Generalized hypergeometric function

Solution to integral involving ${}_{3}F_{2}$

We show the proof of the following result posted by @SrinivasR1729 \[\int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx = 2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right)\]

Proof

Recall that \[{}_{p}F_{q}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};x) = \sum_{j=0}^{\infty} \frac{(a_{1})(a_{2})\cdots(a_{p})}{(b_{1})(b_{2})\cdots(b_{q})} \frac{x^{j}}{j!}\] where the symbol $\displaystyle (x)_{j}$ refers to the Pochhammer polynomial (also called rising factorial): \[(x)_{j} = x(x+1)(x+2)\cdots(x+j-1)\] Hence, \begin{align*} \int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx =& \int_{0}^{1} \frac{\ln(1+w)}{\sqrt{1-w}\sqrt{w}} dw \quad (w \mapsto e^{-w})\\ =& \int_{0}^{1} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} w^k}{k} \frac{1}{\sqrt{1-w}\sqrt{w}} dw\\ =& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{1} w^{k-\frac{1}{2}}(1-w)^{-\frac{1}{2}} dw\\ =& \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} B\left(k+\frac{1}{2},\frac{1}{2}\right)\\ =& \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} B\left(j+\frac{3}{2},\frac{1}{2}\right) \quad (k=j+1)\\ =& \sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\Gamma\left(j+\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(j+2)}\\ =&\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)_{j}}{\Gamma(2) (2)_{j}} \quad \left(\Gamma(t+j) = \Gamma(t)(t)_{j}\right)\\ =& \frac{\pi}{2}\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j+1} \frac{\left(\frac{3}{2}\right)_{j}}{(2)_{j}} \quad \left(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}, \; \Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2}\right)\\ =&\frac{\pi}{2} \sum_{j=0}^{\infty} (-1)^{j} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} }{(2)_{j} (2)_{j}} \quad \left( \frac{(1)_{j}}{(2)_{j}}=\frac{1}{1+j}\right) \\ =&\frac{\pi}{2} \sum_{j=0}^{\infty} (-1)^{j} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} (1)_{j} }{(2)_{j} (2)_{j} (1)_{j}} \\ =&\frac{\pi}{2}\sum_{j=0}^{\infty} \frac{\left(\frac{3}{2}\right)_{j} (1)_{j} (1)_{j} }{(2)_{j} (2)_{j} } \frac{(-1)^{j}}{j!} \quad (j! = (1)_{j}) \\ =&\frac{\pi}{2}{}_{3}F_{2}\left(\frac{3}{2},1,1;2,2;-1\right) \\ =&2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right) \end{align*} Therefore \[\boxed{\int_{0}^{\infty} \frac{\ln(1+e^{-x})}{\sqrt{1-e^{-x}}}e^{-\frac{x}{2}}dx = 2\pi \ln\left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right)}\]