Integral of the day XXXI

Integral involving a partial fraction expansion

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \frac{\pi n}{2(2n-1)}$$ The proof relies on the partial fraction decomposition and some properties of the binomial coefficient.

Proof:

We need the following partial fraction decomposition (inspired by this this exercise ): $$\prod_{k=1}^{n} \frac{1}{x+k^2} = 2\sum_{k=1}^{n}\frac{(-1)^{k-1}k^2}{(x+k^2)(n-k)!(n+k)!}$$

Here is the proof:

Since $\deg 1 \lt \deg (x+1)(x+2^2)\cdots (x+n^2)$ there exists a partial fraction decomposition: $$\frac{1}{(x+1)(x+2^2)\cdots(x+n^2)} = \sum_{k=1}^{n} \frac{A_{k}}{x+k^2}$$ Multiplying by $(x+1)(x+2^2)\cdots (x+n^2)$ $$1= \sum_{k=1}^{n} A_{k}\prod_{i=1 \atop i\neq k}^n (x+i^2)$$ Suppose that $x = -m^2$ with $1\leq m \leq n$ Hence $$\begin{align*} 1= & A_{m}\prod_{i=1 \atop i\neq m}^n (-m^2+i^2) \\ =& A_{m} (-m^2+1)(-m^2+2^2)+\cdots(-m^2+(m-1)^2)(-m^2+(m+1)^2)\cdots (-m^2+n^2)\\ =& A_{m} \left(-(m^2-1)\right)\left(-(m^2-2^2)\right)\cdots\left(-(m^2-(m-1)^2)\right)\left((m+1)^2-m^2\right)\cdots\left(n^2-m^2\right)\\ =& A_{m}(-1)^{m-1} (m-1)(m+1)(m-2)(m+2)\cdots\left(m-(m-1)\right)\left(m+(m-1)\right)\left((m+1)+m\right)\left((m+1)-m\right)\cdots (n-m)(n+m)\\ =& A_{m}(-1)^{m-1}\left[(m-1)(m-2)\cdots 2\cdot 1 \cdot 2 \cdots (n-m)\right]\left[(m+1)(m+2)\cdots \left(m+(m-1)\right)\left(m+(m+1)\right)\cdots (n+m)\right]\\ =& A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[(m+1)(m+2)\cdots \left(m+(m-1)\right)\left(m+(m+1)\right)\cdots (n+m)\right]\\ \end{align*}$$ Multiplying and dividing by $\displaystyle \frac{(2m) m!}{(2m) m!}$ $$\begin{align*} 1 = &\frac{A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[m!(m+1)(m+2)\cdots \left(m+(m-1)\right)(2m)\left(m+(m+1)\right)\cdots (n+m)\right]}{(2m)m!}\\ =& \frac{A_{m} (-1)^{m-1}\left[(m-1)!(n-m)!\right]\left[(n+m)!\right]}{(2m)m!} \end{align*}$$ Multiplying and dividing by $m$ $$\begin{align*} 1 = & \frac{A_{m} (-1)^{m-1}m!(n-m)!(n+m)!}{(2m^2)m!}\\ =& \frac{A_{m} (-1)^{m-1}(n-m)!(n+m)!}{2m^2} \end{align*}$$ Therefore $$A_{m} = \frac{2(-1)^{m-1}m^2}{(n-m)!(n+m)!} \quad 1\leq m \leq n$$ Hence $$\prod_{k=1}^{n}\frac{1}{x+k^2} = 2\sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(x+k^2)(n-k)!(n+k)!}$$ Back to the integral: $$\begin{align*} I = \int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = & \left(\prod_{k=1}^{n}k^2 \right) \int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{1}{x^2+k^2} dx\\ =& 2n!^2 \int_{0}^{\infty} \sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(x^2+k^2)(n-k)!(n+k)!} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k^2}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{x^2+k^2} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{\left(\frac{x}{k}\right)^2+1} dx\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}\int_{0}^{\infty} \frac{1}{w^2+1} dx \quad \left(w \mapsto \frac{x}{k}\right)\\ =& 2n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}\left[\arctan(w)\right]_{0}^{\infty}\\ =& \pi n!^2 \sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!} \end{align*}$$ where we have used the fact that $\displaystyle \prod_{k=1}^{n} k^2 = \left(\prod_{k=1}^{n} k\right)\left(\prod_{k=1}^{n} k\right) = n!^2$

Finally, we have to prove that $$\pi n!^2 \underbrace{\sum_{k=1}^{n} \frac{(-1)^{n-1}k}{(n-k)!(n+k)!}}_{S} = \frac{\pi n}{2(2n-1)}$$ Note that $$\begin{align*} S =\frac{1}{(2n)!}\sum_{k=1}^{n} \frac{(-1)^{k-1}k(2n)!}{(n-k)!(n+k)!} =& \frac{1}{(2n)!}\sum_{k=1}^{n} \binom{2n}{n+k}(-1)^{k-1}k\\ =& \frac{1}{(2n)!}\sum_{j=n+1}^{2n} \binom{2n}{j}(-1)^{n-j-1}(j-n) \quad (j=n+k)\\ =& \frac{1}{(2n)!}\sum_{j=n+1}^{2n} \binom{2n}{2n-j}(-1)^{n-j-1}(j-n) \quad \left(\binom{v}{v-w} = \binom{v}{w} \right)\\ =& \frac{1}{(2n)!}\sum_{m=n-1}^{m=0} \binom{2n}{m}(-1)^{m-n-1}(n-m) \quad (m=2n-j) \\ =& \frac{(-1)^{n-1}}{(2n)!}\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}(n-m)\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}-\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}m \right]\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m}-2n\sum_{m=0}^{n-1} \binom{2n-1}{m-1}(-1)^{m} \right]\\ =& \frac{(-1)^{n-1}}{(2n)!}\left[2n\sum_{m=0}^{n-1} \binom{2n-1}{m}(-1)^{m}-n\sum_{m=0}^{n-1} \binom{2n}{m}(-1)^{m} \right] \quad \left(\binom{v+1}{w} = \binom{v}{w} + \binom{v}{w-1} \right)\\ =& \frac{1}{(2n)!}\left[2n\binom{2n-2}{n-1}- n\binom{2n-1}{n-1} \right] \quad \left(\sum_{j=0}^{m} (-1)^j\binom{v}{j} = (-1)^n\binom{v-1}{m}\right)\\ =& \frac{1}{(2n)!}\left[\frac{1}{2n-1}\frac{(2n)!}{(n-1)!(n-1)!}- \frac{1}{2n}\frac{(2n)!}{(n-1)!(n-1)!} \right]\\ =& \frac{n}{2(2n-1)n!n!} \end{align*}$$ Hence $$I =\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \pi n!^2\sum_{k=1}^{n} \frac{(-1)^{k-1}k}{(n-k)!(n+k)!} = \pi n!^2 S = \frac{\pi n}{2(2n-1)}$$ Therefore, we can conclude $$\boxed{\int_{0}^{\infty} \prod_{k=1}^{\infty} \frac{k^2}{x^2+k^2} dx = \frac{\pi n}{2(2n-1)}}$$

Integral of the day XXX

An integral involving an orthogonal system

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} = \frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})} \quad$$ $$m,n\in \mathbb{Z}_{>0} \textrm{ with } \gcd(m,n) = 1 \textrm{ and } |a|,|b|\leq 1$$ The proof relies on Fourier series and the theory of orthogonal functions.

Proof:

We will use the following Fourier series (Proof in the appendix): $$1+2\sum_{k=1}^{\infty} p^k\cos(kx) = \frac{1-p^2}{1-2p\cos x + p^2} \quad |p|\leq |$$ Let $\displaystyle m,n\in \mathbb{Z}_{>0}$ with $\gcd(m,n) = 1$ and $|a|,|b|\leq 1$ Then $$\begin{align*} \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =& \frac{1}{(1-a^2)(1-b^2)}\int_{0}^{\pi} \frac{1-a^2}{(1-2a\cos(mx)+a^2)}\frac{1-b^2}{(1-2b\cos(mx)+b^2)}dx\\ =& \frac{1}{(1-a^2)(1-b^2)}\int_{0}^{\pi} \left( 1+2\sum_{k=1}^{\infty} a^k\cos(kmx)\right)\left( 1+2\sum_{j=1}^{\infty} b^j\cos(jnx)\right) dx\\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\int_{0}^{\pi}dx + 2\sum_{k=1}^{\infty}a^k\int_{0}^{\pi} \cos(kmx)dx+ 2\sum_{j=1}^{\infty}b^j\int_{0}^{\pi} \cos(jnx)dx + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right] \end{align*}$$ Recall that a system of real functions $$\phi_{0}(x), \phi_{1}(x), \phi_{2}(x),..., \phi_{n}(x),...$$ is said to be orthogonal on the interval $\left[a,b\right]$ if $$\int_{a}^{b} \phi_{n}(x) \phi_{m}(x) dx = 0 \quad n\neq m, m=0,1,2,...$$ We shall also assume that $$\int_{a}^{b} \phi_{n}^{2}(x) dx\neq 0 \quad n=0,1,2,...$$ Consider the system $$1,\cos(x), \cos(2x),...,\cos(nx),...$$ This sytem is orthogonal on the interval $\left[0,\pi\right]$. In fact $$\int_{0}^{\pi} \cos(nx) dx = \left[\frac{\sin(nx)}{n}\right]_{0}^{\pi} = 0 \quad n=1,2,... \tag{1}$$ which means that the functions $\cos(nx)$ and $1$ are orthogonal. Moreover, we have $$\begin{align*} \int_{0}^{\pi} \cos(nx)\cos(mx)dx =& \frac{1}{2} \int_{0}^{\pi} \left[\cos(n+m)x + \cos(n-m)x\right] dx \\ =& \frac{1}{2}\int_{0}^{\pi} \cos(n+m)x dx + \frac{1}{2}\int_{0}^{\pi} \cos(n-m)x dx =0 \quad n\neq m\\ =& 0 \tag{2} \end{align*}$$ Finally, note that $$\int_{0}^{\pi} \cos^2(nx) dx = \frac{1}{2} \int_{0}^{\pi} 1+\cos(2nx) dx = \frac{\pi}{2} \tag{3}$$ Hence, $$\begin{align*} \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =& \frac{1}{(1-a^2)(1-b^2)}\left[\int_{0}^{\pi}dx + 2\sum_{k=1}^{\infty}a^k\underbrace{\int_{0}^{\pi} \cos(kmx)dx}_{=0}+ 2\sum_{j=1}^{\infty}b^j\underbrace{\int_{0}^{\pi} \cos(jnx)dx}_{=0} + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right] \quad \textrm{from (1)} \\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\underbrace{\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx}_{J} \right] \end{align*}$$ From (2) note that $$J = \int_{0}^{\pi} \cos(kmx)\cos(jnx) dx = \begin{cases} 0 \textrm{ if } & km \neq jn \\ \frac{\pi}{2} \textrm{ if }& km = jn \end{cases}$$ However, since $\gcd(m,n)=1$ then $n\nmid m \; \wedge \; m \nmid n$ Therefore $$km = jn \Longleftrightarrow k = in \wedge j=im \quad i\in \mathbb{Z}\setminus\left\{0\right\}$$ Hence if $\displaystyle k = in \wedge j=im$ with $i\in \mathbb{Z}_{>0} \textrm{ and } |a|,|b|\lt 1$ $$\begin{align*} \sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx =& \sum_{i=1}^{\infty}a^{in}b^{im}\int_{0}^{\pi} \cos(inmx)\cos(imnx) dx\\ =& \sum_{i=1}^{\infty}a^{in}b^{im}\int_{0}^{\pi} \cos^2(inmx) dx\\ =& \frac{\pi}{2} \sum_{i=1}^{\infty}a^{in}b^{im} \quad \textrm{ from (3) } \\ =& \frac{\pi}{2} \sum_{i=1}^{\infty}\left(a^{n}b^{m}\right)^{i}\\ =& \frac{\pi}{2} \frac{a^nb^{m}}{(1-a^nb^{m})} \tag{4} \end{align*}$$ Therefore $$\begin{align*} I =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + 4\sum_{k=1}^{\infty}\sum_{j=1}^{\infty}a^kb^j\int_{0}^{\pi} \cos(kmx)\cos(jnx) dx \right]\\ =& \frac{1}{(1-a^2)(1-b^2)}\left[\pi + \frac{2\pi a^nb^{m}}{(1-a^nb^{m})} \right] \quad \textrm{ from (4)} \\ =& \frac{\pi }{(1-a^2)(1-b^2)}\left[\frac{(1-a^nb^{m})}{(1-a^nb^{m})} + \frac{2 a^nb^{m}}{(1-a^nb^{m})} \right]\\ =& \frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})}\\ \end{align*}$$ Hence $$\boxed{ \displaystyle \int_{0}^{\pi} \frac{dx}{(1-2a\cos(mx)+a^2)(1-2b\cos(nx)+b^2)} =\frac{\pi (1+a^nb^{m})}{(1-a^2)(1-b^2)(1-a^nb^{m})} }$$

Appendix

$$1+2\sum_{k=1}^{\infty} p^k\cos(kx) = \frac{1-p^2}{1-2p\cos x + p^2} \quad |p|\lt 1$$ Proof $$\begin{align*} 1+2\sum_{k=1}^{\infty} p^k\cos(kx) =& 1+2\Re\left(\sum_{k=1}^{\infty} p^ke^{ixk}\right)\\ =& 1+2\Re\left(\frac{pe^{ix}}{1-pe^{ix}}\right)\\ =& 1+2\Re\left(\frac{p\cos x +ip\sin x }{1-p\cos x -ip\sin x}\right)\\ =& 1+2\Re\left(\frac{p\cos x +ip\sin x}{1-p\cos x -ip\sin x}\frac{1-p\cos x +ip\sin x}{1-p\cos x+ip\sin x}\right)\\ =& 1+2\Re\left(\frac{p\cos x -p^2 +i(p^2\cos x \sin x +p\sin x (1-p\cos x))}{(1-p\cos)^2+ p^2\sin^2 x}\right)\\ =& 1+\frac{2p\cos x-2p^2}{1-2p\cos x + p^2}\\ =& \frac{1-2p\cos + p^2}{1-2p\cos + p^2}+\frac{2p\cos x-2p^2}{1-2p\cos + p^2}\\ =& \frac{1-p^2}{1-2p\cos + p^2}\\ \end{align*}$$

Integral of the day XXIX

Integral representation of the $\arcsin(a)$ function

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) \quad |a|\leq 1$$ In the proof we use the residue theorem to evaluate it.

Proof:

Recall the logarithmic definition of $\operatorname{arcanh}(x)$ $$\operatorname{arcanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x} \right) \quad |x|\lt 1$$ Hence $$I(a) = \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = 2\int_{0}^{1} \frac{\operatorname{arcanh}(ax)}{x\sqrt{1-x^2}} dx$$ Hence $$\begin{align*} I'(a) = 2\int_{0}^{1} \frac{1}{\sqrt{1-x^2}(1-a^2x^2)} dx = & 2\int_{0}^{\frac{\pi}{2}} \frac{1}{1-a^2\sin^2(w)} dw \quad (x \mapsto \sin(w)) \\ =& 4\int_{0}^{\frac{\pi}{2}} \frac{1}{a^2\cos(2w)-a^2+2} dw \\ =& 2\int_{0}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta \quad (\theta \mapsto 2w)\\ =& \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta \quad \textrm{(The integrand is even)} \end{align*}$$ If we make the following change of variable $$\cos\theta = \frac{z+z^{-1}}{2}$$ $$d\theta = \frac{dz}{zi}$$ $$I'(a) = \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta = \frac{2}{i}\oint_{|z|=1} \frac{dz}{a^2z^2+(4-2a^2)z+a^2}$$ The integrand has two poles: $$z_{1} = \frac{a^2-2+2\sqrt{1-a^2}}{a^2}$$ $$z_{2} = \frac{a^2-2-2\sqrt{1-a^2}}{a^2}$$ Only $z_{1}$ is inside the unit circle. Suppose $|a|\leq 1$: $$\begin{align*} \left| \frac{a^2-2+2\sqrt{1-a^2}}{a^2}\right|\lt 1 \Longrightarrow& -a^2\lt a^2-2+2\sqrt{1-a^2}\lt a^2 \\ \Longrightarrow& -2a^2\lt-2+2\sqrt{1-a^2}\lt0\\ \Longrightarrow& 1-a^2\lt \sqrt{1-a^2}\lt 1\\ \Longrightarrow& a\neq 0 \; \wedge \; a\neq 1 \end{align*}$$ Then, if we add the assumption $|a| \lt 1 \;\wedge\; a\neq 0 \Longrightarrow |z_{1}|\lt1$

By the residue theorem: $$\begin{align*} I'(a) = \int_{-\pi}^{\pi } \frac{1}{a^2\cos \theta -a^2+2} d\theta =& \frac{2}{i}\oint_{|z|=1} \frac{dz}{a^2z^2+(4-2a^2)z+a^2}\\ =& \frac{4\pi}{a^2} \lim_{z \to \frac{a^2-2+2\sqrt{1-a^2}}{a^2}} \frac{1}{\left(z-\frac{a^2-2-2\sqrt{1-a^2}}{a^2}\right)}\\ =& \frac{4\pi}{\left(a^2-2+2\sqrt{1-a^2}-\left(a^2-2-2\sqrt{1-a^2}\right)\right)}\\ =& \frac{\pi}{\sqrt{1-a^2}} \end{align*}$$ Therefore $$I(a) = \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) + C$$ If we let $a \to 0$ then $C \to 0$.

The case when $a = 1$ can be obtained as a limitng case easily.

Therefore $$\boxed{ \int_{0}^{1} \ln\left(\frac{1+ax}{1-ax} \right)\frac{dx}{x\sqrt{1-x^2}} = \pi \arcsin(a) \quad |a|\leq 1}$$

The trilogarithm function

The strong connection between the $\operatorname{Li}_{s}$ function and $\phi$ vol. III

Today we show the proof of this integral posted by @integralsbot $$\int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx = \frac{\zeta(3)}{10}$$ The proof relies on some identities of the trilogarithm and dilogarithm function. We evaluated similar integrals and series here, here.

Proof:

$$\begin{align*} \int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx =& \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)(w+1)}{w(w-1)}dw \quad (w\mapsto e^{2x})\\ =& \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1+}dw + \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)}{w(w-1)}dw \\ =& \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw + \frac{1}{8} \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw - \frac{1}{8}\int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw \\ =& \frac{1}{4} \underbrace{\int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw}_{J} - \frac{1}{8}\underbrace{\int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw}_{K} \end{align*}$$ $$\begin{align*} K = \int_{1}^{\phi^2} \frac{\ln^2(w)}{w}dw =& \int_{0}^{\ln \phi^2 } s^2 ds \quad (s \mapsto \ln(w))\\ =& \left[\frac{s^3}{3}\right]_{0}^{\ln \phi^2}\\ =& \frac{\ln^3(\phi^2)}{3} \end{align*}$$ $$\begin{align*} J = \int_{1}^{\phi^2} \frac{\ln^2(w)}{w-1}dw = & \int_{0}^{\phi} \frac{\ln^2(1+s)}{s} ds \quad (s \mapsto w-1, \phi^2-1 = \phi) \\ \stackrel{IBP}{=}& \ln(s)\ln^2(1+s)\Big|_{0}^{\phi} - 2\int_{0}^{\phi} \frac{\ln(s)\ln(1+s)}{1+s} ds\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln(t-1)\ln(t)}{t}dt \quad (t \mapsto s+1 )\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln\left(t\left(1-\frac{1}{t}\right)\right)\ln(t)}{t}dt\\ =& \ln(\phi)\ln^2(1+\phi) - 2\int_{1}^{\phi^2} \frac{\ln^2(t)}{t}dt - 2\int_{1}^{\phi^2} \frac{\ln\left(1-\frac{1}{t}\right)\ln(t)}{t}dt\\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) -2\int_{1}^{\phi^{-2}} \frac{\ln\left(1-y\right)\ln(y)}{y}dy\\ \stackrel{IBP}{=}& \ln(\phi)\ln^2(1+\phi^2) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(y)\ln(y)\Big|_{1}^{\phi^{-2}}-2\int_{1}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})-2\int_{1}^{0} \frac{\operatorname{Li}_{2}(y)}{y}dy -2\int_{0}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\int_{0}^{1} \frac{\operatorname{Li}_{2}(y)}{y}dy -2\int_{0}^{\phi^{-2}} \frac{\operatorname{Li}_{2}(y)}{y}dy \\ =& \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\operatorname{Li}_{3}(1) -2\operatorname{Li}_{3}(\phi^{-2}) \\ \end{align*}$$ Note that the polylogarithm function is defined by a power series in $z$: $$\operatorname{Li}_{s} (z) = \sum_{k=1}^{\infty} \frac{z^k}{k^s}$$ If we put $s=3$ and $z=1$ : $$\operatorname{Li}_{3} (1) = \sum_{k=1}^{\infty} \frac{1}{k^3} = \zeta(3)$$ Hence $$J = \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) +2\operatorname{Li}_{2}(\phi^{-2})\ln(\phi^{-2})+2\zeta(3) -2\operatorname{Li}_{3}(\phi^{-2})$$ Landen proved the following identity: $$\operatorname{Li}_{3}(\phi^{-2}) = \frac{4}{5}\zeta(3) + \frac{\pi^2}{15} \ln(\phi^{-2})- \frac{1}{12} \ln^3(\phi^{-2})$$ while Euler proved $$\operatorname{Li}_{2} (\phi^{-2}) = \frac{\pi^2}{15} -\ln^2(\phi)$$ Therefore $$J = \ln(\phi)\ln^2(1+\phi) - \frac{2}{3}\ln^3(\phi^2) - 2\ln^2(\phi)\ln(\phi^{-2}) + \frac{2}{5} \zeta(3) + \frac{\ln^3(\phi^{-2})}{6}$$ Additionally, from the identities: $$\ln(\phi)\ln^2(\phi+1) = 4\ln^3(\phi)$$ $$\ln^2(\phi)\ln(\phi^{-2}) = -2\ln^3(\phi)$$ $$\ln^3(\phi^{-2}) = -8\ln^3(\phi)$$ we have $$J = \frac{4}{3}\ln^3(\phi)+ \frac{2}{5} \zeta(3)$$ Hence $$I = \frac{1}{4}J -\frac{1}{8}K = \frac{1}{3}\ln^3(\phi)+ \frac{\zeta(3)}{10} - \frac{1}{3}\ln^3(\phi)$$ Therefore, we can conclude $$\boxed{ \int_{0}^{\ln \phi} \frac{x^2(e^{2x}+1)}{(e^{2x}-1)} dx = \frac{\zeta(3)}{10}}$$

Integral evaluated with the Guassian function

Today we show the proof of this integral posted by $$\int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = \frac{4\pi}{3\sqrt{3}}$$ The proof relies on a special case of the Gaussian function involving one of the roots of the unity, $e^{\frac{i\pi}{3}}$

Proof:

First note that $$\Im \left(\frac{4}{\sqrt{3}(1-i\sqrt{3}-2x)}\right) = \frac{1}{x^2-x+1 }$$ Hence $$\begin{align*} I = \int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = & \Im \left( \frac{4}{\sqrt{3}}\int_{0}^{1} \frac{dx}{(1-i\sqrt{3}-2x)\sqrt[3]{x(1-x)}}\right)\\ =& \Im \left( \frac{4}{\sqrt{3}(1-i\sqrt{3})}\int_{0}^{1} \frac{dx}{\left(1-\frac{2}{1-i\sqrt{3}}x\right)\sqrt[3]{x(1-x)}}\right)\\ \end{align*}$$ Note that $$\frac{2}{1-i\sqrt{3}} = \frac{2}{1-i\sqrt{3}}\left(\frac{1+i\sqrt{3}}{1+i\sqrt{3}}\right) = \frac{1+i\sqrt{3}}{2} = e^{\frac{i\pi}{3}}$$ Hence $$I = \Im \left( \frac{2}{\sqrt{3}}e^{\frac{i\pi}{3}} \underbrace{\int_{0}^{1} \frac{dx}{\left(1-e^{\frac{i\pi}{3}}x\right)\sqrt[3]{x(1-x)}}}_{J}\right)$$ Recall the integral representation of the Gaussian function: $$F(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_{0}^{1} x^{b-1}(1-x)^{c-b-1}(1-xz)^{-a} dx$$ This is called the Pochhammer integral and converges whenever $\quad \Re(c)>\Re(b)>0 \quad \arg\left(|1-z|\right)\lt \pi$

Therefore if we put $\displaystyle a=1, b= \frac{2}{3}$ and $\displaystyle c=\frac{4}{3}$ we have $$\begin{align*} J = \int_{0}^{1}\left(1-e^{\frac{i\pi}{3}} x\right)^{-1} x^{-\frac{1}{3}}(1-x)^{-\frac{1}{3}} dx =& \frac{\Gamma^2\left(\frac{2}{3}\right)}{\Gamma\left(\frac{4}{3}\right)} F\left(1, \frac{2}{3};\frac{4}{3};e^{\frac{i\pi}{3}}\right) \end{align*}$$ The Gaussian hypergeometric function satisfies the following property: $$F\left(3a,\tfrac{1}{3}+a;\tfrac{2}{3}+2a;{\mathrm{e}}^{\frac{\mathrm{i}\pi}{3% }}\right)=\sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi a}{2}}\left(\frac{16}{2% 7}\right)^{(3a+1)/6}\frac{\Gamma\left(\frac{5}{6}+a\right)}{\Gamma\left(\frac{% 2}{3}+a\right)\Gamma\left(\frac{2}{3}\right)},$$ If we put $a = \frac{1}{3}$ $$F\left(1,\tfrac{2}{3};\tfrac{4}{3};{\mathrm{e}}^{\frac{\mathrm{i}\pi}{3% }}\right)=\sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\left(\frac{16}{2 7}\right)^{\frac{1}{3}}\frac{\Gamma\left(\frac{7}{6}\right)}{\Gamma\left(\frac{2}{3}\right)} = \sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2\sqrt[3]{2}}{3}\frac{\Gamma\left(\frac{7}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}$$ Therefore $$\begin{align*} J = \int_{0}^{1}\left(1-e^{\frac{i\pi}{3}} x\right)^{-1} x^{-\frac{1}{3}}(1-x)^{-\frac{1}{3}} dx =& \sqrt{\pi}{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2\sqrt[3]{2}}{3}\frac{\Gamma\left(\frac{7}{6}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{4}{3}\right)} \\ =& \pi{\mathrm{e}}^{\frac{\mathrm{i}\pi }{6}}\frac{2}{3} \end{align*}$$ Hence $$\begin{align*} I = \int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = & \Im \left( \frac{2}{\sqrt{3}}e^{\frac{i\pi}{3}} \int_{0}^{1} \frac{dx}{\left(1-e^{\frac{i\pi}{3}}x\right)\sqrt[3]{x(1-x)}}\right)\\ =& \Im \left(\pi{\mathrm{e}}^{\frac{\mathrm{i}\pi }{2}}\frac{4}{3\sqrt{3}}\right)\\ =& \Im \left(i\pi\frac{4}{3\sqrt{3}}\right)\\ =& \frac{4\pi}{3\sqrt{3}} \end{align*}$$ Therefore, we can conclude $$\boxed{\int_{0}^{1} \frac{dx}{(x^2-x+1)\sqrt[3]{x(1-x)}} = \frac{4\pi}{3\sqrt{3}}}$$

Generalized hypergeometric functions IX

Series invoving a quotient of binomial coefficients

We show the proof of this nice result: $$\sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \frac{(2n)!^3}{n!^3(3n)!}$$ For the proof we use the Saalschütz theorem for hypergeometric functions.

Proof:

We will transform this series to an hypergeometric series

Note that $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ where $$(x)_{m} = x(x+1)(x+2)\cdots (x+m-1)$$ is the Pochhammer polynomial (rising factorial)

Hence $$S= \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \sum_{k=0}^{n} \frac{(n-k+1)_{k}(n-k+1)_{k}(n-k+1)_{k}}{(3n-k+1)_{k}k!k!}$$ Recall that the rising factorial obey the reflection formula: $$(-x)_{k} = (x-k+1)_{k}(-1)^k \tag{*}$$ Therefore $$\begin{align*} S =\sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} =& \sum_{k=0}^{n} \frac{(n-k+1)_{k}(n-k+1)_{k}(n-k+1)_{k}}{(3n-k+1)_{k}k!k!}\\ = & \sum_{k=0}^{n} \frac{(-1)^k(n-k+1)_{k}(-1)^k(n-k+1)_{k}(-1)^k(n-k+1)_{k}}{(-1)^k(3n-k+1)_{k}k!k!}\\ = & \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} k!k!}\\ = & \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} (1)_{k}}\frac{1}{k!} \quad \left(k! = (1)_{k} \right) \end{align*}$$ Recall the generalized hypergeometric function $${}_{p}F_{q} \left[ {a_{1},\cdots a_{p}\atop b_{1}\cdots b_{q}}; z\right] = \sum_{k=0}^{\infty} \frac{(a_{1})_{k}(a_{2})_{k}\cdots (a_{p})_{k}}{(b_{1})_{k}(b_{2})_{k}\cdots (b_{q})_{k}} \frac{z^k}{k!}$$ if one of the $a_{i}$ is a negative integer, for example $a_{i} = -n$ with $n\in \mathbb{N}$ we say that the series "terminates" and $${}_{p}F_{q} \left[ {a_{1},\cdots,-n,\cdots, a_{p}\atop b_{1}\cdots b_{q}}; z\right] = \sum_{k=0}^{n} \frac{(a_{1})_{k}(a_{2})_{k}\cdots(-n)_{k} \cdots (a_{p})_{k}}{(b_{1})_{k}(b_{2})_{k}\cdots (b_{q})_{k}} \frac{z^k}{k!}$$ To show this note that $$(-n)_{k} = (-n)(-n+1)\cdots(-n+k-1)$$ Hence $$(-n)_{k} = 0 \quad \textrm{ if } \quad k\geq n+1$$ From this is easy to see that $$S = \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \sum_{k=0}^{n} \frac{(-n)_{k}(-n)_{k}(-n)_{k}}{(-3n)_{k} (1)_{k}}\frac{1}{k!} = {}_{3}F_{2} \left[ {-n, -n, -n \atop -3n, 1}; 1 \right]$$ This series satisfies the Saalschütz theorem: $${{}_{3}F_{2}}\left[{-n,a,b\atop c,d};1\right]=\frac{{\left(c-a\right)_{n}}{% \left(c-b\right)_{n}}}{{\left(c\right)_{n}}{\left(c-a-b\right)_{n}}},$$ whenever $\displaystyle c+d = a+b+1-n, \quad n=0,1,...$

Therefore $$\begin{align*} S = {}_{3}F_{2} \left[ {-n, -n, -n \atop -3n, 1}; 1 \right] =& \frac{(-2n)_{n}(-2n)_{n}}{(-3n)_{n}(-n)_{n}}\\ =& \frac{(-1)^n(n+1)_{n}(-1)^n(n+1)_{n}}{(-1)^n(2n+1)_{n}(-1)^n(1)_{n}} \quad \textrm{ from (*)} \\ =& \frac{(n+1)_{n}(n+1)_{n}}{(2n+1)_{n}(1)_{n}}\\ =& \frac{\Gamma(2n+1)\Gamma(2n+1)\Gamma(2n+1)}{\Gamma(n+1)\Gamma(n+1)\Gamma(3n+1)n!}\\ =& \frac{(2n)!^3}{n!^3(3n)!} \end{align*}$$ Therefore we can conclude $$\boxed{ \sum_{k=0}^{n} \frac{{\binom{n}{k}}^3}{\binom{3n}{k}} = \frac{(2n)!^3}{n!^3(3n)!}}$$

Integral of the day XXVIII

Integral involving the Gaussian fucntion ${}_{2}F_{1}$

Today we show the proof of this integral posted by @integralsbot $$\int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \quad |a|\lt 1$$ For the proof we will use the generalized binomial theorem, some properties of the Pochhammer symbol (risng facorial) and the duplication formula for the Gamma function

Proof:

Suppose $|a|\lt 1$ $$\begin{align*} I= \int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } =& 2\int_{0}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } \quad \textrm{(The integrand is even)}\\ =& 2\int_{0}^{\infty} \frac{dx}{\left(2a+2\cosh^2 \left(\frac{x}{2}\right)\right)^s } \quad \left( 1+\cosh(x) = 2\cosh^2\left(\frac{x}{2}\right)\right)\\ =& \frac{1}{2^{s-1}}\int_{0}^{\infty} \frac{dx}{\left(a+\cosh^2 \left(\frac{x}{2}\right)\right)^s } \\ =& \frac{1}{2^{s-1}}\int_{0}^{\infty} \sum_{j=0}^{\infty} \binom{-s}{j} \cosh^{2(-s-j)}\left(\frac{x}{2}\right)a^j dx \quad \textrm{(Generalized bionomial theorem)} \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{\infty} \cosh^{2(-s-j)}\left(w\right)dw \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{\infty} \operatorname{sech}^{2(s+j)}\left(w\right)dw \\ =& \frac{1}{2^{s-2}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{1} t^{2(s+j)-1}(1-t^2)^{-\frac{1}{2}} dt \quad (t \mapsto \operatorname{sech}(x) )\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j \int_{0}^{1} s^{(s+j)-1}(1-s)^{-\frac{1}{2}} ds \quad (s \mapsto t^2 )\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j B\left(s+j,\frac{1}{2}\right)\\ \end{align*}$$ Using the fact that $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ and the recursion formula $$(x-n+1)_{n} = (-1)^n(-x)_{n}$$ where $$(x)_{n} = x(x+1)(x+2)\cdots (x+n-1)$$ we have $$\begin{align*} I =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \binom{-s}{j}a^j B\left(s+j,\frac{1}{2}\right)\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(-s-j+1)_{j}}{j!}a^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)}\\ =& \frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(s)_{j}}{j!}(-a)^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)}\\ \end{align*}$$ From the definition of the Pochhammer symbol: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ we have $$I =\frac{1}{2^{s-1}}\sum_{j=0}^{\infty} \frac{(s)_{j}}{j!}(-a)^j \frac{\Gamma(s+j)\sqrt{\pi}}{\Gamma\left(s+j+\frac{1}{2}\right)} = \frac{\Gamma(s)}{2^{s-1}\sqrt{\pi}\Gamma\left(s+\frac{1}{2}\right)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!}$$ From the duplication formula $$\Gamma(2x) = \frac{4^x}{2\sqrt{\pi}} \Gamma(x)\Gamma\left(\frac{1}{2}+x\right)$$ we have $$\begin{align*} I = \frac{\Gamma(s)}{2^{s-1}\sqrt{\pi}\Gamma\left(s+\frac{1}{2}\right)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!} & = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}\sum_{j=0}^{\infty} \frac{(s)_{j}(s)_{j}}{\left(s+\frac{1}{2}\right)_{j}}\frac{(-a)^j}{j!}\\ =& \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \end{align*}$$ Finally, we can conclude $$\boxed{ \int_{-\infty}^{\infty} \frac{dx}{(1+2a+\cosh x)^s } = \frac{2^{s}\Gamma^2(s)}{\Gamma(2s)}{}_{2}F_{1}\left[{s,s \atop s+\frac{1}{2}}; -a\right] \quad |a|\lt 1 }$$