$ Again, making the substitution $$e^{it} = w$$ $$dt = \frac{dw}{wi}$$ and using (1) and (2) we have $$\begin{align*} K = &\Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1}\left(\frac{w+z}{w^2z+w}\right)^{\frac{a}{2}}\frac{w^{-\frac{a}{2}}}{(w+z)^{\frac{a}{2}}(wz+1)^{\frac{a}{2}}w}w^{\left(\frac{b+c-1}{2}-j\right)} dt\right)\\ =& \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1} \frac{w^{\left(\frac{b+c-1}{2}-j-1\right)}}{(wz+1)^a} dw\right) \end{align*}$$ Assuming that $|z|\lt1$ and expanding $\displaystyle \frac{1}{(wz+1)^a}$ with the binomial theorem: $$\begin{align*} K=& \Re \left(\sum_{j=0}^{\infty} \binom{c-1}{j} \frac{1}{2^{c+1}i} \oint_{|w|=1} \sum_{k=0}^{\infty} \binom{-a}{k} w^{\left(\frac{b+c-1}{2}-j+k-1\right)}z^k dw\right)\\ =&\Re \left(\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{c-1}{j} \binom{-a}{k} \frac{z^k}{2^{c+1}i} \oint_{|w|=1} w^{\left(\frac{b+c-1}{2}-j+k-1\right)} dw\right) \end{align*}$$ Applying the residue theorem: $$K = \Re \left(\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{c-1}{j} \binom{-a}{k} \frac{z^k}{2^{c}}\pi \operatorname{Res}\left(w^{\left(\frac{b+c-1}{2}-j+k-1\right)},0\right)\right)$$ The residue is the coefficient of $w^{-1}$ Then $$w^{-1} = w^{\left(\frac{b+c-1}{2}-j+k-1\right)} \Longrightarrow j = k+\frac{b+c-1}{2}$$ Since all the parameters are real or natural numbers we can dropout the $\Re$. Therefore, $$K = \frac{\pi}{2^c}\sum_{j=0}^{\infty} \binom{c-1}{ k+\frac{b+c-1}{2}} \binom{-a}{k} z^{k}$$

Who is this mysterious function? Let's see

Recall the formula that relate the binomial coefficient to the Pochhammer polynomial or rising factorial: $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ Hence $$\begin{align*} K =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \binom{c-1}{ k+\frac{b+c-1}{2}} \binom{-a}{k} z^{k} \\ &= \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \end{align*}$$ Now, using the rule for rising factorial $$(x)_{n+m} = (x)_{n}(x+n)_{m}$$ we have $$\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}} = \left(\tfrac{c-1-b}{2}-k+1\right)_{k}\left(\tfrac{c-1-b}{2}+1\right)_{\tfrac{b+c+1}{2}} \tag{3}$$ Now, using the reflection formula $$(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{4}$$ and the fact that the rising factorial can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)} \tag{5}$$ we have $$\begin{align*} K= &\frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{c-1-b}{2}-k+1\right)_{k+\tfrac{b+c-1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k}\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{ \left(\tfrac{c-1-b}{2}-k+1\right)_{k}\left(\tfrac{c-1-b}{2}+1\right)_{\tfrac{b+c+1}{2}}(-a-k+1)_{k}}{\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \quad \textrm{ from } (3)\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{(-1)^k\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(-a-k+1)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\left(k+\tfrac{b+c-1}{2}\right)!k!} z^{k} \quad \textrm{ from } (4),(5)\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(a)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(k+\tfrac{b+c+1}{2}\right)k!} z^{k}\\ =& \frac{\pi}{2^c}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}\Gamma(c)(a)_{k}}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(\tfrac{b+c+1}{2}\right)\left(\tfrac{b+c+1}{2}\right)_{k}k!} z^{k}\\ =& \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{c+1-b}{2}\right)\Gamma\left(\tfrac{b+c+1}{2}\right)}\sum_{j=0}^{\infty} \frac{\left(\tfrac{b+1-c}{2}\right)_{k}(a)_{k}}{\left(\tfrac{b+c+1}{2}\right)_{k}} \frac{z^{k}}{k!}\\ =& \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right]\\ \end{align*}$$ Therefore $$\boxed{\int_{0}^{\frac{\pi}{2}} \cos\left(bx-a\arctan\left(\frac{z\sin 2x}{1+z\cos 2x}\right)\right)(1+2z\cos2x +z^2)^{-\frac{a}{2}}\cos^{c-1}x dx = \frac{\pi}{2^c}\frac{\Gamma(c)}{\Gamma\left(\tfrac{1-b+c}{2}\right)\Gamma\left(\tfrac{1+b+c}{2}\right)}{{}_{2}F_{1}}\left[{a,\tfrac{1+b-c}{2}\atop \tfrac{1+b+c}{2}};z\right]}$$

Generalized hypergeometric functions VII

Nice series involving the harmonic number $H_{2n}$

Totday we show the proof of this nice series posted by @BenriBot_ccbs $$\sum_{n=1}^{\infty} \frac{(-1)^nH_{2n}}{2n+1} = -\frac{\pi\ln(2)}{8}$$ The proof relies on some properties of the rising factorial and the Gauss hypergometric function ${}_{2}F_{1}$.

Proof

Consider the following series: $$S= \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} x^n$$ where $$(x)_{n} = x(1+x)\cdots (x+n-1)$$ is the rising factorial or Pochhammer polynomial

The rising factorial satisfies the duplication formula: $$(x)_{2n} = 4^n\left(\frac{1}{x}\right)_{n}\left(\frac{1+x}{2}\right)_{n}$$ Hence $$\begin{align*} S= \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} x^n =& \sum_{n=0}^{\infty} \frac{4^n\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{4^n(1)_{n}\left(\frac{3}{2}\right)_{n}} x^n \\ =& \sum_{n=0}^{\infty} \frac{\left(\frac{t}{2}\right)_{n}\left(\frac{1+t}{2}\right)_{n}}{\left(\frac{3}{2}\right)_{n}} \frac{x^n}{n!} \quad \;\; (1)_{n} = n! \\ =& {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2}, x\right) \end{align*}$$ where ${}_{2}F_{1}$ is the Gauss hypergeometric function. ${}_{2}F_{1}$ satisfies: $${}_{2}F_{2} \left(a,\frac{1}{2}+a;\frac{3}{2};z^{2}\right)=\frac{1}{(2-4a)z}\left((1+z)^{1-2a}-(1-z)^{1-2a}\right)$$ If we put $a = \frac{t}{2}$ and $z^2= i^2 = -1$ $$\sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^n = {}_{2}F_{1} \left(\frac{t}{2},\frac{t+1}{2};\frac{3}{2};-1\right) = \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i} \tag{1}$$ The derivative of the rising factorial is: $$\frac{d}{dx} (x)_{n} = (x)_{n} \left[\psi(n+x)-\psi(x)\right]$$ Hence, differentiating (1) $$\frac{d}{dt} \sum_{n=0}^{\infty} \frac{(t)_{2n}}{(2)_{2n}} (-1)^n = \frac{d}{dt} \left[ \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i}\right]$$ $$\begin{align*} \Longrightarrow \sum_{n=0}^{\infty} \frac{(t)_{2n}\left[\psi(2n+t)-\psi(t)\right]}{(2)_{2n}} (-1)^n =& \frac{d}{dt} \left[ \frac{(1+i)^{1-t}-(1-i)^{1-t}}{(2-2t)i}\right]\\ =& \frac{2i\left((1-i)^{1-t}-(1+i)^{1-t}\right)}{(2-2t)^2} - \frac{i\left((1-i)^{1-t}\ln(1-i)-(1+i)^{1-t}\ln(1+i)\right)}{2-2t} \end{align*}$$ Taking the limit as $t \to 1$ $$\begin{align*} \sum_{n=0}^{\infty} \frac{(1)_{2n}\left[\psi(2n+1)+\gamma\right]}{(2)_{2n}} (-1)^n = &\lim_{t \to 1} \left[\underbrace{\frac{2i\left((1-i)^{1-t}-(1+i)^{1-t}\right)}{(2-2t)^2}}_{=0} - \underbrace{\frac{i\left((1-i)^{1-t}\ln(1-i)-(1+i)^{1-t}\ln(1+i)\right)}{2-2t}}_{\textrm{L'Hôpital's rule }}\right] \\ =& -\frac{\pi\ln(2)}{8} \end{align*}$$ Using the recursion formula $$(n+x)(x)_{n} = x(x+1)_{n}$$ Hence $$(2n+1) = \frac{(2)_{2n}}{(1)_{2n}}$$ Therefore $$\sum_{n=0}^{\infty} \frac{\left[\psi(2n+1)+\gamma\right]}{2n+1} (-1)^n = -\frac{\pi\ln(2)}{8}$$ Finally, using the relation of the Harmonic number to the digamma function: $$H_x = \psi(x+1)+\gamma$$ and $\displaystyle H_{0} = 0$

We can conclude $$\boxed{ \sum_{n=1}^{\infty} \frac{(-1)^nH_{2n}}{2n+1} = -\frac{\pi\ln(2)}{8}}$$

Integral of the day XIV

Integral involving the trigamma function $\psi^{(1)}(x)$

Today we show the proof of this result proposed by @Ali39342137 $$\int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right]$$

Proof:

Note that $$1+x+\cdots +x^{n-1}+x^n = \sum_{j=0}^{n} x^j = \frac{1-x^{n+1}}{1-x}$$ Hence $$I= \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \int_{0}^{1} \frac{(1-x)\ln(x)}{1-x^{n+1}} dx = \underbrace{\int_{0}^{1}\frac{\ln(x)}{1-x^{n+1}}dx}_{J} - \underbrace{\int_{0}^{1} \frac{x\ln(x)}{1-x^{n+1}}dx }_{K}$$ Since $0 \lt x \lt 1$ $$\begin{align*} J = \int_{0}^{1}\frac{\ln(x)}{1-x^{n+1}}dx =& \int_{0}^{1} \sum_{j=0}^{\infty} x^{(n+1)j} \ln(x) dx \\ =& \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j} \left(\frac{d}{dt}\Big|_{t=0+} x^t \right) dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j+t} dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\frac{1}{(n+1)j+t+1}\\ =& \sum_{j=0}^{\infty}\left[\frac{d}{dt}\Big|_{t=0+}\frac{1}{(n+1)j+t+1}\right]\\ =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+1\right)^2}\\ \end{align*}$$ $$\begin{align*} K= \int_{0}^{1} \frac{x\ln(x)}{1-x^{n+1}}dx =& \int_{0}^{1} \sum_{j=0}^{\infty}x^{(n+1)j+1}\ln(x)dx\\ =& \sum_{j=0}^{\infty}\int_{0}^{1}x^{(n+1)j+1}\left(\frac{d}{dt}\Big|_{t=0+} x^t \right)dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\int_{0}^{1} x^{(n+1)j+1+t}dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{j=0}^{\infty}\frac{1}{(n+1)j+2+t} \\ =& \sum_{j=0}^{\infty}\left[\frac{d}{dt}\Big|_{t=0+}\frac{1}{(n+1)j+2+t} \right]\\ =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+2\right)^2} \end{align*}$$ Hence $$\begin{align*} \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx =& -\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+1\right)^2} +\sum_{j=0}^{\infty}\frac{1}{\left((n+1)j+2\right)^2}\\ =& -\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{1}{n+1}\right)^2} +\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{2}{n+1}\right)^2} \end{align*}$$ Recall that he polygamma function has the series representation $$\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}}$$ Therefore $$\begin{align*} \int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = & \sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{2}{n+1}\right)^2}-\sum_{j=0}^{\infty}\frac{1}{(n+1)^2\left(j+\frac{1}{n+1}\right)^2} \\ =& \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right] \end{align*}$$ Hence, we can conclude $$\boxed{\int_{0}^{1} \frac{\ln(x)}{x^n+x^{n-1}+\cdots +1} dx = \frac{1}{(n+1)^2}\left[\psi^{(1)}\left(\frac{2}{n+1}\right)+ \psi^{(1)}\left(\frac{1}{n+1}\right)\right]}$$

Confluent hypergeometric function

Singular series posted by @infseriesbot

Today we evaluate this series posted by @infseriesbot $$\left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x}\sum_{n=0}^{\infty} \frac{(2n)!}{n!^4} x^{2n}$$ The proof will rely on some properties of the confluent hypergeometric function and the binomial coefficient

Proof:

We start with the left hand side. Recall that the Pocchhammer symbol or rising factorial can be expressed as quotient of two gamma functions $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Hence $$(2n)! = \Gamma(2n+1) = (1)_{2n}$$ Pochhammer symbols also satisfy the duplication formula: $$(x)_{2n} = 4^n\left(\frac{x}{2}\right)_{n}\left(\frac{1+x}{2}\right)_{n}$$ Hence $$(2n)! = 4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}$$ Note also that $\displaystyle (1)_{n} = n!$. Hence, $$\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n =\sum_{n=0}^{\infty} \frac{4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}}{(1)_{n}(1)_{n}} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{(1)_{n}} \frac{(4x)^n}{n!} = {}_{1}F_{1}\left(\frac{1}{2};1;4x\right)$$ The right hand side is the confluent hypergometric function ${}_{1}F_{1}(a;b;x)$. This function is also denoted with $M(a;b;x)$:

The confluent hypergometric function satisfies the Kummer's second theorem or second transformation: $${}_1F_1(a,2a,x)= e^{x/2}\, {}_0F_1 \left(; a+\tfrac{1}{2}; \tfrac{x^2}{16} \right)$$ Therefore $$\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n= {}_{1}F_{1}\left(\frac{1}{2};1;4x\right) =e^{2x}\, {}_0F_1 \left(; 1; x^2 \right)$$ From the definition of ${}_{0}F_{1}$: $$e^{2x}\, {}_0F_1 \left(; 1; x^2 \right) = e^{2x} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2}$$ Hence $$\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n = e^{2x} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2}$$ Squaring both sides $$\left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x} \left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 \tag{1}$$ We can evaluate the right hand side with the Cauchy product: $$\begin{align*}\left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 = &\sum_{n=0}^{\infty}\sum_{j=0}^{n} \frac{1}{j!^2(n-j)!^2} x^{2n}\\ =& \sum_{n=0}^{\infty}\frac{1}{n!^2}\sum_{j=0}^{n} \frac{n!^2}{j!^2(n-j)!^2} x^{2n}\\ =& \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\sum_{j=0}^{n} {\binom{n}{j}}^2 \tag{2} \end{align*}$$ From the properties of the binomial coefficient we know that: $$\sum_{j=0}^{J}{\binom{n}{j}}^2 = \binom{2n}{n} \quad J\geq n \tag{3}$$ Therefore $$\left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 = \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\sum_{j=0}^{n} {\binom{n}{j}}^2 = \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\binom{2n}{n} = \sum_{n=0}^{\infty}\frac{(2n)!}{n!^4}x^{2n} \tag{4}$$ Hence from (1),(2),(3) and (4) we can conclude: $$\boxed{\left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x}\sum_{n=0}^{\infty} \frac{(2n)!}{n!^4} x^{2n}}$$

Integral of the day XIII

Integral involving $\zeta(2)$

Today we show the proof of this result posted by @integralsbot $$\int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \zeta(2)$$

Proof:

Consider the following integral: $$\Phi(a) = \int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx \quad 0\lt a\lt 2$$ Differentiating under the integral sing and using the fact that $\displaystyle \frac{1}{2\cos x + a }$ is an even function : $$\Phi'(a) = \int_{0}^{\pi} \frac{1}{2\cos x + a } dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{1}{2\cos x + a } dx$$ If we make the substitution: $$\cos x = \frac{z+z^{-1}}{2}$$ $$dx = \frac{dz}{zi}$$ we transform this integral in a contour integral round the unit complex circle $$\Phi'(a) = \frac{1}{2}\int_{-\pi}^{\pi} \frac{1}{2\cos x + a } dx = -i\oint_{|z|=1} \frac{dz}{az^2+4z+a}$$ The function $$f(z) = \frac{1}{az^2+4z+a}$$ has two poles at $\displaystyle z = \frac{\sqrt{4-a^2}-2}{a}$ and $\displaystyle z = \frac{-\sqrt{4-a^2}-2}{a}$

The condition that $a>0$ assures that the only pole inside the unit complex circle is $\displaystyle z = \frac{\sqrt{4-a^2}-2}{a}$

By the Cauchy integral formula:

$$\Phi'(a) = -i\oint_{|z|=1} \frac{dz}{az^2+4z+a} = \frac{2\pi}{a\left(\frac{\sqrt{4-a^2}-2}{a} + \frac{\sqrt{4-a^2}+2}{a}\right)} = \frac{\pi}{\sqrt{4-a^2}}$$ Integrating with respect to $a$: $$\begin{align*} \Phi(a) = \pi\int \frac{da}{\sqrt{4-a^2}} + C =& \frac{\pi}{2} \int \frac{da}{\sqrt{1-\frac{a^2}{4}}} + C\\ =& \pi \int \frac{dw}{\sqrt{1-w^2}} + C \quad \left(w^2 \mapsto \frac{a^2}{4}\right) \\ =& \pi\arcsin(w) + C\\ =& \pi\arcsin\left(\frac{a}{2}\right) + C \end{align*}$$ Therefore $$\Phi(a) = \int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx = \pi\arcsin\left(\frac{a}{2}\right) + C$$ If we let $a\to 0+$ then $\Phi(0)=0$, $\arcsin(0)=0$ and $C=0$

If we put $a=1$ $$\Phi(1) = \int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \frac{\pi^2}{6} = \zeta(2)$$ Hence, we can conclude $$\boxed{\int_{0}^{\pi} \frac{\ln\left(1+a\frac{\cos x}{2}\right)}{\cos x} dx = \pi\arcsin\left(\frac{a}{2}\right) \quad 0\lt a \lt 2}$$ $$\boxed{\int_{0}^{\pi} \frac{\ln\left(1+\frac{\cos x}{2}\right)}{\cos x} dx = \zeta(2)}$$

Integral of the day XII

Fun integral posted by @integralsbot

Today we evaluate this fun integral posted by @integralsbot $$\int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = \frac{2\pi \ln(1-a^2)}{1-a^2} \quad |a|<1$$ To proof this result we will use Fourier series and contour integration.

Proof

Firt, we need the following Fourier series: $$\sum_{k=1}^{\infty} \frac{a^k\cos(kx)}{k} = -\frac{1}{2} \ln (1-2a\cos x + a^2) \quad x\in(0,2\pi), \quad a^2\leq 1 \tag{1}$$ The proof is almost straight: take principal branch of the $\ln(z)$ function, then: $$\begin{align*} \sum_{k=1}^{\infty} \frac{a^k\cos(kx)}{k} =& \frac{1}{2}\sum_{k=1}^{\infty} \frac{a^k}{k}(e^{ikx} + e^{-ikx})\\ =& \frac{1}{2}\sum_{k=1}^{\infty}\frac{(ae^{ix})^k}{k} + \frac{1}{2}\sum_{k=1}^{\infty}\frac{(ae^{-ix})^k}{k} \\ =& -\frac{1}{2}\ln\left(1-ae^{ix}\right) -\frac{1}{2}\ln\left(1-ae^{-ix}\right)\\ =& -\frac{1}{2}\ln\left((1-ae^{ix})(1-ae^{-ix})\right)\\ =& -\frac{1}{2}\ln\left(1-a(e^{-ix}+e^{ix})+a^2\right)\\ =& -\frac{1}{2}\ln(1-2a\cos(x) +a^2) \end{align*}$$ Back to the integral: $$\begin{align*} \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = -2\int_{0}^{\pi} \sum_{k=1}^{\infty}\frac{a^k\cos(kx)}{k(1-2a\cos x +a^2)} dx\\ = -2 \sum_{k=1}^{\infty}\frac{a^k}{k}\int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \tag{2} \end{align*}$$ Now note that $$f(x) = \frac{\cos(kx)}{1-2a\cos x +a^2}$$ is an even function.

Hence $$\int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx = \frac{1}{2}\Re{\left(\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx\right)} \tag{3}$$ If we make the substitution $$\cos x = \frac{z+z^{-1}}{2}$$ $$e^{ix} = z$$ $$dx = \frac{dz}{zi}$$ Our integral is transformed in a contour integral round the unit complex circle: $$\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx = i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz \tag{4}$$

The function $\displaystyle f(z) = \frac{z^{k}}{(z-a)(az-1)}$ has two poles at $z=a$ and $\displaystyle z=\frac{1}{a}$.

The conditions in (1) assure that $z=a$ with $|a|<1$ is the only pole inside the unit circle.

Then, by the Cauhcy integral formula $$i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz = 2\pi \frac{ a^k}{1-a^2} \tag{5}$$ Therefore $$\begin{align*} \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx =& -2\sum_{k=1}^{\infty}\frac{a^k}{k}\int_{0}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \quad \textrm{from (2) }\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\int_{-\pi}^{\pi} \frac{\cos(kx)}{1-2a\cos x +a^2}dx \\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(\int_{-\pi}^{\pi} \frac{e^{ixk}}{1-2a\cos x +a^2}dx\right)} \quad \textrm{from (3) }\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(i\oint_{|z|=1} \frac{z^{k}}{(z-a)(az-1)}dz\right)} \quad \textrm{from (4)}\\ =& -\sum_{k=1}^{\infty}\frac{a^k}{k}\Re{\left(2\pi \frac{ a^k}{1-a^2}\right)} \quad \textrm{from (5)}\\ =& \frac{2\pi}{1-a^2}\left(-\sum_{k=1}^{\infty}\frac{a^{2k}}{k}\right)\\ =& \frac{2\pi \ln(1-a^2)}{1-a^2} \end{align*}$$ Hence, we can conclude $$\boxed{ \int_{0}^{\pi} \frac{\ln (1-2a\cos x + a^2)}{1-2a\cos x +a^2} dx = \frac{2\pi \ln(1-a^2)}{1-a^2} \quad |a|<1 }$$

Integral of the day XI

Integral involving the trigamma constant $\psi^{(1)}\left(\frac{1}{3}\right)$

Today we show the proof of the following result proposed by @Ali3934213 $$\int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \frac{1}{9} \left[-\pi^2+\pi\sqrt{3}\ln(3) +\psi^{(1)}\left(\frac{1}{3}\right) \right]$$ The proof relies on some properties of the polylogarithm and the trigamma function.

Proof

$$I = \int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \int_{0}^{1} \frac{\ln(x+1)}{x^3+1} dx+ \underbrace{\int_{1}^{\infty} \frac{\ln(x+1)}{x^3+1} dx}_{J}$$ $$J = \int_{1}^{\infty} \frac{\ln(x+1)}{x^3+1} dx \stackrel{w \mapsto \frac{1}{x}}{=} \int_{0}^{1} \frac{\ln\left(\frac{w+1}{w}\right)w}{w^3+1} dw= \int_{0}^{1} \frac{\ln\left(w+1\right)w}{w^3+1} dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw$$ Hence $$\begin{align*} I =& \int_{0}^{1} \frac{\ln(x+1)}{x^3+1} dx + \int_{0}^{1} \frac{\ln\left(w+1\right)w}{w^3+1} dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw\\ =& \int_{0}^{1} \frac{\ln(w+1)(w+1)}{1+w^3}dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw \\ =& \underbrace{\int_{0}^{1} \frac{\ln(w+1)}{w^2-w+1}dw}_{K} - \underbrace{\int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw}_{L}\\ \end{align*}$$ Note that $$\Im\left(\frac{2}{\sqrt{3}}\frac{1}{w-\frac{1}{2}-i\frac{\sqrt{3}}{2}}\right) = \frac{1}{w^2-w+1}$$ Hence $$\begin{align*} K = \int_{0}^{1} \frac{\ln(w+1)}{w^2-w+1}dw =& \Im \left( \frac{2}{\sqrt{3}}\int_{0}^{1} \frac{\ln(w+1)}{w-\frac{1}{2}-i\frac{\sqrt{3}}{2}} dw \right)\\ =&\Im \left( \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(t+\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{t} dt \right)\\ =& \Im \left( \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)\right)}{t} dt \right)\\ =& \Im \left( \frac{2\ln\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{1}{t} dt + \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)}{t} dt\right)\\ =& \Im \left( \frac{2\ln\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{\sqrt{3}}\left(\ln\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right)-\ln\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right) + \Im\left(\frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)}{t} dt\right) \\ =& \frac{\pi\ln(3)}{3\sqrt{3}}+ \Im \left( \frac{2}{\sqrt{3}}\int_{\frac{1}{2}+\frac{i}{2\sqrt{3}}}^{\frac{i}{\sqrt{3}}} \frac{\ln\left(1-s\right)}{s} ds\right) \quad \left( s \mapsto -\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}\right)\\ =& \frac{\pi\ln(3)}{3\sqrt{3}} - \frac{2}{\sqrt{3}}\Im\left(\operatorname{Li}_{2}\left(\frac{i}{\sqrt{3}}\right)-\operatorname{Li}_{2}\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)\right) \end{align*}$$ The polylogarithm function satisfies: $$\Im\left(\operatorname{Li}_{2}\left(\frac{i}{\sqrt{3}}\right)-\operatorname{Li}_{2}\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)\right) = -\frac{\pi^2}{18\sqrt{3}} + \frac{1}{12\sqrt{3}}\psi^{(1)}\left(\frac{1}{3}\right)$$ Hence $$K = \frac{\pi\ln(3)}{3\sqrt{3}}+\frac{\pi^2}{27} -\frac{1}{18}\psi^{(1)}\left(\frac{1}{3}\right)$$ Now, for $L$; $$\begin{align*} L= \int_{0}^{1} \frac{\ln(w)w}{1+w^3}dw = \int_{0}^{1} \frac{\left(\frac{d}{dt}\Big|_{t=0+} w^t \right)w}{1+w^3}dw = &\frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \frac{w^{t+1}}{1+w^3} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^nw^{3n}w^{t+1} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} w^{t+1+3n} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} (-1)^n\frac{1}{(3n+2+t)} \\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(3n+2)^2} \\ =& \sum_{n=0}^{\infty} \frac{1}{(6n+5)^2} - \sum_{n=0}^{\infty} \frac{1}{(6n+2)^2}\\ =& \frac{1}{36}\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{5}{6}\right)^2} - \frac{1}{36} \sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{1}{3}\right)^2}\\ \end{align*}$$ Recall the series representation of the polygamma function: $$\psi^{(m)}(z) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(z+k)^{m+1}}$$ Hence $$L = \frac{1}{36} \left[ \psi^{(1)}\left(\frac{5}{6}\right) -\psi^{(1)}\left(\frac{1}{3}\right)\right]$$ Therefore $$I = K-L = \frac{\pi\ln(3)}{3\sqrt{3}}+\frac{\pi^2}{27} -\frac{1}{36}\psi^{(1)}\left(\frac{1}{3}\right) -\frac{1}{36}\psi^{(1)}\left(\frac{5}{6}\right)$$ Now using the identity $$\psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{5}{6}\right) = 4\psi^{(1)}\left(\frac{2}{3}\right)$$ and the reflection formula $$\psi^{(1)}(1-z) + \psi^{(1)}(z) = \frac{\pi^2}{\sin^2(\pi z)}$$ with $\displaystyle z = \frac{2}{3}$ we have $$\psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{2}{3}\right) = \frac{4}{3}\pi^2$$ Therefore $$\boxed{\int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \frac{1}{9} \left[-\pi^2+\pi\sqrt{3}\ln(3) +\psi^{(1)}\left(\frac{1}{3}\right) \right]}$$