오늘의 계산 49 - A Zeta-related Integral

Problem 1. Evaluate the following integral.^[1] \begin{equation}\label{prob:wts} \int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^{2\pi u} - 1} \; du \end{equation}

Proof. Let $I(a)$ denote the given integral. Then we have \begin{align*}I(0) & = 2 \int_{0}^{\infty} \frac{u \log u}{e^{2\pi u} - 1} \; du \\ & = \frac{1}{2\pi^2} \int_{0}^{\infty} \frac{x (\log x - \log (2\pi))}{e^{x} - 1} \; dx \qquad (x = 2\pi u) \\ & = \frac{1}{2\pi^2} \left( \left. \frac{d}{ds}\zeta(s)\Gamma(s) \right|_{s=2} - \zeta(2) \log (2\pi) \right) \\ & = \frac{1}{2\pi^2} \left(\zeta'(2) + \zeta(2)\psi_0(2) - \zeta(2) \log (2\pi) \right). \end{align*} This can be simplified further. Using functional identity of the zeta function, we can show that $$ 12 \zeta(2) \zeta'(-1) = \zeta'(2) + \zeta(2) \psi_0(2) - \zeta(2) \log(2\pi). $$ Thus we have $$I(0) = \zeta'(-1). $$ Also, $$\begin{align*} I'(a) &= \int_{0}^{\infty} \frac{2au}{u^2+a^2} \frac{du}{e^{2\pi u}-1} \\ &= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} e^{-2\pi nu} \left(\int_{0}^{\infty} \sin(ux)e^{-ax}\;dx\right)\;du \\ &= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} \left(\int_{0}^{\infty} \sin(xu)e^{-2\pi nu}\;du\right)e^{-ax}\;dx \\ &= a \int_{0}^{\infty} \left(\sum_{n=1}^{\infty} \frac{2x}{x^2+4\pi^2 n^2}\right)e^{-ax}\;dx \\ &= a \int_{0}^{\infty} \left(\frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}\right)e^{-ax}\;dx \end{align*}$$ Proceeding, \begin{align*}I'(a) &= a \left[ \left(\log\sinh\left(\frac{x}{2}\right)-\log x \right)e^{-ax} \right]_{0}^{\infty} + a^2 \int_{0}^{\infty} \left(\log\sinh\left(\frac{x}{2}\right)-\log x \right)e^{-ax} \; dx \\ &= a\log 2 + a^2 \int_{0}^{\infty} \left(\frac{x}{2} + \log\left(1 - e^{-x}\right)-\log2 + \log a -\log ax \right)e^{-ax} \; dx \\ &= \frac{1}{2} + a(\gamma + \log a) + a^2 \int_{0}^{\infty} e^{-ax} \log\left(1 - e^{-x}\right) \; dx \\ &= \frac{1}{2} + a(\gamma + \log a) - a^2 \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n} e^{-(a+n)x} \; dx \\ &= \frac{1}{2} + a(\gamma + \log a) - a \sum_{n=1}^{\infty} \frac{a}{n(n+a)} \\ &= \frac{1}{2} + a(\gamma + \log a) - a (\gamma + \psi_0 (a+1) ) \\ &= \frac{1}{2} + a\left( \log a - \psi_0 (a+1) \right). \end{align*} Thus integrating, we have the following result for \eqref{prob:wts}. $$\begin{align*} I(a) &= I(0) + \int_{0}^{a} I'(t) \; dt \\ &= \zeta'(-1) + \frac{a}{2} + \int_{0}^{a} t \left( \log t - \psi_0 (t+1) \right) \; dt \\ &= \zeta'(-1) + \frac{a}{2} - \frac{a^2}{4} + \frac{a^2}{2}\log a - \int_{0}^{a} t \psi_0 (t+1) \; dt \\ &= \zeta'(-1) + \frac{a}{2}\left(1 - \frac{a}{2}\right) + \frac{a^2}{2}\log a - a \log\Gamma(a+1) + \int_{0}^{a} \log\Gamma(t+1) \; dt. \end{align*}$$

Almost same argument proves the following identity.

Problem 2. Show the following identity.^[2] \begin{equation}\label{prob2:wts} \int_{0}^{1} \frac{\log(1-x)}{x} \frac{2z}{\log^{2}x+(2\pi z)^{2}} \; dx = -\log\left(\frac{z!\, e^{z}}{z^{z}\sqrt{2\pi z}}\right). \end{equation}

Proof. Using the substitution $x = e^{-2\pi u}$, we have \begin{align*} \int_{0}^{1}\frac{\log(1-x)}{x}\frac{2z}{\log^{2}x+(2\pi z)^{2}}\; dx &=\frac{1}{\pi}\int_{0}^{\infty}\frac{z}{u^{2}+z^{2}}\log(1-e^{-2\pi u})\; du \\ &=-\frac{1}{\pi}\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{e^{-2\pi n u}}{n}\int_{0}^{\infty}\cos (ut) e^{-zt}\; dt du\\ &=-\frac{1}{\pi}\int_{0}^{\infty}e^{-zt}\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\infty}\cos (tu) e^{-2\pi n u}\; du dt\\ &=-\int_{0}^{\infty}\frac{e^{-zt}}{t}\left(\sum_{n=1}^{\infty}\frac{2t}{t^{2}+(2\pi n)^{2}}\right) dt\;\\ &=-\int_{0}^{\infty}\frac{e^{-zt}}{t}\left(\frac{1}{2}\coth\left(\frac{t}{2}\right)-\frac{1}{t}\right)\; dt. \end{align*} Now let $I(z)$ denote this integral. Then \begin{align*} I'(z) &= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{t}{2}\right) - \frac{1}{t} \right) e^{-zt} \; dt \\ &= \left[ \left( \log \sinh\left(\frac{t}{2}\right) - \log t + \log 2 \right) e^{-zt} \right]_{0}^{\infty} \\ &\quad + z \int_{0}^{\infty} \left( \log \sinh\left(\frac{t}{2}\right) - \log t + \log 2 \right) e^{-zt} \; dt \\ &= z \int_{0}^{\infty} \left( \frac{t}{2} - \log t + \log \left(1 - e^{-t}\right) \right) e^{-zt} \; dt \\ &= \frac{1}{2z} + \gamma + \log z + z \int_{0}^{\infty} e^{-zt} \log \left(1 - e^{-t}\right) \; dt \\ &= \frac{1}{2z} + \gamma + \log z - z \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(z+n)t}}{n} \; dt \\ &= \frac{1}{2z} + \gamma + \log z - z \sum_{n=1}^{\infty} \frac{1}{n(n+z)} \\ &= \frac{1}{2z} + \log z - \psi_0 (z+1). \end{align*} Therefore, we have \begin{align*} I(z) &=-\int_{z}^{\infty} I'(w) \; dw \\ &=-\left[ \frac{1}{2}\log w + w \log w - w - \log(w!) \right]_{z}^{\infty} \\ &= \log\sqrt{2\pi} + \frac{1}{2}\log z + z \log z - z - \log (z!) \\ &= -\log \left( \frac{z! e^z}{z^z \sqrt{2\pi z}} \right). \end{align*}

References

1. Evaluate $\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du$ in "Math StackExchange"
2. An Integral in "AoPS Forum"