積分botの問題を解いてみた3

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{x^{\alpha }}{1+2x\cos \pi \beta +x^2}dx\end{array}$$
$$\begin{array}{rcl}& =& {\int }_0^{\mathrm{\infty }}\frac{x^{\alpha }}{(x+\cos \pi \beta {)}^2+{\sin }^2\pi \beta }dx\\ & =& -\frac{\mathrm{Im}}{\sin \pi \beta }{\int }_0^{\mathrm{\infty }}{x}^{\alpha }{\int }_0^{\mathrm{\infty }}{e}^{-t(x+e^{i\pi \beta })}dtdx\\ & =& -\frac{\mathrm{Im}}{\sin \pi \beta }{\int }_0^{\mathrm{\infty }}{e}^{-t{e}^{i\pi \beta }}{\int }_0^{\mathrm{\infty }}{x}^{\alpha }{e}^{-tx}dxdt\\ & =& -\mathrm{Im}\frac{\mathrm{\Gamma }(\alpha +1)}{\sin \pi \beta }{\int }_0^{\mathrm{\infty }}{t}^{-\alpha -1}{e}^{-t{e}^{i\pi \beta }}dt\\ & =& -\mathrm{Im}\frac{\mathrm{\Gamma }(1+\alpha )\mathrm{\Gamma }(-\alpha )}{\sin \pi \beta }{e}^{i\pi \alpha \beta }\\ & =& \frac{\pi \sin \pi \alpha \beta }{\sin \pi \alpha \sin \pi \beta }\end{array}$$

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-ab\cosh x+iac\sinh x-x/2}dx\end{array}$$
$$\begin{array}{rcl}& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}(\frac{b}{\sqrt{b^2+c^2}}\cosh x-\frac{ic}{\sqrt{b^2+c^2}}\sinh x)-x/2}dx\end{array}$$
$$\cos \theta =\frac{b}{\sqrt{b^2+c^2}},\sin \theta =\frac{c}{\sqrt{b^2+c^2}}$$
$$\begin{array}{rcl}& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}(\cos \theta \cos ix+\sin \theta \sin ix)-x/2}dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}\cos (ix-\theta )-x/2}dx\\ & =& {\int }_{-\mathrm{\infty }-i\theta }^{\mathrm{\infty }-i\theta }{e}^{-a\sqrt{b^2+c^2}\cosh x-(x+i\theta )/2}dx\\ & =& e^{-i\theta /2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}\cosh x-x/2}dx\\ & =& 2e^{-i\theta /2}{\int }_0^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}\cosh x}\cosh \frac{x}{2}dx\\ & =& \sqrt{2}{e}^{-i\theta /2}{e}^{-a\sqrt{b^2+c^2}}{\int }_0^{\mathrm{\infty }}{e}^{-a\sqrt{b^2+c^2}t}\frac{dt}{\sqrt{t}}(\cosh x-1=t)\\ & =& \sqrt{\frac{2\pi }{a\sqrt{b^2+c^2}}}{e}^{-a\sqrt{b^2+c^2}-\frac{i}{2}\arccos \frac{b}{\sqrt{b^2+c^2}}}\\ & =& \sqrt{\frac{2\pi }{a\sqrt{b^2+c^2}}}{e}^{-a\sqrt{b^2+c^2}-\frac{i}{2}\arctan \frac{c}{b}}\end{array}$$

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}{e}^{-\beta \sqrt{{\alpha }^2+x^2}}\sqrt{\frac{\sqrt{{\alpha }^2+x^2}-\alpha }{{\alpha }^2+x^2}}\sin \gamma xdx\end{array}$$
$$\begin{array}{rcl}& =& \sqrt{\alpha }{\int }_0^{\mathrm{\infty }}{e}^{-\alpha \beta \cosh x}\sin (\alpha \gamma \sinh x)\sqrt{\cosh x-1}dx(x=\alpha \sinh x)\\ & =& \sqrt{2\alpha }{\int }_0^{\mathrm{\infty }}{e}^{-\alpha \beta \cosh x}\sin (\alpha \gamma \sinh x)\sinh \frac{x}{2}dx\\ & =& -\sqrt{\frac{\alpha }{2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha \beta \cosh x-x/2}\sin (\alpha \gamma \sinh x)dx\\ & =& -\sqrt{\frac{\alpha }{2}}\mathrm{Im}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\alpha \beta \cosh x+i\alpha \beta \sinh x-x/2}dx\\ & =& \sqrt{\frac{\pi }{\sqrt{{\beta }^2+{\gamma }^2}}}{e}^{-\alpha \sqrt{{\beta }^2+{\gamma }^2}}\sin \frac{\arctan \frac{\beta }{\gamma }}{2}\\ & =& \sqrt{\frac{\pi }{2}}{e}^{-\alpha \sqrt{{\beta }^2+{\gamma }^2}}\sqrt{\frac{\sqrt{{\beta }^2+{\gamma }^2}-\beta }{{\beta }^2+{\gamma }^2}}\end{array}$$

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan e^x\tanh \frac{x}{2}}{x\cosh x}dx\end{array}$$
$$\begin{array}{rcl}& =& {\int }_0^{\mathrm{\infty }}\frac{1-e^{-x}}{x}\frac{\arctan e^x+\arctan e^{-x}}{\cosh x(1+e^{-x})}dx\\ & =& \pi {\int }_0^{\mathrm{\infty }}{\int }_0^1\frac{e^{-ux}}{(1+e^{-x})(1+e^{-2x})}{e}^{-x}dudx\\ & =& \pi {\int }_0^1{\int }_0^1\frac{v^u}{(1+v)(1+v^2)}dvdu\end{array}$$
$$\begin{array}{rcl}& =& \frac{\pi }{2}{\int }_0^1{\int }_0^1{v}^u(\frac{1}{1+v}+\frac{1-v}{1+v^2})dvdu\\ & =& \frac{\pi }{2}{\int }_0^1{\int }_0^1{v}^u\sum _{n\ge 0}(-1{)}^n{v}^n+(-1{)}^n(1-v)v^{2n}dvdu\\ & =& \frac{\pi }{2}{\int }_0^1\sum _{n\ge 0}\frac{(-1{)}^n)}{n+u+1}+\frac{(-1{)}^n}{2n+u+1}-\frac{(-1{)}^n}{2n+u+2}du\\ & =& \frac{\pi }{2}\prod _{n\ge 0}\frac{(2n+2{)}^2}{(2n+1)(2n+3)}+\frac{\pi }{8}{\int }_0^1-\psi (\frac{u+1}{4})+\psi (\frac{u+2}{4})+\psi (\frac{u+3}{4})-\psi (\frac{u+4}{4})du\\ & =& \frac{\pi }{2}\ln \frac{\pi }{2}-\frac{\pi }{2}\ln \frac{\pi }{4}\\ & =& \frac{\pi }{2}\ln 2\end{array}$$