\begin{align*} \int_{-\infty}^{\infty}\Gamma\left(1-ix\right)\Gamma\left(1+ix\right)dx & =\int_{-\infty}^{\infty}ix\Gamma\left(1-ix\right)\Gamma\left(ix\right)dx\\ & =\int_{-\infty}^{\infty}ix\frac{\pi}{\sin\left(i\pi x\right)}dx\\ & =\int_{-\infty}^{\infty}\frac{x\pi}{\sinh\left(\pi x\right)}dx\\ & =\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x}{\sinh\left(x\right)}dx\cmt{\pi x\rightarrow x}\\ & =\frac{2}{\pi}\int_{0}^{\infty}\frac{x}{\sinh\left(x\right)}dx\\ & =\frac{4}{\pi}\int_{0}^{\infty}\frac{x}{e^{x}-e^{-x}}dx\\ & =\frac{4}{\pi}\int_{0}^{\infty}\frac{xe^{-x}}{1-e^{-2x}}dx\\ & =\frac{4}{\pi}\int_{0}^{\infty}xe^{-x}\sum_{k=0}^{\infty}e^{-2kx}dx\cmt{\because0<x\rightarrow\left|e^{-2x}\right|<1}\\ & =\frac{4}{\pi}\sum_{k=0}^{\infty}\int_{0}^{\infty}xe^{-\left(2k+1\right)x}dx\cmt{\text{積分と総和の順序変更}}\\ & =\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}\int_{0}^{\infty}xe^{-x}dx\cmt{\left(2k+1\right)x\rightarrow x}\\ & =\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}1!\\ & =\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)^{2}}\\ & =\frac{4}{\pi}\left(\sum_{k=0}^{\infty}\frac{1}{k^{2}}-\sum_{k=0}^{\infty}\frac{1}{\left(2k\right)^{2}}\right)\\ & =\frac{4}{\pi}\left(1-\frac{1}{2^{2}}\right)\zeta\left(2\right)\\ & =\frac{3}{\pi}\cdot\frac{\pi^{2}}{6}\\ & =\frac{\pi}{2} \end{align*}