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@infseriesbot
さんがツイートした
こちらの定理
の解説です.
以下の等式が成り立ちます.
$\displaystyle\int_0^\infty\frac{\cos\sqrt{x}}{e^{2\pi\sqrt{x}}-1}\mathrm{d}x=1-\frac{e}{\left(e-1\right)^2}$
解説
\begin{align*}
&\int_0^\infty\frac{\cos\sqrt{x}}{e^{2\pi\sqrt{x}}-1}\mathrm{d}x\\
=&\int_0^\infty\cos\sqrt{x}\frac{e^{-2\pi\sqrt{x}}}{1-e^{-2\pi\sqrt{x}}}\mathrm{d}x\\
=&\int_0^\infty\cos\sqrt{x}\sum_{k=1}^\infty e^{-2\pi k\sqrt{x}}\mathrm{d}x\\
=&\int_0^\infty\sum_{k=1}^\infty\left(e^{-2\pi k\sqrt{x}}\frac{\frac{1}{\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+\left(1+\left(2\pi k\right)^2\right)\cos\sqrt{x}}{1+\left(2\pi k\right)^2}-e^{-2\pi k\sqrt{x}}\frac{1}{\sqrt{x}}\frac{\left(1+\left(2\pi k\right)^2\right)\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\mathrm{d}x\\
=&\int_0^\infty\sum_{k=1}^\infty\left(e^{-2\pi k\sqrt{x}}\frac{\frac{1}{\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)-2\pi k\sin\sqrt{x}+\left(2\pi k\right)^2\cos\sqrt{x}+\cos\sqrt{x}+2\pi k\sin\sqrt{x}}{1+\left(2\pi k\right)^2}-e^{-2\pi k\sqrt{x}}\frac{1}{\sqrt{x}}\frac{\left(-1+\left(2\pi k\right)^2\right)\left(-2\pi k\cos\sqrt{x}\right)+2\left(2\pi k\right)^2\sin\sqrt{x}+\left(1-\left(2\pi k\right)^2\right)\sin\sqrt{x}+2\left(-2\pi k\cos\sqrt{x}\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\mathrm{d}x\\
=&\int_0^\infty\sum_{k=1}^\infty\left(e^{-2\pi k\sqrt{x}}\frac{\frac{1}{\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)-2\pi k\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+\cos\sqrt{x}+2\pi k\sin\sqrt{x}}{1+\left(2\pi k\right)^2}-e^{-2\pi k\sqrt{x}}\frac{1}{\sqrt{x}}\frac{-2\pi k\left(\left(-1+\left(2\pi k\right)^2\right)\cos\sqrt{x}-4\pi k\sin\sqrt{x}\right)-\left(-1+\left(2\pi k\right)^2\right)\sin\sqrt{x}-4\pi k\cos\sqrt{x}}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\mathrm{d}x\\
=&\int_0^\infty\sum_{k=1}^\infty\left(\frac{2\frac{1}{2\sqrt{x}}e^{-2\pi k\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+2\sqrt{x}\left(-2\pi k\frac{1}{2\sqrt{x}}\right)e^{-2\pi k\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+2\sqrt{x}e^{-2\pi k\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\cos\sqrt{x}+2\pi k\frac{1}{2\sqrt{x}}\sin\sqrt{x}\right)}{1+\left(2\pi k\right)^2}-\frac{2\left(-2\pi k\frac{1}{2\sqrt{x}}\right)e^{-2\pi k\sqrt{x}}\left(\left(-1+\left(2\pi k\right)^2\right)\cos\sqrt{x}-4\pi k\sin\sqrt{x}\right)+2e^{-2\pi k\sqrt{x}}\left(-\left(-1+\left(2\pi k\right)^2\right)\frac{1}{2\sqrt{x}}\sin\sqrt{x}-4\pi k\frac{1}{2\sqrt{x}}\cos\sqrt{x}\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\mathrm{d}x\\
=&\int_0^\infty\sum_{k=1}^\infty\left(\frac{2\left(\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}\right)e^{-2\pi k\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+2\sqrt{x}\left(\frac{\mathrm{d}}{\mathrm{d}x}e^{-2\pi k\sqrt{x}}\right)\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)+2\sqrt{x}e^{-2\pi k\sqrt{x}}\left(\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin\sqrt{x}\right)-2\pi k\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos\sqrt{x}\right)\right)}{1+\left(2\pi k\right)^2}-\frac{2\left(\frac{\mathrm{d}}{\mathrm{d}x}e^{-2\pi k\sqrt{x}}\right)\left(\left(-1+\left(2\pi k\right)^2\right)\cos\sqrt{x}-4\pi k\sin\sqrt{x}\right)+2e^{-2\pi k\sqrt{x}}\left(\left(-1+\left(2\pi k\right)^2\right)\left(\frac{\mathrm{d}}{\mathrm{d}x}\cos\sqrt{x}\right)-4\pi k\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin\sqrt{x}\right)\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\mathrm{d}x\\
=&\left[\sum_{k=1}^\infty\left(\frac{2\sqrt{x}e^{-2\pi k\sqrt{x}}\left(\sin\sqrt{x}-2\pi k\cos\sqrt{x}\right)}{1+\left(2\pi k\right)^2}-\frac{2e^{-2\pi k\sqrt{x}}\left(\left(-1+\left(2\pi k\right)^2\right)\cos\sqrt{x}-4\pi k\sin\sqrt{x}\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\right)\right]_{x=0}^\infty\\
=&\sum_{k=1}^\infty\frac{2\left(-1+\left(2\pi k\right)^2\right)}{\left(1+\left(2\pi k\right)^2\right)^2}\\
=&\left(\frac{1}{2\pi}\right)^2\sum_{k=1}^\infty\frac{2\left(-\left(\frac{1}{2\pi}\right)^2+k^2\right)}{\left(\left(\frac{1}{2\pi}\right)^2+k^2\right)^2}\\
=&\left(\frac{1}{2\pi}\right)^2\sum_{k=1}^\infty\left(\frac{\left(\frac{i}{2\pi}+k\right)^2+\left(-\frac{i}{2\pi}+k\right)^2}{\left(-\frac{i}{2\pi}+k\right)^2\left(\frac{i}{2\pi}+k\right)^2}\right)\\
=&\left(\frac{1}{2\pi}\right)^2\sum_{k=1}^\infty\left(\frac{1}{\left(-\frac{i}{2\pi}+k\right)^2}+\frac{1}{\left(\frac{i}{2\pi}+k\right)^2}\right)\\
=&\left(\frac{1}{2\pi}\right)^2\sum_{n=0}^\infty\left(\frac{1}{\left(1-\frac{i}{2\pi}+n\right)^2}+\frac{1}{\left(1+\frac{i}{2\pi}+n\right)^2}\right)\\
=&\left(\frac{1}{2\pi}\right)^2\left(\psi^\left(1\right)\left(1-\frac{i}{2\pi}\right)+\psi^\left(1\right)\left(1+\frac{i}{2\pi}\right)\right)\\
=&\left(\frac{1}{2\pi}\right)^2\left(\psi^\left(1\right)\left(1-\frac{i}{2\pi}\right)+\psi^\left(1\right)\left(\frac{i}{2\pi}\right)-\frac{1}{\left(\frac{i}{2\pi}\right)^2}\right)\\
=&\left(\frac{1}{2\pi}\right)^2\left(\frac{\pi^2}{\sin^2\pi\frac{i}{2\pi}}+\frac{1}{\left(\frac{1}{2\pi}\right)^2}\right)\\
=&\left(\frac{1}{2i\sinh\frac{1}{2}}\right)^2+1\\
=&1-\left(\frac{1}{e^\frac{1}{2}-e^{-\frac{1}{2}}}\right)^2\\
=&1-\frac{e}{\left(e-1\right)^2}
\end{align*}
なので,$\displaystyle\int_0^\infty\frac{\cos\sqrt{x}}{e^{2\pi\sqrt{x}}-1}\mathrm{d}x=1-\frac{e}{\left(e-1\right)^2}$です.$\blacksquare$