Integral of the day XVII

Integral involving the complete Gamma function $\Gamma$

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

Proof

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

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