ラプラス変換を用いて解ける積分(積分botの問題を解いてみた4)

$$\begin{array}{r}{\int }_0^{\pi }\arctan \frac{\ln \sin x}{x}dx\end{array}$$
$$\begin{array}{rcl}& =& {\int }_0^{\pi }\frac{\pi }{2}-\arctan \frac{x}{\ln \sin x}dx\end{array}$$
$$\begin{array}{r}{\int }_0^{\pi }\arctan \frac{x}{\ln \sin x}dx\end{array}$$
$$\begin{array}{rcl}& =& -\mathrm{Im}{\int }_0^{\pi }\ln (-\ln \sin x+ix)dx\\ & =& -\mathrm{Im}\frac{d}{ds}{\int }_0^{\pi }\frac{dx}{(-\ln \sin x+ix{)}^s}\\ & =& -\mathrm{Im}\frac{d}{ds}\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\pi }{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-(-\ln \sin x+ix)t}dtdx\\ & =& \frac{d}{ds}\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{\int }_0^{\pi }{\sin }^{t}x\sin txdxdt\end{array}$$
ここで
$${\int }_0^{\pi }{\sin }^{u-1}x\sin vxdx=\frac{\pi \sin \frac{\pi v}{2}}{2^{u-1}uB(\frac{u+v+1}{2},\frac{u-v+1}{2})}$$
を用いる。
これは留数定理を使って証明できるが今回は省略する。
$$\begin{array}{r}\frac{d}{ds}\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{\int }_0^{\pi }{\sin }^{t}x\sin txdxdt\end{array}$$
$$\begin{array}{rcl}& =& \frac{d}{ds}\frac{\pi }{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}\sin \frac{\pi t}{2}{2}^{-t}dt\\ & =& \mathrm{Im}\frac{d}{ds}\frac{\pi }{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t(\ln 2-\frac{i\pi }{2})}dt\\ & =& \mathrm{Im}\frac{d}{ds}\frac{\pi }{(\ln 2-\frac{i\pi }{2}{)}^s}\\ & =& \pi \arctan \frac{\pi }{2\ln 2}\end{array}$$
$$\begin{array}{r}\therefore {\int }_0^{\pi }\arctan \frac{\ln \sin x}{x}dx=-\pi \arctan \frac{2\ln 2}{\pi }\end{array}$$

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin (s\arctan x)}{(1+x^2{)}^{s/2}(e^{\pi x}+1)}dx\end{array}$$
$$\begin{array}{rcl}& =& -\frac{\mathrm{Im}}{\mathrm{\Gamma }(s)}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{e^{\pi x}+1}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t(1+xi)}dtdx\\ & =& -\frac{\mathrm{Im}}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-tix}}{e^{\pi x}+1}dxdt\\ & =& -\frac{\mathrm{Im}}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}{\int }_0^{\mathrm{\infty }}\frac{e^{-tix}{e}^{-\pi x}}{1+e^{-\pi x}}+\frac{e^{tix}}{1+e^{-\pi x}}dxdt\\ & =& -\frac{\mathrm{Im}}{\mathrm{\Gamma }(s)}\sum _{n\ge 0}(-1{)}^n{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}{\int }_0^{\mathrm{\infty }}{e}^{-x(ti+\pi (n+1))}+e^{-x(-ti+\pi n)}dx\\ & =& -\frac{\mathrm{Im}}{\mathrm{\Gamma }(s)}\sum _{n\ge 0}(-1{)}^n{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}(\frac{1}{\pi (n+1)+ti}+\frac{1}{\pi n-ti})dt\\ & =& \frac{1}{\mathrm{\Gamma }(s)}\sum _{n\ge 0}(-1{)}^n{\int }_0^{\mathrm{\infty }}{t}^s{e}^{-t}(\frac{t}{{\pi }^2(n+1{)}^2+t^2}-\frac{t}{{\pi }^2{n}^2+t^2})dt\\ & =& {\int }_0^{\mathrm{\infty }}{t}^s{e}^{-t}\sum _{n\in Z}\frac{(-1{)}^{n-1}}{{\pi }^2{n}^2+t^2}dt\\ & =& -\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}\frac{t^{s-1}{e}^{-}t}{\sinh t}dt\\ & =& -\frac{2}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}\frac{e^{-2t}}{1-e^{-2t}}dt\\ & =& -\frac{2}{\mathrm{\Gamma }(s)}\sum _{n\ge 1}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-2nt}dt\\ & =& -\frac{1}{2^{s-1}}\sum _{n\ge 1}\frac{1}{n^s}\\ & =& \eta (s)-\zeta (s)\end{array}$$

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}(1-\frac{\cos ax}{e^x})\frac{dx}{x^{1+r}}\end{array}$$
$$\begin{array}{rcl}& =& \mathrm{Re}{\int }_0^{\mathrm{\infty }}(1-e^{-(ai+1)x})\frac{dx}{x^{1+r}}\\ & =& \mathrm{Re}(ai+1){\int }_0^{\mathrm{\infty }}{\int }_0^1{e}^{-(ai+1)xy}\frac{dydx}{x^r}\\ & =& \mathrm{Re}{\int }_0^1\frac{\mathrm{\Gamma }(1-r)}{(ai+1{)}^r{y}^{1-r}}dy\\ & =& \mathrm{Re}(ai+1{)}^r\frac{\mathrm{\Gamma }(1-r)}{r}\\ & =& \frac{\mathrm{\Gamma }(1-r)}{r}(1+a^2{)}^{r/2}\cos (r\arctan a)\end{array}$$

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos (s\arctan \left(ax\right))}{(1+x^2)(1+(ax{)}^2{)}^{s/2}}dx\end{array}$$
$$\begin{array}{rcl}& =& \frac{\mathrm{Re}}{\mathrm{\Gamma }(s)}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+x^2}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-(1+axi)t}dtdx\\ & =& \frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos axt}{1+x^2}dxdt\\ & =& \frac{\pi }{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-(1+a)t}dt\\ & =& \frac{\pi }{(1+a{)}^s}\end{array}$$

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}{t}^{u-1}{e}^{-at}\cos \left(bt\right)dt\end{array}$$
$$\begin{array}{rcl}& =& \frac{\mathrm{\Gamma }(u)}{(a^2+b^2{)}^{u/2}}\cos (u\arctan \frac{b}{a})\end{array}$$
u→iu
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}{t}^{ui-1}{e}^{-at}\cos \left(bt\right)dt\end{array}$$
$$\begin{array}{rcl}& =& \mathrm{\Gamma }(ui)e^{-\frac{iu}{2}\ln (a^2+b^2)}\cosh (u\arctan \frac{b}{a})\end{array}$$
$$\begin{array}{r}\therefore {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cos (\frac{r}{2}\ln (1+a^2{x}^2))\cosh (r\arctan \left(ax\right))\frac{dx}{1+x^2}\end{array}$$
$$\begin{array}{rcl}& =& \frac{\mathrm{Re}}{\mathrm{\Gamma }(ir)}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+x^2}{\int }_0^{\mathrm{\infty }}{t}^{ir-1}{e}^{-t}\cos \left(axt\right)dtdx\\ & =& \frac{\mathrm{Re}}{\mathrm{\Gamma }(ir)}{\int }_0^{\mathrm{\infty }}{t}^{ir-1}{e}^{-t}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos \left(axt\right)}{1+x^2}dxdt\\ & =& \mathrm{Re}\frac{\pi }{\mathrm{\Gamma }(ir)}{\int }_0^{\mathrm{\infty }}{t}^{ir-1}{e}^{-(1+a)t}dt\\ & =& \mathrm{Re}\frac{\pi }{(1+a{)}^{ir}}\\ & =& \pi \cos (r\ln (1+a))\end{array}$$