積分botの問題を解いてみた3

\begin{eqnarray} \int_{0}^{\infty}\frac{x^\alpha} {1+2x\cos\pi\beta+x^2}dx\\ \end{eqnarray}
\begin{eqnarray} &=&\int_{0}^{\infty} \frac{x^\alpha}{(x+\cos\pi\beta)^2+\sin^2\pi\beta}dx \\ &=&-\frac{\Im}{\sin\pi\beta}\int_{0}^{\infty}x^\alpha\int_{0}^{\infty}e^{-t(x+e^{i\pi\beta})}dtdx \\ &=&-\frac{\Im}{\sin\pi\beta} \int_{0}^{\infty}e^{-te^{i\pi\beta}} \int_{0}^{\infty}x^\alpha e^{-tx}dxdt\\ &=&-\Im\frac{\Gamma(\alpha+1)}{\sin\pi\beta} \int_{0}^{\infty}t^{-\alpha-1}e^{-te^{i\pi\beta}}dt\\ &=&-\Im\frac{\Gamma(1+\alpha)\Gamma(-\alpha)}{\sin\pi\beta} e^{i\pi\alpha\beta}\\ &=&\frac{\pi\sin\pi\alpha\beta} {\sin\pi\alpha\sin\pi\beta} \end{eqnarray}

\begin{eqnarray} \int_{-\infty}^{\infty} e^{-ab\cosh x+iac\sinh x-x/2}dx \end{eqnarray}
\begin{eqnarray} &=&\int_{-\infty}^{\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{eqnarray}
$$\cos\theta=\frac{b}{\sqrt{b^2+c^2}},\sin\theta=\frac{c}{\sqrt{b^2+c^2}}$$
\begin{eqnarray} &=&\int_{-\infty}^{\infty} e^{-a\sqrt{b^2+c^2}(\cos\theta\cos ix+\sin\theta\sin ix)-x/2}dx\\ &=&\int_{-\infty}^{\infty} e^{-a\sqrt{b^2+c^2}\cos(ix-\theta)-x/2}dx\\ &=&\int_{-\infty-i\theta}^{\infty-i\theta} e^{-a\sqrt{b^2+c^2}\cosh x-(x+i\theta)/2}dx\\ &=&e^{-i\theta/2}\int_{-\infty}^{\infty}e^{-a\sqrt{b^2+c^2}\cosh x-x/2}dx\\ &=&2e^{-i\theta/2}\int_{0}^{\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}^{\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{eqnarray}

\begin{eqnarray} \int_{0}^{\infty}e^{-\beta\sqrt{\alpha^2+x^2}}\sqrt{\frac{\sqrt{\alpha^2+x^2}-\alpha}{\alpha^2+x^2}}\sin\gamma xdx \end{eqnarray}
\begin{eqnarray} &=&\sqrt{\alpha}\int_{0}^{\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}^{\infty}e^{-\alpha\beta\cosh x}\sin(\alpha\gamma\sinh x) \sinh\frac{x}{2}dx\\ &=&-\sqrt{\frac{\alpha}{2}} \int_{-\infty}^{\infty}e^{-\alpha\beta\cosh x-x/2}\sin(\alpha\gamma\sinh x)dx\\ &=&-\sqrt{\frac{\alpha}{2}} \Im \int_{-\infty}^{\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{eqnarray}

\begin{eqnarray} \int_{-\infty}^{\infty} \frac{\arctan e^x \tanh \frac{x}{2}}{x\cosh x}dx \end{eqnarray}
\begin{eqnarray} &=&\int_{0}^{\infty}\frac{1-e^{-x}}{x}\frac{\arctan e^x+\arctan e^{-x}}{\cosh x(1+e^{-x})}dx\\ &=&\pi\int_{0}^{\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{eqnarray}
\begin{eqnarray} &=&\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≥0}(-1)^nv^n+(-1)^n(1-v)v^{2n}dvdu\\ &=&\frac{\pi}{2}\int_{0}^{1} \sum_{n≥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≥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{eqnarray}