Beta function

Derivative of Dirichlet Beta function

We prove the following infinite product which turned out to be a modified version of the Dirichlet Beta function derivative. $$\frac{1}{\sqrt[3]{3}}\frac{\sqrt[5]{5}}{\sqrt[7]{7}}\frac{\sqrt[9]{9}}{\sqrt[11]{11}}\frac{\sqrt[13]{13}}{\sqrt[15]{15}}\ldots =\left(\frac{\pi}{e^\gamma \Gamma(\frac{3}{4})^4}\right)^\frac{\pi}{4}$$ Proof

Take logarithm in both sides $$-\frac{\ln(3)}{3}+\frac{\ln(5)}{5}-\frac{\ln(7)}{7}+\frac{\ln(9)}{9}-\frac{\ln(11)}{11}+\frac{\ln(13)}{13}-... = \frac{\pi}{4}\left[\ln(\pi)-\gamma -4\ln\Gamma\left(\frac{3}{4}\right)\right]$$ Then, we want to prove $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)} = \frac{\pi}{4}\left[\ln(\pi)-\gamma -4\ln\Gamma\left(\frac{3}{4}\right)\right]$$ Consider the Dirichlet beta function $$\beta(v) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^{v}}$$ Then $$\beta'(v) = \sum_{n=1}^{\infty} \frac{(-1)^{n}\ln(2n-1)}{(2n-1)^{v}}$$ Therefore we want to find $$-\beta'(1) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\ln(2n-1)}{(2n-1)}$$ There are various ways to find the derivative, one of the most straigforward is using the Kummer's Fourier series for the log-gamma function: $$\ln \Gamma(t) = \frac{1}{2}\ln\left(\frac{\pi}{\sin \pi t}\right)+\left[\gamma+\ln(2\pi)\right]\left(\frac{1}{2}-t\right) +\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\ln n}{n} \sin(2\pi n t)$$ It turned out that if $\displaystyle t=\frac{1}{4}$ then $$\sin\left(\frac{\pi n}{2}\right)= \left\{\begin{array}{lr} \sin\left(\frac{\pi}{2}\right) = 1 & \textrm{ if } n=1 \\ \sin\left(\pi \right) = 0 & \textrm{ if } n=2 \\ \sin\left(\frac{3\pi}{2}\right) = -1 & \textrm{ if } n=3 \\ \sin\left(2\pi \right) = 0 & \textrm{ if } n=4 \\ \vdots & \vdots \\ \end{array} \right\}$$ Then $\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n}\sin\left(\frac{\pi n}{2}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)}$ Therefore, $$\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)} =& \pi\left[ \ln \Gamma\left(\frac{1}{4}\right)- \frac{1}{2}\ln\left(\frac{\pi}{\sin \frac{\pi}{4} }\right)-\left[\gamma+\ln(2\pi)\right]\left(\frac{1}{4}\right) \right]\\ =& \pi\left[ \ln(\pi)- \frac{1}{2}\ln(2) -\ln\Gamma\left(\frac{3}{4}\right) - \frac{1}{2} \ln(\pi) -\frac{1}{4} \ln(2) -\frac{1}{4} \gamma -\frac{\ln (2\pi)}{4} \right]\\ =& \frac{\pi}{4} \left[\ln(\pi)-\gamma-4\Gamma\left(\frac{3}{4}\right)\right] \end{align*}$$ Therefore $$\boxed{\frac{1}{\sqrt[3]{3}}\frac{\sqrt[5]{5}}{\sqrt[7]{7}}\frac{\sqrt[9]{9}}{\sqrt[11]{11}}\frac{\sqrt[13]{13}}{\sqrt[15]{15}}\ldots =\left(\frac{\pi}{e^\gamma \Gamma(\frac{3}{4})^4}\right)^\frac{\pi}{4}}$$

Application of Ramanujan's master theorem

Mellin transform of the Bessel function

We show the following result $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{s} J_{s}(x)dx = \Gamma(s)$$ Proof
We will use the Ramanujan's master theorem to prove this result: Assume that the function $f$ has an expansion of the form $$f(x) = \sum_{k=0}^{\infty} \left(\frac{\phi(k)}{k!}\right)(-x)^{k}$$ for some analytic function $\phi(k)$, then the Mellin transform of $f(x)$ is given by $$\{\mathscr{M}f(x)\} = \int_{0}^{\infty } x^{s-1} f(x) dx = \Gamma(s)\phi(-s)$$ Back to our problem, recall the expansion series of the Bessel function: $$J_{s}(x) = \left(\frac{x}{2}\right)^{s} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}$$ Then $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{s-1} J_{s}(x)dx=\int_{0}^{\infty} \left(\frac{x}{2}\right)^{2s-1} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}dx$$ let $\displaystyle w=\frac{x^2}{4}$ then $\displaystyle dw= \frac{x}{2}dx$, therefore $$\int_{0}^{\infty} \left(\frac{x}{2}\right)^{2s-1} \sum_{j=0}^{\infty} \frac{(-x^2/4)^j}{j!\Gamma(1+j+s)}dx = \int_{0}^{\infty} w^{s-1} \sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}dw$$ Clearly $\displaystyle f(w)=\sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}$ and $\displaystyle \phi(j) =\frac{1}{\Gamma(1+j+s)}$ By Ramanujan's master theorem : $$\int_{0}^{\infty} w^{s-1} \sum_{j=0}^{\infty} \frac{(-w)^j}{j!\Gamma(1+j+s)}dw = \Gamma(s)\phi(-s) =\frac{\Gamma(s)}{\Gamma(1)} =\Gamma(s)$$ Then $$\boxed{ \int_{0}^{\infty} \left(\frac{x}{2}\right)^{s} J_{s}(x)dx = \Gamma(s)}$$

Integral of the day

Integral involving the compostion of sin function with itself

We prove the following integral: $$\int_{0}^{\frac{\pi}{2}} \sin\sin xdx = 1-\frac{1}{3!!^2}+\frac{1}{5!!^2}-\frac{1}{7!!^{2}}+...$$ Proof
Recall the series expansion of $\sin(x)$ $$\sin x = \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}x^{2n+1} \quad \forall x\in \mathbb{R}$$ Therefore $$\begin{align*} \int_{0}^{\frac{\pi}{2}} \sin\sin xdx =& \int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}\sin^{2n+1}(x)dx\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{\frac{\pi}{2}}\sin^{2n+1}(x)dx\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{1}w^{2n+1}(1-w^2)^{-\frac{1}{2}}dw \quad (w\mapsto \sin x)\\ =& \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \int_{0}^{1} \frac{t^{n}(1-t)^{-\frac{1}{2}}}{2}dt \quad (t \mapsto w^2)\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{B(n+1,\frac{1}{2})}{2}\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{2\Gamma(n+1+\frac{1}{2})}\\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{n! \sqrt{\pi}}{2\Gamma(n+1+\frac{1}{2})} \quad (\Gamma(\frac{1}{2}) = \sqrt{\pi}))\\ \end{align*}$$ Now recall the Legrende duplication formula $$\sqrt{\pi} \Gamma(2z) = 2^{2z-1} \Gamma(z)\Gamma(z+\frac{1}{2})$$ Applying the formula to $\Gamma(n+1+\frac{1}{2})$ $$\begin{align*} \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{n! \sqrt{\pi}}{2\Gamma(n+1+\frac{1}{2})} = &\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{2^{2n}n! \Gamma(n+1)}{\Gamma(2n+2)} \\ =&\sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!} \frac{2^{2n}n!^2 }{(2n+1)!}\\ =&\sum_{n=0}^{\infty } (-1)^n \frac{2^{2n}n!^2 }{(2n+1)!^2}\\ =&\sum_{n=0}^{\infty } (-1)^n \frac{1 }{(2n+1)!!^2} \end{align*}$$ $$\boxed{\int_{0}^{\frac{\pi}{2}} \sin\sin xdx = \sum_{n=0}^{\infty } (-1)^n \frac{1 }{(2n+1)!!^2}}$$

Feynman's trick II

Formula involving integrals of logarithm function

We prove the following formula involving the integral of powers of the function $\ln(1-x)$. $$\int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k)\right]$$ Some examples for the first naturals: $$\begin{align*} &n=1& \int_{0}^{1} \frac{\ln(1-x)\ln(x)}{x} dx &= \zeta(3)\\ &n=2& \int_{0}^{1} \frac{\ln^2(1-x)\ln(x)}{x} dx &= -3\zeta(4)+\zeta(2)\zeta(2) = -\frac{\pi^4}{180} \\ &n=3& \int_{0}^{1} \frac{\ln^3(1-x)\ln(x)}{x} dx & = 12\zeta(5)-6\zeta(2)\zeta(3) = 12\zeta(5)-\pi^2\zeta(3)\\ &n=4& \int_{0}^{1} \frac{\ln^4(1-x)\ln(x)}{x} dx & =-60\zeta(6)+24\zeta(2)\zeta(4)+12\zeta(3)\zeta(3) = 12\zeta^2(3)-\frac{2\pi^6}{105}\\ &n=5& \int_{0}^{1} \frac{\ln^5(1-x)\ln(x)}{x} dx &= 360\zeta(7) -120\zeta(2)\zeta(5)-120\zeta(3)\zeta(4) = -\frac{4\pi^4\zeta(3)}{3} - 20\pi^2\zeta(5) +360\zeta(7) \end{align*}$$ Proof $$\begin{align*} \int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = & \int_{0}^{1} \frac{\left[\frac{d^n}{dt^n}\Big|_{t=0+} (1-x)^t\right] \left[\frac{d}{ds}\Big|_{s=0+} x^s\right]}{x} dx\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \int_{0}^{1} \frac{ (1-x)^t x^s}{x} dx \\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \int_{0}^{1} (1-x)^t x^{s-1} dx\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} B(t+1,s) \\ \end{align*}$$ Now, recall the series expansion of the beta function (proof in Appendix 1): $$B(x,y) = \sum_{j=0}^{\infty} \frac{(1-y)_{j}}{j!(x+j)} \quad y>0$$ Therefore $$\begin{align*} \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} B(t+1,s) = & \frac{d^n}{dt^n}\Big|_{t=0+}\frac{d}{ds}\Big|_{s=0+} \sum_{j=0}^{\infty} \frac{(1-s)_{j}}{j!(t+1+j)}\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[\lim_{s \to 0+}\frac{d}{ds} \sum_{j=0}^{\infty} \frac{(1-s)_{j}}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[\lim_{s \to 0+} \sum_{j=0}^{\infty} \frac{(1-s)_{j}(\psi^{(0)}(1-s) -\psi^{(0)}(1-s+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{(1)_{j}(\psi^{(0)}(1) -\psi^{(0)}(1+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{\Gamma(j+1)(-\gamma -\psi^{(0)}(1+j))}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{-\Gamma(j+1)H_{j}}{j!(t+1+j)}\right]\\ =& \frac{d^n}{dt^n}\Big|_{t=0+}\left[ \sum_{j=0}^{\infty} \frac{-H_{j}}{(t+1+j)}\right]\\ =& \lim_{t \to 0+} \sum_{j=0}^{\infty}\frac{d^n}{dt^n}\frac{-H_{j}}{(t+1+j)}\\ =& \lim_{t \to 0+} \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(t+1+j)^{n+1}}\\ =& \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(1+j)^{n+1}} \end{align*}$$ Now we know that $\displaystyle H_{j+1} = \frac{1}{j+1} + H_{j}$ therefore $$\begin{align*} \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j}}{(1+j)^{n+1}} = & \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{(-1)^{n+1}n!}{(1+j)^{n+2}} \\ =& (-1)^{n+1}n!\left[\sum_{j=0}^{\infty}\frac{H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{1}{(1+j)^{n+2}}\right] \end{align*}$$ If $m=j+1$ then $$\begin{align*} (-1)^{n+1}n!\left[\sum_{j=0}^{\infty}\frac{H_{j+1}}{(1+j)^{n+1}} - \sum_{j=0}^{\infty}\frac{1}{(1+j)^{n+2}}\right] = & (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \sum_{m=1}^{\infty}\frac{1}{m^{n+2}}\right]\\ =& (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \zeta(n+2) \right] \end{align*}$$ Now, recall this beutiful result proven by Euler (Proof in Appendix 2): $$\sum_{n=1}^{\infty} \frac{H_{n}}{n^r} = \left(1+\frac{r}{2}\right) \zeta(r+1) - \frac{1}{2} \sum_{k=1}^{r-2}\zeta(k+1)\zeta(r-k)$$ Then $$\begin{align*} (-1)^{n+1}n!\left[\sum_{m=1}^{\infty}\frac{H_{m}}{m^{n+1}} - \zeta(n+2) \right] =& (-1)^{n+1}n!\left[\left(1+\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) - \zeta(n+2) \right]\\ =& (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) \right] \end{align*}$$ Therefore $$\boxed{ \int_{0}^{1} \frac{\ln^n(1-x)\ln(x)}{x} dx = (-1)^{n+1}n!\left[\left(\frac{n+1}{2}\right) \zeta(n+2) - \frac{1}{2} \sum_{k=1}^{n-1}\zeta(k+1)\zeta(n+1-k) \right]}$$

Apendix 1: Series expansion for the beta function

From the definition of the beta function: $$B(x,y) = \int_{0}^{1} t^{x-1}(1-t)^{y-1}dt$$ Recall the series expansion of $(1-t)^{y-1}$ $$(1-t)^{y-1} = \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}$$ $$\begin{align*} \int_{0}^{1} t^{x-1}(1-x)^{y-1}dt =& \int_{0}^{1} t^{x-1}\sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}t^{j}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \int_{0}^{\infty} t^{x+j-1}dt\\ =& \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j} \frac{1}{(x+j)} \end{align*}$$ Then we have $$\begin{align*} \sum_{j=0}^{\infty} (-1)^{j} \binom{y-1}{j}\frac{1}{(x+j)} =& \sum_{j=0}^{\infty}\binom{j-y}{j}\frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{(j-y)!}{j!(-y)!}\frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{\Gamma(j-y+1)}{j!\Gamma(1-y)} \frac{1}{(x+j)}\\ =& \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)} \end{align*}$$ where the last equality is the definnition of the Pochhammer polynomials. Therefore $$\boxed{B(x,y) = \sum_{j=0}^{\infty}\frac{(1-y)_{j}}{j!(x+j)}}$$

Appendix 2: Proof of Euler's theorem for harmonic series

I found a beautiful proof of this result which relies on the residue theorem. I really liked this proof because it shows that the famous result due to Euler is just a residue of certain kernel function. I present here the original proof in the spirit of Flajolet et al. [1] but I complete the omitted parts by the authors in the original paper.

We need the following Lemma due to Cauchy and Lindelöf.

Lemma. Let $\xi$ be a kernel function and let $r(s)$ be a rational function with $\mathscr{O}(s^{-2})$ at infinity. Then $$\underbrace{\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\}}_{\textrm{infinite series}} = -\underbrace{\sum_{\beta } \left\{\operatorname{Res}\left(r(s)\xi(s), \beta \right) \Big| \beta \textrm{ is a pole of } r(s) \right\}}_{\textrm{finite series}}$$

We will apply the lemma to the kernel $\displaystyle \xi(s) = (\psi(-s)+\gamma)^2$

Propositon. If $\displaystyle \xi(s) = (\psi(-s)+\gamma)^2$ then
$$\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\} = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)$$
Proof

The kernel $\xi(s)$ has a countable number of pole at the positive integers.
Let $\displaystyle n\in \mathbb{N}$, then the function $\displaystyle (\psi(-s)+\gamma)$ has the following expansion at $s \to n$.

$$\psi(-s)+\gamma = \frac{1}{(s-n)}+ H_{n} + \sum_{k=1}^{\infty}\left[(-1)^K H_{n}^{k+1} - \zeta(k+1)\right](s-n)^{k}$$ Therefore
$$\begin{align*} (\psi(-s)+\gamma)^2 =& \left[\frac{1}{(s-n)}+ H_{n} + \sum_{k=1}^{\infty}\underbrace{\left[(-1)^K H_{n}^{k+1} - \zeta(k+1)\right]}_{A_{k}}(s-n)^{k}\right]^2\\ =& \left(\frac{1}{(s-n)}+ H_{n}\right)^2 + 2\left(\frac{1}{(s-n)}+ H_{n}\right)\sum_{k=1}^{\infty}A_{k}(s-n)^{k} + \left(\sum_{k=1}^{\infty}A_{k}(s-n)^{k}\right)^2\\ =& \frac{1}{(s-n)^2}+\frac{2H_{n}}{(s-n)} + H^2_{n} + 2\left(\frac{1}{(s-n)}+ H_{n}\right)\sum_{k=1}^{\infty}A_{k}(s-n)^{k} + \left(\sum_{k=1}^{\infty}A_{k}(s-n)^{k}\right)^2\\ =& \frac{1}{(s-n)^2}+\frac{2H_{n}}{(s-n)} + H^2_{n} + 2A_{1} + \mathscr{O}(s-n) \end{align*}$$ Now, suppose $r(s)$ is a rational function that can be expanded with its power series (in reality, we have assumed something weaker: that $r(s)$ is $\mathcal{O}(s^{-2}$)). Then for $s \to n$: $$r(s) = r(n)+ r'(n)(s-n)+\frac{r''(n)}{2!}(s-n)^2+\frac{r'''(n)}{3!}(s-n)^3+...$$ Therefore $$\begin{align*} r(s)(\psi(-s)+\gamma)^2 =& r(n)(\psi(-s)+\gamma)^2+ r'(n)(s-n)(\psi(-s)+\gamma)^2+\frac{r''(n)}{2!}(s-n)^2(\psi(-s)+\gamma)^2+...\\ =& \frac{r(n)}{(s-n)^2}+\frac{2H_{n}r(n)}{(s-n)} + H^2_{n}r(n) + 2A_{1}r(n) + \mathscr{O}(s-n)r(n) + \frac{r'(n)}{(s-n)}+2H_{n}r'(n) + H^2_{n}r'(n)(s-n) + 2A_{1}r'(n)(s-n) + \mathscr{O}(s-n)r'(n)(s-n) + ... \\ =& H^2_{n}r(n)+ 2A_{1}r(n)+2H_{n}r'(n)+\frac{r''(n)}{2!}+\frac{1}{s-n}\underbrace{\left[2H_{n}r(n)+r'(n)\right]}_{\textrm{residue at } n} +\frac{r(n)}{(s-n)^2} + \mathscr{O}(s-n) \end{align*}$$ $$\therefore \operatorname{Res}\left(r(s)(\psi(-s)+\gamma)^2, n\right) = 2H_{n}r(n)+r'(n)$$ Summing each $n \in \mathbb{N}\setminus\left\{0\right\}$ $$\boxed{\sum_{n=1}^{\infty} \operatorname{Res}\left(r(s)(\psi(-s)+\gamma)^2, n\right) = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)}$$ or, equivalently $$\boxed{\sum_{\alpha } \left\{\operatorname{Res}\left(r(s)\xi(s), \alpha\right) \Big| \alpha \textrm{ is a pole of } \xi \textrm{ that is not a pole of } r(s) \right\} = 2\sum_{n=1}^{\infty}H_{n}r(n)+ \sum_{n=1}^{\infty}r'(n)}$$

We are ready to prove Euler's theorem

Theorem (Euler). For integer $\geq 2$, $$\sum_{n=1}^{\infty} \frac{H_{n}}{n^q} = \left(1+\frac{q}{2}\right)\zeta(q+1) -\frac{1}{2}\sum_{k=1}^{q-2} \zeta(k+1)\zeta(q-k)$$ Proof If we apply the last result to the function $\displaystyle r(s)= \frac{1}{s^q}$ we have $$\sum_{n=1}^{\infty} \operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, n\right) = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)$$ By the first lemma, $$\displaystyle -\sum_{\beta} \left\{\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, \beta \right) \Big| \beta \textrm{ is a pole of } \frac{1}{s^q} \right\} = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)$$ Given that $\displaystyle r(s)= \frac{1}{s^q}$ has a single pole at $s=0$ of order $q$ we have: $$-\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q}, 0\right) =2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)$$

To find the residue we can expand $\displaystyle \frac{(\psi(-s)+\gamma)^2}{s^q}$:
First, recall the series expansion of $\psi(-s)+\gamma$ at $s=0$:
$$\psi(-s)+\gamma = \frac{1}{s} - \sum_{k=1}^{\infty} \zeta(k+1)s^{k}$$ Therefore $$\begin{align*} (\psi(-s)+\gamma)^2 =& \left[\frac{1}{s} -\sum_{k=1}^{\infty} \zeta(k+1)s^{k}\right]^2 \\ =& \frac{1}{s^2} -2\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1} + \left(\sum_{k=1}^{\infty} \zeta(k+1)s^{k}\right)^2\\ =& \frac{1}{s^2} -2\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1} + \sum_{k=1}^{\infty}\sum_{j=1}^{k} \zeta(j+1)\zeta(k-j+2)s^{k+1}\\ \end{align*}$$ Then $$\begin{align*} \frac{(\psi(-s)+\gamma)^2}{s^q} = \frac{1}{s^{2+q}} -2\underbrace{\sum_{k=1}^{\infty} \zeta(k+1)s^{k-1-q}}_{A} + \underbrace{\sum_{k=1}^{\infty}\sum_{j=1}^{k} \zeta(j+1)\zeta(k-j+2)s^{k+1-q}}_{B} \end{align*}$$ From here is easy to show that the residue in A is the coefficient of $\displaystyle s^{k-1-q}=s^{-1}$ $$\Longrightarrow k-1-q=-1 \Longrightarrow k=q$$ and the residue in B is coefficient of $\displaystyle s^{k+1-q}=s^{-1}$ where $$\Longrightarrow k+1-q=-1 \Longrightarrow k=q-2$$ Therefore $$\operatorname{Res}\left(\frac{(\psi(-s)+\gamma)^2}{s^q},0\right) = -2\zeta(q+1) + \sum_{j=1}^{q-2} \zeta(j+1)\zeta(q-j)$$ Then $$2\zeta(q+1) - \sum_{j=1}^{q-2} \zeta(j+1)\zeta(q-j) = 2\sum_{n=1}^{\infty}\frac{H_{n}}{n^q} -q\zeta(q+1)$$ $$\boxed{\therefore\sum_{n=1}^{\infty} \frac{H_{n}}{n^q} = \left(1+\frac{q}{2}\right)\zeta(q+1) -\frac{1}{2}\sum_{k=1}^{q-2} \zeta(k+1)\zeta(q-k)}$$

[1] Philippe Flajolet, Bruno Salvy. Euler Sums and Contour Integral Representations. [Research Report] RR-2917, INRIA. 1996. ffinria-00073780f

Cauchy integral formula

Triple exponential integral

We show the proof of the following integral: $$\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = \pi e$$ On the way we will find this beautiful result: $$\int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx =2\pi e$$ Proof $$\begin{align*} \int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = &\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x} \left[ \frac{ e^{ie^{\cos x}\sin\sin x} + e^{-ie^{\cos x}\sin\sin x}}{2}\right]dx \\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}\cos\sin x+ie^{\cos x}\sin\sin x} + e^{e^{\cos x}\cos\sin x-ie^{\cos x}\sin\sin x}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}(\cos\sin x+i\sin\sin x)} + e^{e^{\cos x}(\cos\sin x-i\sin\sin x)}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x}e^{i\sin x}} + e^{e^{\cos x}e^{-i\sin x}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{\cos x + i\sin x}} + e^{e^{\cos x-i\sin x}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx\\ =& \frac{1}{2}\int_{0}^{\pi } \left[ \sum_{n=0}^{\infty} \frac{(e^{e^{ix}})^n}{n!} + \sum_{0}^{\infty} \frac{(e^{e^{-ix}})^{n}}{n!} \right]dx\\ =& \frac{1}{2}\int_{0}^{\pi } \sum_{n=0}^{\infty} \frac{(e^{e^{ix}})^n+(e^{e^{-ix}})^{n}}{n!} dx\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\int_{0}^{\pi } (e^{e^{ix}})^n+(e^{e^{-ix}})^{n}dx\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!} \left[\int_{0}^{\pi } (e^{e^{ix}})^ndx+ \int_{0}^{\pi }(e^{e^{-ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\left[\int_{0}^{\pi } (e^{e^{ix}})^ndx- \int_{\pi}^{0 }(e^{e^{-ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\left[\int_{0}^{\pi } (e^{e^{ix}})^ndx+ \int_{-\pi}^{0 }(e^{e^{ix}})^{n}dx\right]\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\int_{-\pi}^{\pi } (e^{e^{ix}})^ndx \\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!}\oint_{\gamma} \frac{e^{zn}}{zi} dz \quad \left( \textrm{where } \gamma(x) = e^{ix} \quad x\in [-\pi,\pi]\right)\\ =& \frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{n!} 2\pi \quad (\textrm{Cauchy integral formula}) \\ =& \pi e \end{align*}$$ $$\boxed{\int_{0}^{\pi} e^{e^{\cos x}\cos\sin x}\cos(e^{cos x} \sin \sin x) dx = \pi e }$$ Corollary $$\boxed{ \int_{0}^{\pi } e^{e^{e^{ix}}} + e^{e^{e^{-ix}}}dx =2\pi e}$$

Riemann zeta function

Series involving the zeta function, e and pi

We show the proof of this beutiful series posted by @infseriesbot. $$\exp\left(\frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...\right)=\sqrt{\frac{2\pi}e}$$ To prove this result we use two concepts:

1. The Taylor series expansion of the log-Gamma function $\ln \Gamma(x)$
2. The Euler's reflection formula for the Gamma function $\Gamma(x)$

Lets start with the left hand side $$\begin{align*} \frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...=&\sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)\zeta(2n)\\ =&\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)(2n+1)} \end{align*}$$ What is the series of the right hand side? This series is related to the expansion series of the log-Gamma function: $$\ln(\Gamma(z))=-\gamma(z-1)+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k} (z-1)^{k} \quad |z-1|<1$$ where $\gamma$ is the Euler-Mascheroni constant. If we let $z=1+t$ and then $z=1-t$ in the expansion series we obtain two equations: $$\begin{align*} \ln(\Gamma(t+1))&=\gamma t+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k} t^{k} \\ \ln(\Gamma(1-t))&=-\gamma t+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k} t^{k} \end{align*}$$ Adding both equations: $$\begin{align*} \ln(\Gamma(t+1))+\ln(\Gamma(1-t))&=\sum_{k=1}^{\infty}\frac{\zeta(2k)}{2k} t^{2k} \label{eq:3} \end{align*}$$ Integrating both sides: $$\begin{align*} \int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt &=\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k} \int_{0}^{1}t^{2k} dt \\ \Longrightarrow \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt & = \sum_{k=1}^{\infty}\frac{\zeta(2k)}{2k(2k+1)} \tag{1} \end{align*}$$ This is the series we were looking for in terms of a definite integral. Hopefully we can transform it to a maneuverable one. Here is where the Euler's reflection formula for the Gamma function is useful: $$\begin{align*} \Gamma(t)\Gamma(1-t)=\frac{\pi}{\sin(\pi t)} \quad t\notin \mathbb{Z} \end{align*}$$ Therefore $$\begin{align*} \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt & = \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))(\Gamma(1-t))\right)dt\\ & =\frac{1}{2}\int_{0}^{1}\left(ln(t\cdot\Gamma(t))(\Gamma(1-t))\right)dt\\ & =\frac{1}{2}\int_{0}^{1}\ln\left( \frac{t \cdot \pi}{\sin(\pi t)}\right)dt\\ & = \frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}-\frac{1}{2}\int_{0}^{1}\ln(\sin(t\pi))dt \tag{2} \end{align*}$$ We just have to prove that $\int_{0}^{1}\ln(\sin(t\pi))dt=-\ln(2)$ $$\begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& = \int_{0}^{\frac{1}{2}}\ln(\sin(2\pi t))dt\\ & =\ln(2)+ \int_{0}^{\frac{1}{2}}(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\cos(\pi t))dt \quad (\sin(2t)=2\sin(t)\cos(t)) \\ \end{align*}$$ We also know that $$\begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& = 2\int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt \quad \text{(using $-\sin(\theta)=\sin(-\theta)$)}\\ &= \int_{0}^{\frac{1}{2}}\ln(\sin(2\pi t))dt \end{align*}$$ Therefore $$\begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& \ln(2)+ \int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\cos(\pi t))dt \\ & = \ln(2)+ \int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\sin(\pi w))dw \quad \text{($w=\frac{1}{2}-t$)}\\ & = \ln(2)+ 2\int_{0}^{1}\ln(\sin(\pi t))dt+2\int_{0}^{1}\ln(\sin(\pi w))dw \end{align*}$$ $$\begin{align*} \Longrightarrow \int_{0}^{1}\ln(\sin(t\pi))dt & = \ln(2)+ \int_{0}^{1}\ln(\sin(t\pi))dt+\int_{0}^{1}\ln(\sin(t\pi))dt \\ \Longrightarrow \int_{0}^{1}\ln(\sin(\pi t))dt & = -\ln(2) \\ \tag{3} \end{align*}$$ We conclude from results (1),(2) and (3) our claim: $$\begin{align*} \frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...&= \frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}-\int_{0}^{1}\ln(\sin(t\pi))dt\\ =&\frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}+\frac{1}{2}\ln(2) \\ \end{align*}$$ $$\boxed{ \exp\left(\frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...\right)=\sqrt{\frac{2\pi}e}}$$

Feynman's trick

Integral involving the Feynman's integral trick

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— 級数bot (@infseriesbot) June 16, 2021

Today we show the proof of this result which involves the famous Feynman's integral trick twice. The post of @infseriesbot has a typo (the sign is wrong). First, we differentiate the parameter $a$ under the integral sign: $$\begin{align*} \frac{d}{da} \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx =& \int_{0}^{\infty} \frac{d}{da} \frac{(1-\cos a x)\ln(x)}{x^2}dx\\ =&\int_{0}^{\infty} \frac{ \sin a x \ln (x)}{x} dx\\ =&\int_{0}^{\infty} \frac{ \sin w \ln (\frac{w}{a})}{w} dw \quad (w \mapsto ax)\\ =&\int_{0}^{\infty} \frac{ \sin w \ln (w)}{w} dw-\ln (a)\int_{0}^{\infty} \frac{ \sin w }{w} dw \\ =&\int_{0}^{\infty} \frac{\sin w \ln (w)}{w} dw-\frac{\pi \ln(a)}{2}\\\ =& \int_{0}^{\infty} \frac{ \sin w \frac{d}{dt} w^{t} \big|_{t=0+}}{w} dw-\frac{\pi \ln(a)}{2}\\ =& \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} w^{t-1} \sin w dw -\frac{\pi \ln(a)}{2} \end{align*}$$ Here we are using the fact that $\displaystyle \frac{d}{dt} w^{t} \big|_{t=0+} =\ln(w)$ and the fact that $\displaystyle \int_{0}^{\infty} \frac{\sin(w)}{w}dw = \frac{\pi}{2}$ (a famous result obtained through complex analysis). Note that the first integral is the Mellin transform of $\sin w$, so we could use the following expansion series of $\sin$ (a consequence of the Euler's formula and the de Moivre theorem) in oder to apply the Ramanujan's master theorem: $$\sin w = \sum_{n=0}^{\infty}\frac{(-w)^n[-\sin\left(\frac{n\pi}{2}\right)]}{n!}$$ Then, by Ramanujan's master theorem: $$\begin{align*} \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} w^{t-1} \sin w dw= &\frac{d}{dt}\Big|_{t=0+} \left[\Gamma(t) \sin\left(\frac{t\pi}{2}\right)\right] \\ =& \lim_{t \to 0+}\Gamma(t)\left[\frac{\pi}{2}\cos\left(\frac{t\pi}{2}\right) +\psi^0(t)\sin\left(\frac{\pi t}{2} \right) \right]\\ =& \lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} \right] \end{align*}$$ Update: I found a cool way to calculate this limit:
Note that $$\begin{align*} \frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} = &\frac{\pi}{2}\frac{1}{t}-\frac{\pi^3}{2^3}\frac{t}{2!}+\frac{\pi^5}{2^5}\frac{t^3}{4!}-\frac{\pi^7}{2^7}\frac{t^5}{6!}+... \\ =& \frac{\pi}{2}\frac{1}{t}+\mathcal{O}(t) \end{align*}$$ Also note that $$\begin{align*}\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} =&\psi^0(t)\frac{\pi}{2}-\psi^0(t)\frac{\pi^3}{2^3}\frac{t^2}{3!}+\psi^0(t)\frac{\pi^5}{2^5}\frac{t^4}{5!}-\psi^0(t)\frac{\pi^7}{2^7}\frac{t^6}{7!}+...\\ =& \psi^0(t)\frac{\pi}{2}-t\psi^0(t)\left(\frac{\pi^3}{2^3}\frac{t}{3!}-\frac{\pi^5}{2^5}\frac{t^3}{5!}+\frac{\pi^7}{2^7}\frac{t^5}{7!}-...\right)\\ =& \psi^0(t)\frac{\pi}{2}-t\psi^0(t)\mathcal{O}(t) \end{align*}$$ Therefore $$\begin{align*}\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} =& \frac{\pi}{2}\frac{1}{t}+\psi^0(t)\frac{\pi}{2}+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t) \\ =& \frac{\pi}{2}\left(\frac{1}{t}+ \psi^0(t)\right)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t)\\ =&\frac{\pi}{2}\psi^{0}(t+1)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t)\\ \end{align*}$$ $$\lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\frac{\cos\left(\frac{t\pi}{2}\right)}{t} +\psi^0(t)\frac{\sin\left(\frac{\pi t}{2} \right)}{t} \right] =\lim_{t \to 0+}\Gamma(t+1)\left[\frac{\pi}{2}\psi^{0}(t+1)+\mathcal{O}(t)-t\psi^0(t)\mathcal{O}(t) \right]=-\frac{\pi}{2}\gamma$$
where $\displaystyle \lim_{t \to 0+} \psi^0(t+1) =-\gamma$ and $\displaystyle \lim_{t \to 0+} t\psi^0(t) =-1$

Therefore
$$\begin{align*} \frac{d}{da} \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx =& -\frac{\pi}{2} \gamma -\frac{\pi \ln(a)}{2}\\ \Longrightarrow \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx = &-\frac{\pi}{2} \int a\gamma +a\ln(a) da + C\\ =& -\frac{a\pi\gamma}{2} - \frac{a\pi \ln(a)}{2} + \frac{\pi a}{2} + C \end{align*}$$ Note that if $a=0$ then $C=0$ therefore $$\boxed{ \int_{0}^{\infty} \frac{(1-\cos a x)\ln(x)}{x^2}dx = -\frac{\pi a}{2}(\gamma +\ln(a)-1)}$$

Continued fractions

Continued fraction expansion for $\coth\left(z\right)$

We will show the proof for this continued fraction posted by @infseriesbot. $$\frac{e+1}{e-1} = \coth\left(\frac{1}{2}\right) = 2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cdots}}}$$ Update We add the proof of this result, which is bascially the same function evaluated at other point: $$0 = \coth\left(-\frac{i\pi}{2}\right) = 2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cdots}}}$$
Recall the continued fraction representation for ratios of confluent hypergeometric functions of the type $_{0}F_{1}(c,z)$: Let $(a_{n})$ be a sequence of complex numbers defined by $$a_{n} = \frac{1}{(c+n-1)(c+n)}$$ where $c$ is a complex constant such that $c\notin \mathbb{Z^{-}}\cup \{0\}$. Then $$1+\mathcal{K}_{n=1}^{\infty} \left(\frac{a_{n}z}{1}\right) = \frac{_{0}F_{1} (c,z)}{_{0}F_{1} (c+1,z)}$$
Now, for this particular case, we will use the hypergemetric representation of $\sin(z)$ and $\cos(z)$ $$\sin(z)= z \, {_{0}F_{1}}\left(\frac{3}{2},-\frac{z^2}{4}\right)$$ $$\sin(z)= {_{0}F_{1}}\left(\frac{1}{2},-\frac{z^2}{4}\right)$$ Then $$\begin{align*} \tan(x) = &\frac{\sin(x)}{\cos(x)} = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}{1+\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}\\ =& \frac{x(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...)}{1+\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}\\ =& \frac{x\cdot _{0}F_{1}\left(\frac{3}{2},-\frac{x^2}{4}\right)}{ _{0}F_{1}\left(\frac{1}{2},-\frac{x^2}{4}\right)}\\ =& \frac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}} \end{align*}$$ If we evaluate at $x=\frac{iz}{2}$ $$\begin{align*} \tan\left(\frac{iz}{2}\right) = &\frac{iz^2}{2-\cfrac{-z^2}{6-\cfrac{-z^2}{10-\cfrac{-z^2}{14-\cdots}}}}\\ =&\frac{iz}{2+\cfrac{z^2}{6+\cfrac{z^2}{10+\cfrac{z^2}{14-\cdots}}}}\\ \Longrightarrow \coth\left(\frac{z}{2}\right)z = &\cot\left(\frac{iz}{2}\right)iz = 2+\cfrac{z^2}{6+\cfrac{z^2}{10+\cfrac{z^2}{14+\cdots}}} \end{align*}$$ If we evaluate at $z=1$ $$\boxed{ \frac{e+1}{e-1} = \coth\left(\frac{1}{2}\right) = 2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cdots}}}}$$ Since the result is valid for al the complex plane, if we evaluate at $z=-i\pi$ then $\displaystyle \coth\left(-\frac{i\pi}{2}\right) =0$ and $$\boxed{ 0 = \coth\left(-\frac{i\pi}{2}\right) = 2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cdots}}}}$$

Residue theorem

Integral involving the Residue theorem

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— 級数bot (@infseriesbot) June 13, 2021

Today we present here the proof of this nice integral. The solution involves Complex Analysis, geometric series, even functions and trigonometry $$\begin{align*} \int_{0}^{\infty} \frac{\cos \sqrt{x} }{e^{2\pi \sqrt{x}}-1}dx = & \int_{0}^{\infty}\left[ \sum_{n=1}^{\infty} e^{-2\pi n \sqrt{x}} \cos \sqrt{x}\right] dx\\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}e^{-2\pi n \sqrt{x}} \cos \sqrt{x} dx \\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}e^{-2\pi n \sqrt{x}} \left(\frac{e^{i\sqrt{x}}+e^{-i\sqrt{x}}}{2}\right) dx \\ =& \sum_{n=1}^{\infty} \int_{0}^{\infty}\frac{e^{\sqrt{x}(i-2\pi n)}+e^{\sqrt{x}(-i-2\pi n)}}{2} dx \\ =& \frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{\sqrt{x}(i-2\pi n)}dx+\frac{1}{2}\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{\sqrt{x}(-i-2\pi n)}dx\\ =& \frac{1}{2}\sum_{n=1}^{\infty} \tfrac{e^{\sqrt{x}(i-2\pi n)}(-1+(i-2\pi n) \sqrt{x})}{(i-2\pi n)^2} \Big|_0^\infty+\frac{1}{2}\sum_{n=1}^{\infty} \tfrac{e^{\sqrt{x}(-i-2\pi n)}(-1+(-i-2\pi n) \sqrt{x})}{(-i-2\pi n)^2}\Big|_{0}^{\infty}\\ =& \sum_{n=1}^{\infty} \frac{1}{(i-2\pi n)^2}+ \sum_{n=1}^{\infty} \frac{1}{(-i-2\pi n)^2}\\ =& \sum_{n=1}^{\infty} \left[\frac{1}{(i-2\pi n)^2}+\frac{1}{(-i-2\pi n)^2}\right]\\ =& 2+\sum_{n=0}^{\infty} \left[\frac{1}{(i-2\pi n)^2}+\frac{1}{(-i-2\pi n)^2}\right]\\ =& 1+\frac{1}{2}\sum_{n=-\infty}^{\infty} \frac{1}{(i-2\pi n)^2} + \frac{1}{2}\sum_{n=-\infty}^{\infty} \frac{1}{(-i-2\pi n)^2}\\ =& 1- \frac{1}{2}\operatorname{Res}\left(\frac{\pi \cot \pi z}{(i-2\pi z)^2}, \frac{i}{2\pi} \right)- \frac{1}{2}\operatorname{Res}\left(\frac{\pi \cot \pi}{(-i-2\pi z)^2}, -\frac{i}{2\pi} \right)\\ =&1- \frac{1}{2}\lim_{z\to \frac{i}{2\pi}} \left[\frac{d}{dz} \frac{(z-\frac{i}{2\pi})^2(\pi \cot \pi z)}{(i-2\pi z)^2}\right]- \frac{1}{2}\lim_{z\to -\frac{i}{2\pi}} \left[\frac{d}{dz} \frac{(z+\frac{i}{2\pi})^2(\pi \cot \pi z)}{(-i-2\pi z)^2}\right]\\ =&1- \frac{1}{8\pi}\lim_{z\to \frac{i}{2\pi}} \frac{d}{dz}\cot \pi z- \frac{1}{8\pi}\lim_{z\to -\frac{i}{2\pi}} \frac{d}{dz}\cot \pi z\\ =&1+ \frac{1}{8}\csc^2 \left(\frac{i}{2}\right) + \frac{1}{8}\csc^2\left( -\frac{i}{2}\right)\\ =& 1-\frac{1}{8}\left( \frac{2i}{e^{-\frac{1}{2}}-e^{\frac{1}{2}}}\right)^2 -\frac{1}{8}\left( \frac{2i}{e^{\frac{1}{2}}-e^{-\frac{1}{2}}}\right)^2 \\ =& 1- \frac{e}{( e-1)^2} \end{align*}$$

Gaussian integral

Generalization of Gaussian integral

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— 級数bot (@infseriesbot) June 9, 2021

We are going to prove this result posted by @infeseriesbot that is a generalization of the Gaussian integral $(s=\frac{1}{2})$. $$\int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx = \pi^{1-s} \quad \operatorname{Re}(s)>0$$ It was previously discussed in Twitter a few months ago and turned out that the proof is a consquence of the Ramanujan's master theorem. However, other proofs were provided using the bracketts method, the log-gamma function and contour integration. In the proof we also find the values of $s$ that guarantee the convergence of the integral $(\operatorname{Re}(s)>0)$.

Proof. Recall the Ramanujan's master theorem: If a function $f(x)$ has an expansion of the form $$f(x) = \sum_{k=0}^{\infty} \frac{\phi(k)}{k!} (-x)^k$$ for some function (say analytic or integrable) $\varphi(k)$, then the Mellin transform of $f(x)$ is given by $$\{\mathscr{M}f(x)\}(t) = \int_{0}^{\infty} x^{t-1} f(x) dx = \Gamma(t)\varphi(-t)$$ In this particular case $$\begin{align*} \int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx =& 2\int_{0}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx \\ =& \int_{0}^{\infty}\sum_{n=0}^{\infty} w^{\frac{1}{2}-1} \frac{(-w)^n}{n!^{2s}} dw \quad (w \mapsto x^2)\\ =& \int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } \frac{(-w)^n}{n!^{2s}} dw \\ =& \int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } (n!)^{1-2s} \frac{(-w)^n}{n!} dw \end{align*}$$ this is a Mellin transform $\displaystyle \{\mathscr{M}f(w)\}\left(\frac{1}{2}\right) = \int_{0}^{\infty} w^{\frac{1}{2}-1} f(w)dw$ $$\textrm{ where } f(w) = \sum_{n=0}^{\infty } (n!)^{1-2s}\frac{(-w)^n}{n!} \quad \textrm{and } \; \phi(-t) = (-t)!^{1-2s} = \left[\Gamma(1-t)\right]^{1-2s}$$ By the Ramanujan's master theorem $$\int_{0}^{\infty} w^{\frac{1}{2}-1} \sum_{n=0}^{\infty } (n!)^{1-2s}\frac{(-w)^n}{n!} dw = \Gamma\left(\frac{1}{2}\right) \left[\Gamma\left(\frac{1}{2}\right)\right]^{1-2s} = \pi^{1-s}$$ The result is valid for $\operatorname{Re}(s)>0$. To show this, lets review the convergence of $\displaystyle \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx$. We can use the ratio test for the sequence $\displaystyle \left(\frac{(-x^2)^n}{n!^{2s}}\right)_{n}$ $$\begin{align*} \lim_{n\to \infty}\left| \frac{\frac{(-x^2)^{n+1}}{(n+1)!^{2s}}}{\frac{(-x^2)^{n}}{n!^{2s}}}\right| =& \lim_{n\to \infty} \left| \frac{n!^{2s}}{(n+1)!^{2s}}\right||x^2| \\ =& \lim_{n\to \infty} \left| \frac{n!^{2s}}{(n+1)!^{2s}}\right||x^2| \\ =& \lim_{n \to \infty } \exp\left(2\operatorname{Re}(s\ln n!)-2\operatorname{Re}(s\ln(n+1)!)\right) |x^2|\\ =& \lim_{n \to \infty } \exp\left(2\operatorname{Re}(s) \left(\ln n!-\ln(n+1)!\right)\right) |x^2|\\ =& \exp\left( \lim_{n \to \infty } \operatorname{Re}(s) 2\left(\ln n!-\ln(n+1)!\right)\right) |x^2|\\ =& \exp \left(\operatorname{Re}(s) \lim_{n \to \infty}2\ln\left(\frac{n!}{(n+1)!}\right) \right) |x^2|\\ =& \exp \left(\operatorname{Re}(s) (-\infty) \right) |x^2| \end{align*}$$ $$\lim_{n\to \infty}\left| \frac{\frac{(-x^2)^{n+1}}{(n+1)!^{2s}}}{\frac{(-x^2)^{n}}{n!^{2s}}}\right| = \begin{cases} 0 \quad \textrm{ if } \operatorname{Re}(s) >0 \\ |x^2| \quad \textrm{ if } \operatorname{Re}(s) =0\\ \infty \quad \textrm{ if } \operatorname{Re}(s) <0 \end{cases}$$ Therefore $$\sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx <\infty \quad \textrm{if} \begin{cases} \operatorname{Re}(s)>0\\ \operatorname{Re}(s)=0 \wedge |x|<1 \end{cases}$$ Then, we have obtained the desired result: $$\boxed{ \int_{-\infty}^{\infty} \sum_{n=0}^{\infty } \frac{(-x^2)^n}{n!^{2s}} dx = \pi^{1-s} \quad \operatorname{Re}(s)>0}$$