\begin{align*} \int_{0}^{\frac{1}{2}}\Gamma\left(1-x\right)\Gamma\left(1+x\right)dx & =\int_{0}^{\frac{1}{2}}x\Gamma\left(1-x\right)\Gamma\left(x\right)dx\\ & =\int_{0}^{\frac{1}{2}}x\frac{\pi}{\sin\left(\pi x\right)}dx\\ & =\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin\left(x\right)}dx\cmt{\pi x\rightarrow x}\\ & =\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{2ix}{e^{ix}-e^{-ix}}dx\\ & =\frac{2i}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{xe^{-ix}}{1-e^{-2ix}}dx\\ & =\frac{2i}{\pi}\int_{0}^{\frac{\pi}{2}}xe^{-ix}\sum_{k=0}^{\infty}e^{-2kix}dx\cmt{\because0<x\rightarrow\left|e^{-2x}\right|<1}\\ & =\frac{2i}{\pi}\sum_{k=0}^{\infty}\int_{0}^{\frac{\pi}{2}}xe^{-\left(2k+1\right)ix}dx\cmt{\text{積分と総和の順序変更}}\\ & =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left[\frac{1}{-\left(2k+1\right)i}xe^{-\left(2k+1\right)ix}-\frac{1}{\left(-\left(2k+1\right)i\right)^{2}}e^{-\left(2k+1\right)ix}\right]_{0}^{\frac{\pi}{2}}\\ & =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left[\frac{i}{2k+1}xe^{-\left(2k+1\right)ix}+\frac{1}{\left(2k+1\right)^{2}}e^{-\left(2k+1\right)ix}\right]_{0}^{\frac{\pi}{2}}\\ & =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left\{ \frac{i}{2k+1}\cdot\frac{\pi}{2}e^{-\left(2k+1\right)i\frac{\pi}{2}}+\frac{1}{\left(2k+1\right)^{2}}\left(e^{-\frac{2k+1}{2}\pi i}-1\right)\right\} \\ & =\frac{2i}{\pi}\sum_{k=0}^{\infty}\left\{ \frac{i}{2k+1}\cdot\frac{\pi}{2}\left(-1\right)^{k+1}i+\frac{1}{\left(2k+1\right)^{2}}\left(\left(-1\right)^{k+1}i-1\right)\right\} \\ & =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}\\ & =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}}-\sum_{k=1}^{\infty}\frac{1}{\left(2k\right)^{2}}\\ & =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\zeta\left(2\right)\left(1-\frac{1}{4}\right)\\ & =i\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(-1\right)^{k}+\frac{2}{\pi}\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{\left(2k+1\right)^{2}}-\frac{2i}{\pi}\zeta\left(2\right)\cdot\frac{3}{4}\\ & =\frac{\pi}{4}i+\frac{2}{\pi}C-\frac{3i}{2\pi}\cdot\frac{\pi^{2}}{6}\cmt{C\text{はカタラン定数}}\\ & =\frac{\pi}{4}i+\frac{2}{\pi}C-\frac{\pi}{4}i\\ & =\frac{2}{\pi}C \end{align*}