Integral of the day X

Integral involving the Glaisher–Kinkelin constant

We show the proof of this weird integral posted by @integralsbot $$\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx = \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right)$$ where $A$ is the Glaisher–Kinkelin constant

Proof

Recall the logarithmic representation of $\operatorname{arctanh}(x)$: $$\operatorname{arctanh}(x) = \ln\left(\sqrt{\frac{1+x}{1-x}}\right)$$ Hence $$\begin{align*} I=\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx =& \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(2\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right)dx\\ =&\ln(2)\int_{0}^{1} \frac{\ln(x)}{1-x^2}dx + \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\sqrt{\frac{1+x}{1-x}}\right)\right)dx\\ =&\ln(2) \underbrace{\int_{0}^{1} \frac{\ln(x)}{1-x^2}dx}_{J} + \underbrace{\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\operatorname{arctanh}(x)\right)dx}_{K}\\ \end{align*}$$ $$J = \int_{0}^{1} \frac{\ln(x)}{1-x^2}dx = \frac{1}{2}\underbrace{\int_{0}^{1}\frac{\ln(x)}{x+1}dx}_{L} - \frac{1}{2}\underbrace{\int_{0}^{1}\frac{\ln(x)}{1-x}dx}_{M}$$ $$L = \int_{0}^{1}\frac{\ln(x)}{x+1}dx \stackrel{IBP}{=} \ln(x)\ln(x+1)\Big|_{0}^{1} + \int_{-1}^{0} \frac{\ln(1-x)}{x}dx = \operatorname{Li}_{2}(-1) = -\frac{\pi^2}{12}$$ $$M = \int_{0}^{1}\frac{\ln(x)}{1-x}dx = -\int_{1}^{0} \frac{\ln(1-w)}{w} dw = -\operatorname{Li}_{2}(1) = -\frac{\pi^2}{6}$$ $$\therefore J = \frac{1}{2}L+\frac{1}{2}M = -\frac{\pi^2}{24} -\frac{\pi^2}{12} = -\frac{\pi^2}{8}$$ Now, for $K$: $$\begin{align*} K = \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\operatorname{arctanh}(x)\right)dx =& \int_{0}^{\infty} \ln\left(\operatorname{tanh}(w)\right) \ln(w)dw \quad ( w \mapsto \operatorname{arctanh}(x))\\ =&\int_{0}^{\infty} \ln\left(\frac{e^{w}-e^{-w}}{e^{w}+e^{-w}}\right) \ln(w)dw \\ =&\int_{0}^{\infty} \ln\left(\frac{1-e^{-2w}}{1+e^{-2w}}\right) \ln(w)dw \\ =&\underbrace{\int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw}_{C} - \underbrace{\int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw }_{D} \end{align*}$$ $$\begin{align*} C = \int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw =& -\int_{0}^{\infty} \sum_{n=1}^{\infty}\frac{ e^{-2nw}}{n} \ln(w)dw\\ =& -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} \ln(w)dw\\ =& -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} \left(\frac{d}{dt}\Big|_{t=0+} w^t \right)dw\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} w^tdw\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{n^{t+2}2^{t+1}}\int_{0}^{\infty} e^{-s} s^tds\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{2^{t+1}n^{t+2}}\Gamma(t+1)\\ =& -\sum_{n=1}^{\infty} \left[\frac{1}{2n^2}\frac{d}{dt}\Big|_{t=0+}\frac{1}{(2n)^{t}} \Gamma(t+1)\right]\\ =& \sum_{n=1}^{\infty}\frac{\ln(2n)+\gamma}{2n^2}\\ =& \frac{\ln(2)}{2} \sum_{n=1}^{\infty}\frac{1}{n^2} + \frac{1}{2} \sum_{n=1}^{\infty}\frac{\ln(n)}{n^2} + \frac{\gamma}{2} \sum_{n=1}^{\infty}\frac{1}{n^2}\\ =& \frac{\ln(2)}{2} \zeta(2) - \frac{1}{2} \zeta'(2) + \frac{\gamma}{2} \zeta(2)\\ =& \frac{\ln(2)\pi^2}{12} - \frac{1}{2} \zeta'(2) + \frac{\gamma\pi^2}{12} \end{align*}$$ Recall $$-\zeta'(2) = 2\pi^2 \ln(A) - \gamma\frac{\pi^2}{6} - \ln(2)\frac{\pi^2}{6} -\ln(\pi)\frac{\pi^2}{6} \tag{*}$$ whre $A$ is the Glaisher–Kinkelin constant. Hence $$C = \int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw = \pi^2\ln(A) - \frac{\pi^2\ln(\pi)}{12}$$ In a similar way: $$\begin{align*} D = \int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw =& \int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}e^{-2nw}}{n} \ln(w)dw\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} \ln(w)dw\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} \left(\frac{d}{dt}\Big|_{t=0+} w^t\right)dw\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} w^tdw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2^{t+1}n^{t+2}}\int_{0}^{\infty} e^{-s} s^{t}ds\\ =&\sum_{n=0}^{\infty} \left[ \frac{(-1)^{n+1}}{2n^2} \frac{d}{dt}\Big|_{t=0+} \frac{\Gamma(t+1)}{2^{t}n^{t}}\right]\\ =&\sum_{n=0}^{\infty} \frac{(-1)^{n}(\ln(2n)+\gamma)}{2n^2}\\ =&\frac{\ln(2)}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n^2} + \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}\ln(n)}{n^2} + \frac{\gamma}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n^2}\\ =&\frac{-\ln(2)}{2}\eta(2) +\frac{1}{2}\eta'(2) - \frac{\gamma}{2}\eta(2)\\ =&-\frac{\ln(2)\pi^2}{24} + \frac{1}{2}\eta'(2) - \frac{\gamma\pi^2}{24}\\ \end{align*}$$ Using the relation $$\eta(v) = (1-2^{1-v})\zeta(v)$$ and using (*) we have $$\eta'(2) = -\pi^2\ln(A) +\frac{\gamma\pi^2}{12} + \frac{\pi^2\ln(\pi)}{12} + \frac{\pi^2\ln(2)}{6}$$ Hence $$D = \int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw= -\frac{\pi^2\ln(A)}{2} + \frac{\ln(\pi)\pi^2}{24} + \frac{\ln(2)\pi^2}{24}$$ Therefore $$K = C-D =\pi^2\ln(A) - \frac{\pi^2\ln(\pi)}{12} + \frac{\pi^2\ln(A)}{2} - \frac{\ln(\pi)\pi^2}{24} - \frac{\ln(2)\pi^2}{24} = \frac{3\pi^2}{2} \ln(A)-\frac{\pi^2\ln(\pi)}{8}-\frac{\ln(2)\pi^2}{24}$$ Therefore $$\begin{align*} I = \ln(2)J+K =-\frac{\pi^2\ln(2)}{8} + \frac{3\pi^2}{2} \ln(A)-\frac{\pi^2\ln(\pi)}{8}-\frac{\ln(2)\pi^2}{24} =& \frac{\pi^4}{24}\left[-4\ln(2)+36\ln(A)-3\ln(\pi) \right]\\ =& \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right) \end{align*}$$ Hence, we can conclude $$\boxed{\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx = \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right)}$$

Integral of the day IX

Another singular integral of @integralsbot

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

Proof

Firt, if we make the substitution $$w = \sqrt{\frac{1-x}{1+x}} \Longrightarrow x = \frac{1-w^2}{1+w^2} \Longrightarrow dx = -\frac{4w}{(1+w^2)^2}dw$$ Hence $$\begin{align*} I =& \int_{0}^{1} \frac{x^2}{1+x^2} \sqrt{\frac{1-x}{1+x}}dx = -\int_{0}^{1} \frac{2w^2(w-1)^2(1+w)^2}{(1+w^2)^2(1+w^4)}dw\\ =& 4\underbrace{\int_{0}^{1} \frac{1}{1+w^2}dw}_{J} -4\underbrace{\int_{0}^{1} \frac{1}{(1+w^2)^2} dw}_{K} - 2 \underbrace{\int_{0}^{1} \frac{w^2}{1+w^4}dw}_{Q} \end{align*}$$ $$J = \int_{0}^{1} \frac{1}{1+w^2}dw = \arctan(w)\Big|_{0}^{1} = \frac{\pi}{4}$$ $$\begin{align*} K =& \int_{0}^{1} \frac{1}{(1+w^2)^2} dw \\ =& \int_{0}^{\frac{\pi}{4}} \frac{\sec^2(r)}{(1+\tan^2(r))^2} dr \quad (w \mapsto \tan(r))\\ =& \int_{0}^{\frac{\pi}{4}} \cos^2(r) dr \quad ( 1+\tan^2(r) = \sec^2(r))\\ =& \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\cos(2r) dr + \frac{1}{2}\int_{0}^{\frac{\pi}{4}}dr \quad \left(\cos^2(r)=\frac{\cos(2r)+1}{2}\right)\\ =& \frac{1}{4}\int_{0}^{\frac{\pi}{2}}\cos(t) dt + \frac{\pi}{8} \quad (t \mapsto 2r)\\ =& \frac{1}{4} + \frac{\pi}{8} \end{align*}$$ Now for $Q$, note that $$\underbrace{\int_{0}^{\infty} \frac{w^2}{1+w^4} dw}_{L} - \underbrace{\int_{1}^{\infty} \frac{w^2}{1+w^4} dw}_{M} = \underbrace{\int_{0}^{1} \frac{w^2}{1+w^4} dw}_{Q}$$ $$L = \int_{0}^{\infty} \frac{w^2}{1+w^4} dw = \frac{1}{2}\int_{0}^{\infty} \frac{\sqrt{s}}{1+s^2} ds\quad (s^2 \mapsto w^4)$$ Recall the integral representation of the $\sec(x)$ function: $$\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{t^{2x/\pi}}{t^2+1}dt \quad |x|< \frac{\pi}{2}$$ If we put $\displaystyle x = \frac{\pi}{4}$ $$\sec\left(\frac{\pi}{4}\right) = \frac{2}{\pi} \int_{0}^{\infty} \frac{\sqrt{t}}{t^2+1}dt \Longrightarrow \int_{0}^{\infty} \frac{\sqrt{t}}{t^2+1}dt = \frac{\pi}{2} \sec\left(\frac{\pi}{4}\right) = \frac{\pi}{\sqrt{2}}$$ Then $$L =\frac{1}{2}\int_{0}^{\infty} \frac{\sqrt{s}}{1+s^2} ds = \frac{\pi}{2\sqrt{2}}$$ For $M$: $$M=\int_{1}^{\infty} \frac{w^2}{1+w^4} dw = \int_{0}^{1} \frac{1}{1+r^4} dr \quad \left(r \mapsto \frac{1}{w}\right)$$ Hence $$M= \int_{0}^{1} \frac{1}{1+r^4} dr = \frac{1}{2}\int_{0}^{1} \frac{(1+r^2)+(1-r^2)}{1+r^4} dr = \frac{1}{2} \underbrace{\int_{0}^{1} \frac{1+r^2}{1+r^4}dr}_{F} + \frac{1}{2}\underbrace{\int_{0}^{1} \frac{1-r^2}{1+r^4}dr}_{G}$$ $$\begin{align*} F= \int_{0}^{1} \frac{1+x^2}{1+x^4}dx =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}dx\\ =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}dx \\ =& \int_{0}^{1} \frac{\frac{1}{x^2}+1}{\left(x-\frac{1}{x}\right)^2+2}dx\\ =& \int_{0}^{\infty} \frac{1}{w^2+2}dx \quad \left(w \mapsto x-\frac{1}{x}\right) \\ =& \frac{\sqrt{2}}{2}\int_{0}^{\infty} \frac{1}{r^2+1}dx \quad \left(r^2 \mapsto \frac{w^2}{2}\right)\\ =& \frac{\pi}{2\sqrt{2}} \end{align*}$$ $$\begin{align*} G = \int_{0}^{1} \frac{1-x^2}{1+x^4}dx =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx\\ =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\frac{1}{x^2}+x^2}dx \\ =& \int_{0}^{1} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2}dx\\ =& -\int_{2}^{\infty} \frac{1}{w^2-2} dx \quad \left(w \mapsto x+\frac{1}{x}\right) \\ =& \frac{1}{2\sqrt{2}}\int_{2}^{\infty} \left[ \frac{1}{w+\sqrt{2}}-\frac{1}{w-\sqrt{2}} \right] dx\\ =& \frac{1}{2\sqrt{2}} \left[\ln(w+\sqrt{2}) - \ln(w-\sqrt{2})\right]_{2}^{\infty}\\ =& \frac{1}{\sqrt{2}} \left[\ln\left(\sqrt{\frac{\frac{w}{\sqrt{2}}+1}{\frac{w}{\sqrt{2}}-1}}\right) \right]_{2}^{\infty}\\ =& \frac{1}{\sqrt{2}} \left[\operatorname{arccoth}\left(\frac{w}{\sqrt{2}}\right)\right]_{2}^{\infty}\\ =& \frac{\operatorname{arccoth}(\sqrt{2})}{\sqrt{2}} \end{align*}$$ Hence $$M = \frac{1}{2}F+\frac{1}{2}G = \frac{\pi}{4\sqrt{2}}+ \frac{\operatorname{arccoth}(\sqrt{2})}{2\sqrt{2}}$$ $$Q = L - M = \frac{\pi}{4\sqrt{2}} -\frac{\operatorname{arccoth}(\sqrt{2})}{2\sqrt{2}}$$ Therefore $$I = 4J - 4K - 2Q = \pi-1 - \frac{\pi}{2} - \frac{\pi}{2\sqrt{2}} + \frac{\operatorname{arccoth}(\sqrt{2})}{\sqrt{2}}$$ Finally, using the following identity: $$\operatorname{arccoth}(x) = \operatorname{sign}(x)\operatorname{arcsinh}\left(\frac{1}{\sqrt{x^2-1}}\right)$$ We have $$\operatorname{arccoth}(\sqrt{2})= \operatorname{arcsinh}(1) = \ln(1+\sqrt{2})$$ Therefore $$\boxed{\int_{0}^{1} \frac{x^2}{1+x^2} \sqrt{\frac{1-x}{1+x}}dx = \frac{\ln(1+\sqrt{2})}{\sqrt{2}} + \left(1-\frac{1}{\sqrt{2}}\right)\frac{\pi}{2}-1}$$

Generalized hypergeometric functions VI

Singular integral involving the Gauss hypergeometric function

Today we show the proof of this singular integral posted by @integralsbot $$\int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!}$$ The proof relies on some properties of the Gauss hypergeometric function $F(a,b;c;x)$

Proof

$$\begin{align*} I = \int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = &\int_{a}^{\infty} a^{m+n}\left(\frac{x}{a}-1\right)^m\left(\frac{x}{a}-\sqrt{\frac{x^2 }{a^2}-1}\right)^n dx\\ =& \int_{1}^{\infty} a^{m+n+1}\left(w-1\right)^m\left(w-\sqrt{w^2-1}\right)^n dw \quad \left(w \mapsto \frac{x}{a} \right)\\ =& \int_{1}^{0} a^{m+n+1}\left(\frac{t^2-2t+1}{2t}\right)^mt^n \frac{(t^2-1)}{2t^2} dt \quad (t \mapsto w-\sqrt{w^2-1})\\ =& \frac{a^{m+n+1}}{2^{m+1}}\int_{0}^{1} (t-1)^{2m} t^{n-m-2}(1-t^2) dt\\ =& \frac{a^{m+n+1}}{2^{m+1}}\int_{0}^{1} t^{n-m-2}(1-t^2)\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} t^{j} dt\\ =& \frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\int_{0}^{1} t^{j+n-m-2}(1-t^2) dt\\ =& \frac{a^{m+n+1}}{2^{m+2}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\int_{0}^{1} s^{\frac{j+n-m-3}{2}}(1-s) ds \quad (s \mapsto t^2)\\ =& \frac{a^{m+n+1}}{2^{m+2}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}B\left(\frac{j+n-m-1}{2},2\right)\\ =& \frac{a^{m+n+1}}{2^{m}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n-1)(j-m+n+1)}\\ =& \frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n-1)}-\frac{a^{m+n+1}}{2^{m+1}}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{1}{(j-m+n+1)}\\ \end{align*}$$ Now using the recursion formula for Pochhammer symbols (rising factorial): $$(n+x)(x)_{n} = x(x+1)_{n}$$ Then: $$\frac{1}{j-m+n-1} = \frac{(n-m-1)_{j}}{(n-m-1)(n-m)_{j}}$$ $$\frac{1}{j-m+n+1} = \frac{(n-m+1)_{j}}{(n-m+1)(n-m+2)_{j}}$$ Hence $$I =\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j}\frac{(n-m-1)_{j}}{(n-m)_{j}}-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}\sum_{j=0}^{2m}\binom{2m}{j}(-1)^{j} \frac{(n-m+1)_{j}}{(n-m+2)_{j}}$$ Now, recall that if either of the parameters in the Gauss hypergeometric function is a nonpositive integer, then the function reduces to a polynomial function of degree n: $$F(a,-n;c;x) = \sum_{j=0}^{n}\binom{n}{j} \frac{(a)_{j}}{(c)_{j}}(-x)^j$$ Then $$I =\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}F(n-m-1,-2m;n-m;1)-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}F(n-m+1,-2m;n-m+2;1)$$ Using the formula: $$F(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\quad c>a+b$$ $$\begin{align*} I =&\frac{a^{m+n+1}}{2^{m+1}(n-m-1)}\frac{\Gamma(n-m)\Gamma(2m+1)}{\Gamma(n+m)}-\frac{a^{m+n+1}}{2^{m+1}(n-m+1)}\frac{\Gamma(n-m+2)\Gamma(2m+1)}{\Gamma(n+m+2)}\\ =& \frac{a^{m+n+1}}{2^{m+1}}\left[\frac{\Gamma(n-m-1)\Gamma(2m+1)}{\Gamma(n+m)}-\frac{\Gamma(n-m+1)\Gamma(2m+1)}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\left[\frac{\Gamma(n-m-1)\Gamma(2m+1)\Gamma(n+m+2)-\Gamma(n-m+1)\Gamma(2m+1)\Gamma(n+m)}{\Gamma(n+m)\Gamma(n+m+2}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\left[\frac{\Gamma(n-m-1)\Gamma(n+m)(n+m+1)(n+m)-\Gamma(n-m+1)\Gamma(n+m)}{\Gamma(n+m)\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\left[\frac{\Gamma(n-m-1)(n+m+1)(n+m)-\Gamma(n-m-1)(n-m)(n-m-1)}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m+1}}\Gamma(2m+1)\Gamma(n-m-1)\left[\frac{2(2m+1)n}{\Gamma(n+m+2)}\right]\\ =& \frac{a^{m+n+1}}{2^{m}}\frac{n\Gamma(2m+2)\Gamma(n-m-1)}{\Gamma(n+m+2)}\\ =& \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!}\\ \end{align*}$$ Hence, we can conclude $$\boxed{\int_{a}^{\infty} (x-a)^m\left(x-\sqrt{x^2-a^2}\right)^n dx = \frac{a^{m+n+1}}{2^{m}}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!} }$$

Residue theorem VII

Nice integral involving ${}_1F_{1}$

Today we show the proof of this mysterious integral posted by @infseriesbot. $$\int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =\frac{\pi \Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)} {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]$$ This integral resembles another one we previously solved using the Cauchy integral formula and it turned out that this also can be solved using contour integration.

Proof

First, note that the function $$f(x) = e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}$$ Is an even function...

Hence we can write: $$I = \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx = \frac{1}{2} \int_{-\pi }^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx$$ Expand the integrand using the Euler's formula: $$\begin{align*} I = \frac{1}{2}\int_{-\pi }^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =& \frac{1}{2}\int_{-\pi }^{\pi}e^{-z\cos x}\left(\frac{e^{iax-iz\sin x}+e^{-iax+iz\sin x}}{2}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{ia-z(\cos x -i\sin x)} + e^{-iax +z(-\cos x +iz\sin x)}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-z(\cos x -i\sin x)} + e^{-iax}e^{+z(-\cos x +iz\sin x)}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx\\ =& \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\underbrace{\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}}_{A}dx \end{align*}$$ Expand $A$ with the binomial theorem: $$A = \left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b} = \sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}$$ Hence $$I = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\left(e^{\frac{ix}{2}}+e^{-\frac{ix}{2}}\right)^{2b}dx = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}dx$$ Now, do the following substitution: $$e^{ix} = w$$ $$dx = \frac{dw}{wi}$$ Our integral is transformed in a contour integral round the unit circle: $$\begin{align*} I = \frac{1}{4}\int_{-\pi }^{\pi}\left( e^{iax} e^{-ze^{ix}} + e^{-iax}e^{+ze^{-ix}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} e^{ix(b-j)}dx =& \frac{1}{4}\oint_{|w|=1} \frac{\left( w^a e^{-zw} + w^{-a}e^{+zw^{-1}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-j)}}{wi}dw\\ =&\frac{1}{4i}\oint_{|w|=1} \left( w^{a-1} e^{-zw} + w^{-a-1}e^{zw^{-1}}\right)\sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-j)}dw\\ =& \frac{1}{4i}\underbrace{\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(a+b-j-1)}e^{-zw}dw}_{J} + \frac{1}{4i}\underbrace{\oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-a-j-1)}e^{zw^{-1}}dw}_{K} \tag{1} \end{align*}$$ Expand $e^{-zw}, e^{-zw^{-1}}$ in $J$ and $K$ and apply the residue theorem: $$J = \oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(a+b-j-1)}e^{-zw}dw = \oint_{|w|=1} \underbrace{ \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{2b}{j} \frac{w^{(a+b-j+k-1)}(-z)^k}{k!}}_{g(w)}dw = 2\pi i \operatorname{Res}(g,0)$$ $$K = \oint_{|w|=1} \sum_{j=0}^{\infty} \binom{2b}{j} w^{(b-a-j-1)}e^{zw^{-1}}dw = \oint_{|w|=1} \underbrace{\sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \binom{2b}{j} \frac{w^{(b-a-j-k-1)}(-z)^n}{k!}}_{h(w)}dw = 2\pi i\operatorname{Res}(h,0)$$ The functions $g(w),h(w)$ have a singularity at $w=0$.

To find the residue we need to know the coefficient of $w^{-1}$

For $J$: $$w^{-1} = w^{a+b-j+k-1 } \Longrightarrow j = a+b+k$$ For $K$: $$w^{-1} = w^{b-a-j-k-1} \Longrightarrow j = b-a-k$$ Therefore $$J = 2\pi i\operatorname{Res}(g,0) = 2\pi i\sum_{k=0}^{\infty} \binom{2b}{a+b+k} \frac{(-z)^k}{k!}$$ $$K = 2\pi i\operatorname{Res}(h,0) = 2\pi i \sum_{k=0}^{\infty} \binom{2b}{b-a-k} \frac{(-z)^k}{k!}$$ Hence, from (1): $$I = \frac{1}{4i}J + \frac{1}{4i}K = \frac{\pi}{2}\sum_{k=0}^{\infty} \binom{2b}{a+b+k} \frac{(-z)^k}{k!} + \frac{\pi}{2} \sum_{k=0}^{\infty} \binom{2b}{b-a-k} \frac{(-z)^k}{k!}$$ Using the property $\displaystyle \binom{n}{n-m} = \binom{n}{m}$: $$\binom{2b}{b-a-k} = \binom{2b}{2b-a-b-k} = \binom{2b}{a+b+k}$$ Therefore $$I = \pi\sum_{k=0}^{\infty}\binom{2b}{a+b+k} \frac{(-z)^k}{k!} \tag{2}$$ From the formula that relates the binomial coefficient to the Pochhammer symbol $\displaystyle \binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$ : $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{a+b+k}}{(a+b+k)!} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+k+1)} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+1)(a+b+1)_{k}}$$ Now, from the rule $\displaystyle (x)_{n+m} = (x)_{n}(x+n)_{m}$ we have $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{a+b+k}}{\Gamma(a+b+k+1)(a+b+1)_{k}} = \frac{(b-a-k+1)_{k}(b-a+1)_{a+b}}{\Gamma(a+b+k+1)(a+b+1)_{k}}$$ From definition $$(b-a+1)_{a+b} = \frac{\Gamma(2b+1)}{\Gamma(b-a+1)}$$ Therefore $$\binom{2b}{a+b+k} = \frac{(b-a-k+1)_{k}\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)(a+b+1)_{k}}$$ Hence, from (2): $$\begin{align*} I = \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx = & \pi \sum_{j=0}^{\infty}\binom{2b}{a+b+k} \frac{(-z)^k}{k!}\\ =& \pi\sum_{j=0}^{\infty} \frac{(b-a-k+1)_{k}\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)(a+b+1)_{k}} \frac{(-z)^{k}}{k!}\\ = & \frac{\pi\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)}\sum_{k=0}^{\infty} \frac{(b-a-k+1)_{k}}{(a+b+1)_{k}} \frac{(-1)^{k}z^k}{k!}\\ =& \frac{\pi\Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)}\sum_{k=0}^{\infty} \frac{(a-b)_{k}}{(a+b+1)_{k}} \frac{z^k}{k!} \quad ((b-a-k+1)_{k}(-1)^k = (a-b)_{k}) \end{align*}$$ The last series satisfy the definition of ${}_{1}F_{1}$ also called Confluent hypergeometric function: $$\sum_{k=0}^{\infty} \frac{(a-b)_{k}}{(a+b+1)_{k}} \frac{z^k}{k!} = {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]$$ Hence, we can conclude $$\boxed{ \int_{0}^{\pi}e^{-z\cos x}\cos(ax-z\sin x)\left(2\cos \frac{x}{2}\right)^{2b}dx =\frac{\pi \Gamma(2b+1)}{\Gamma(a+b+1)\Gamma(b-a+1)} {}_{1}F_{1}\left[{a-b\atop 1+a+b};z\right]}$$

Generalized hypergeometric functions V

Nice infnite series involving binomial coefficients

Today we show the proof of this nice series posted by @infseriesbot $$\sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = \frac{(a+b)!(b+c)!(c+a)!}{a!b!c!(a+b+c)!}$$ The proof will rely on some properties of the rising factorial and the generalized hypergeometric function.

Proof

First, we need to recall some properties of the binomial coefficient and the rising factorial: The following formula $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \tag{1}$$ relates the binomial coefficient to the Pochhammer polynomial (rising factorial) where $$(x)_{n}= x(x+1)(x+2)\cdots(x+n-1)$$ is the definition for the rising factorial.

It is easy to show this. Just note that $$\binom{v}{m} = \frac{v!}{m!(v-m)!} = \frac{(v-m+1)(v-m+2)\cdots(v-1)\cdot v}{m!}$$ The numerator in the right hand side is the pochhammer polynomial $(v-m+1)_{n}$ also called the falling factorial.

Pochhammer polynomials obey the reflection formula: $$(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{2}$$ Note that, by definition $$(x-n+1)_{n} = \underbrace{(x-n-1)(x-n)(x-n+1)\cdots (x-1)\cdot x }_{\textrm{n}-terms}$$ Hence $$(-1)^n(x-n+1)_{n} =(-x)(-x+1)\cdots (-x+n-1) = (-x)_{n}$$ Hence, back to our series: $$\begin{align*} S = \sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = &\sum_{k=0}^{\infty} \frac{\frac{(a-k+1)_{k}}{k!}\frac{(b-k+1)_{k}}{k!}\frac{(c-k+1)_{k}}{k!} }{\frac{(a+b+c-k+1)_{k}}{k!}} \quad (\textrm{from } (1))\\ =&\sum_{k=0}^{\infty} \frac{(a-k+1)_{k}(b-k+1)_{k}(c-k+1)_{k} }{k!k!(a+b+c-k+1)_{k}}\\ =& \sum_{k=0}^{\infty} \frac{(-1)^k(-a)_{k}(-1)^k(-b)_{k}(-1)^k(-c)_{k} }{k!k!(-1)^k(-a-b-c)_{k}} \quad (\textrm{from } (2))\\ =& \sum_{k=0}^{\infty} \frac{(-a)_{k}(-b)_{k}(-c)_{k} }{(1)_{k}(-a-b-c)_{k}}\frac{1}{k!} \quad ((1)_{k} = k!)\\ =& {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) \end{align*}$$ Now we will make use of the Saalschütz theorem (proof in the Appendix): $${{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},$$ where $g+h = e+f+1-n \quad n=0,1,2,..$.

Note that in our case $-n = -a$ and $g=1$ so the series satisfy the hypothesis. This kind of series are called Saalschutzian.

Therefore $$S = {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) = \frac{(1+a)_{c}(1+b)_{c}}{(1)_{c}(1+a+b)_{c}}$$ Now recall that Pochhammer polynomials can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Hence $$S = {{}_{3}F_{2}}\left({-a,-b,-c\atop 1,-a-b-c};1\right) = \frac{\frac{\Gamma(c+a+1)}{\Gamma(a+1)}\frac{\Gamma(c+b+1)}{\Gamma(b+1)}}{\Gamma(c+1)\frac{\Gamma(c+a+b+1)}{\Gamma(a+b+1)}} = \frac{\Gamma(c+a+1)\Gamma(c+b+1)\Gamma(a+b+1)}{\Gamma(c+1)\Gamma(a+1)\Gamma(b+1)\Gamma(a+b+c+1)}$$ Using $\Gamma(n+1) = n!$, we can conclude $$\boxed{ \sum_{k=0}^{\infty} \frac{\binom{a}{k}\binom{b}{k}\binom{c}{k}}{\binom{a+b+c}{k}} = \frac{(a+b)!(b+c)!(c+a)!}{a!b!c!(a+b+c)!} \quad a,b,c\in \mathbb{N}_{0}}$$

Appendix.

Now, we will provide a proof for the Saalschütz theorem: $${{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},$$ where $g+h = e+f+1-n \quad n=0,1,2,..$

Proof

Consider the Euler's identity for ${}_{2}F_{1}$: $${}_{2}F_{1}(g-e,g-f;g;z) = (1-z)^{e+f-g}{}_{2}F_{1}(e,f;g;z)$$ Note that the left hand side is: $${}_{2}F_{1}(g-e,g-f;g;z) = \sum_{n=0}^{\infty} \frac{(g-e)_{n}(g-f)_{n}}{(g)_{n}}\frac{z^n}{n!} \tag{1}$$ while for the right hand side, if we expand $\displaystyle (1-z)^{e+f-g}$: $$(1-z)^{e+f-g} = \sum_{k=0}^{\infty}\binom{e+f-g}{k}(-1)^kz^k = \sum_{k=0}^{\infty}\frac{(g-e-f)_{k}}{k!}z^k$$ Hence $$(1-z)^{e+f-g}{}_{2}F_{1}(e,f;g;z) = \sum_{j=0}^{\infty} \frac{(e)_{j}(f)_{j}}{(g)_{j}}\frac{z^j}{j!}\sum_{k=0}^{\infty}\frac{(g-e-f)_{k}}{k!}z^k$$ By the identity theorem for power series the coefficients of the left hand side should be equal to the coefficients of the right hand side i.e. $n=k+j$ So, if we put $k = n-j$ $$\begin{align*} \sum_{j=0}^{n} \frac{(e)_{j}(f)_{j}(g-e-f)_{n-j}}{(g)_{j}j!(n-j)!} =& \frac{(g-e-f)_{n}}{n!} \sum_{j=0}^{n} \frac{(e)_{j}(f)_{j}(-n)_{j}}{j!(g)_{j}(1-g+e+f-n)_{j}}\\ =& \frac{(g-e-f)_{n}}{n!} {}_{3}F_{2}(e,f,-n;g,1+e+f-g-n;1) \tag{2} \end{align*}$$ From (1) and (2) we can conclude $${{}_{3}F_{2}}\left({-n,e,f\atop g,h};1\right)=\frac{{\left(g-e\right)_{n}}{% \left(g-f\right)_{n}}}{{\left(g\right)_{n}}{\left(g-e-f\right)_{n}}},$$ where $g+h = e+f+1-n \quad n=0,1,2,..$

Integral of the day VII

The strong connection between the $\operatorname{Li}_{2}$ function and $\phi$ vol. II

Today we show the proof of this pair of integrals posted by @infseriesbot $$\int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx = \frac{\pi^2}{12} + \frac{3}{4} \ln^2(\phi)$$ $$\int_{0}^{\infty} \frac{\arctan\left(\phi x\right)}{1+x^2}dx= \frac{\pi^2}{6} - \frac{3}{4} \ln^2(\phi)$$ In general, we found that $$\int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a) \quad a\neq 1,-1$$

Proof:

Consider the integral $$\Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx$$ Differentiating under the integral sign: $$\Phi'(a) = \int_{0}^{\infty} \frac{x}{(1+x^2)(1+a^2x^2)}dx$$ Applying partial fraction decomposition $$\Phi'(a) = \int_{0}^{\infty} \frac{x}{(1+x^2)(1+a^2x^2)}dx = \frac{1}{2(a^2-1)}\int_{0}^{\infty}\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx$$ Note that $$\int\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx = \int \frac{2a^2x}{1+a^2x^2} dx - \int \frac{2x}{1+x^2} dx = \ln(1+a^2x^2)-\ln(1+x^2) + C$$ Therefore $$\Phi'(a) = \frac{1}{2(a^2-1)}\int_{0}^{\infty}\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx = \frac{1}{2(a^2-1)} \left[\ln\left(\frac{1+a^2x^2}{1+x^2}\right)\right]_{0}^{\infty}$$ Note that $$\lim_{x\to \infty} \ln\left(\frac{1+a^2x^2}{1+x^2}\right) = \ln\left(\lim_{x \to \infty} \frac{\frac{1}{x^2}+a^2}{\frac{1}{x^2}+1}\right) = \ln(a^2)$$ Hence $$\Phi'(a) = \frac{1}{2(a^2-1)} \left[\ln\left(\frac{1+a^2x^2}{1+x^2}\right)\right]_{0}^{\infty} = \frac{\ln(a^2)}{2(a^2-1)}$$ Given that for $a\in \mathbb{C}$ in general $$\ln(a^n) \neq n\ln(a)$$ we have to take care when evaluating the right hand side.

Integrating

$$\Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \int\frac{\ln(a^2)}{2(a^2-1)}da + C$$ $$\frac{1}{2}\int \frac{\ln(a^2)}{a^2-1}da = \frac{1}{4}\underbrace{\int\frac{\ln(a^2)}{a-1}da }_{J} -\frac{1}{4} \underbrace{\int \frac{\ln(a^2)}{a+1}da}_{K}$$ For $J$: $$\begin{align*} J = \int\frac{\ln(a^2)}{a-1}da \stackrel{IBP}{=}& -\ln(1-a)\ln(a^2)+2\int \frac{\ln(1-a)}{a} +C \\ =&-\ln(1-a)\ln(a^2)-2\operatorname{Li}_{2}(a)+C \end{align*}$$ $$\therefore J = -\ln(1-a)\ln(a^2)-2\operatorname{Li}_{2}(a)+C$$ Now, for $K$: $$K = \int \frac{\ln(a^2)}{a+1}da \stackrel{IBP}{=} \ln(1+a)\ln(a^2)-2\int \frac{\ln(1+a)}{a}da$$ $$\int\frac{\ln(a+1)}{a}da \stackrel{a=-w}{=} \int \frac{\ln(1-w)}{w}dw = -\operatorname{Li}_{2}(w)+ C = -\operatorname{Li}(-a) + C$$ $$\therefore K = \int \frac{\ln(a^2)}{a+1}da= \ln(a+1)\ln(a^2) + 2\operatorname{Li}_{2}(-a)+ C$$ Hence $$\Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx =\frac{1}{2}\int \frac{\ln(a)}{a^2-1}da = \frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a)+C$$ To find $C$ note that if $a \to 0 \Longrightarrow C \to 0$: Hence $$\Phi(a) = \int_{0}^{\infty} \frac{\arctan(\phi x)}{1+x^2}dx =\frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a)$$ If we put $a = \phi$

and using the following identities and properties of the dilogarithm function and the golden ratio: $$\ln(\phi)\ln(\phi+1)= 2\ln^2(\phi)$$ $$\ln(1-\phi)\ln(\phi) = -\frac{\ln^2(\phi)}{2}+\frac{i\pi \ln(\phi)}{2}$$ $$\operatorname{Li}_{2}(-\phi) = -\frac{\pi^2}{10}-\ln^2(\phi)$$ $$\operatorname{Li}_{2}(\phi^{-1}) = \frac{\pi^2}{10} -\ln^2(\phi)$$ $$\operatorname{Li}_{2}(\phi) +\operatorname{Li}_{2}(\phi^{-1}) = -\frac{\pi^2}{6}-\frac{\ln^2(-\phi)}{2}$$ $$\operatorname{Li}_{2}(\phi) = \frac{7\pi^2}{30} + \frac{\ln^2(\phi)}{2} -i\pi \ln(\phi)$$ Hence $$\int_{0}^{\infty} \frac{\arctan(\phi x)}{1+x^2}dx = \underbrace{\frac{1}{2}\ln(1-\phi)\ln(\phi)}_{-\frac{\ln^2(\phi)}{2}+\frac{i\pi\ln(\phi)}{2}}+\underbrace{\frac{1}{2}\operatorname{Li}_{2}(\phi)}_{ \frac{7\pi^2}{60} + \frac{\ln^2(\phi)}{4} -\frac{i\pi \ln(\phi)}{2}} \underbrace{- \frac{1}{2}\ln(\phi+1)\ln(\phi)}_{-\ln^2(\phi)} \underbrace{- \frac{1}{2}\operatorname{Li}_{2}(-\phi)}_{\frac{\pi^2}{20}+\frac{\ln^2(\phi)}{2}}$$ $$\boxed{\therefore \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \frac{\pi^2}{6} - \frac{3}{4}\ln^2(\phi)}$$ If now we put $a = \phi^{-1}$

and using th following identities in addition to the others: $$\ln(1-\phi^{-1})\ln(\phi^{-1}) = \ln^2(\phi)$$ $$\ln(\phi^{-1}+1)\ln(\phi^{-1}) = -2\ln^2(\phi)$$ $$\operatorname{Li}_{2}(-\phi^{-1}) = -\frac{\pi^2}{15} + \frac{1}{2}\ln^2(\phi)$$ we have $$\Phi( \phi^{-1}) = \int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx =\underbrace{\frac{1}{2}\ln(1-\phi^{-1})\ln(\phi^{-1})}_{\frac{1}{2}\ln^2(\phi)}+\underbrace{\frac{1}{2}\operatorname{Li}_{2}(\phi^{-1})}_{\frac{\pi^2}{20}-\ln^2(\phi)}\underbrace{ - \frac{1}{2}\ln(\phi^{-1}+1)\ln(\phi^{-1})}_{\ln^2(\phi)} \underbrace{- \frac{1}{2}\operatorname{Li}_{2}(-\phi^{-1}) }_{\frac{\pi^2}{30}-\frac{1}{4}\ln^2(\phi)}$$ $$\boxed{\therefore \int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx = \frac{\pi^2}{12} +\frac{3}{4}\ln^2(\phi)}$$