πとγがでてくる定積分

\begin{align*} \int_{0}^{\infty}\frac{\sin\left(x\right)\log\left(x\right)}{x}dx & =-\left[\frac{\partial}{\partial y}\int_{0}^{\infty}\frac{\sin\left(x\right)}{x^{y}}dx\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\int_{0}^{\infty}\sin\left(x\right)\mathcal{L}_{t}\left[H_{1}\left(t\right)t^{y-1}\right]\left(x\right)dx\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\int_{0}^{\infty}\sin\left(x\right)\frac{1}{\Gamma\left(y\right)}\int_{0}^{\infty}t^{y-1}e^{-tx}dtdx\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\int_{0}^{\infty}t^{y-1}\int_{0}^{\infty}\sin\left(x\right)e^{-tx}dxdt\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\int_{0}^{\infty}t^{y-1}\mathcal{L}_{x}\left[H_{1}\left(x\right)\sin\left(x\right)\right]\left(t\right)dt\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\int_{0}^{\infty}t^{y-1}\frac{1}{1+t^{2}}dt\right]_{y=1}\cmt{\because0<\Re\left(t\right)}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\cdot\frac{1}{2}\int_{0}^{\infty}s^{\frac{y}{2}-1}\frac{1}{1+s}ds\right]_{y=1}\cmt{s=t^{2}}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\cdot\frac{-1}{2}\int_{1}^{0}\left(\frac{1}{u}-1\right)^{\frac{y}{2}-1}u^{-1}du\right]_{y=1}\cmt{u=\frac{1}{1+s}}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\cdot\frac{1}{2}\int_{0}^{1}u^{1-\frac{y}{2}-1}\left(1-u\right)^{\frac{y}{2}-1}du\right]_{y=1}\\ & =-\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\cdot\frac{1}{2}B\left(\frac{y}{2},1-\frac{y}{2}\right)\right]_{y=1}\\ & =-\frac{1}{2}\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\Gamma\left(\frac{y}{2}\right)\Gamma\left(1-\frac{y}{2}\right)\right]_{y=1}\\ & =-\frac{1}{2}\left[\frac{\partial}{\partial y}\frac{1}{\Gamma\left(y\right)}\pi\sin^{-1}\left(\frac{y}{2}\pi\right)\right]_{y=1}\\ & =-\frac{\pi}{2}\left[-\frac{\Gamma'\left(y\right)}{\Gamma^{2}\left(y\right)}\sin^{-1}\left(\frac{y}{2}\pi\right)-\frac{1}{\Gamma\left(y\right)}\sin^{-2}\left(\frac{y}{2}\pi\right)\cos\left(\frac{y}{2}\pi\right)\frac{\pi}{2}\right]_{y=1}\\ & =\frac{\pi}{2}\Gamma'\left(1\right)\\ & =\frac{\pi}{2}\Gamma\left(1\right)\psi\left(1\right)\\ & =\frac{\pi}{2}\Gamma\left(1\right)\left(-\gamma\right)\\ & =-\frac{\pi\gamma}{2} \end{align*}