Evaluar

Question

¿La siguiente integral definida tiene un valor conocido "forma cerrada"?

$$\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du,$$

¿o puede ver cualquier persona una manera de integrarlo?

Answer 1

El resultado consiste en la Glaisher constante denominado aquí por $A$, por lo que dudo que no existe una prueba simple para esta fórmula

Para el comienzo $$ \int\limits_{0}^{+\infty}\frac{u\log(a^2+u^2)}{e^{u}-1}du= \int\limits_{0}^{+\infty}\frac{u\log(a^2)}{e^{u}-1}du+ \int\limits_{0}^{+\infty}\frac{u\log(1+e^{-2}u^2)}{e^{u}-1}du= $$ $$ \log(a^2)\int\limits_{0}^{+\infty}\frac{u}{e^{u}-1}du+ \int\limits_{0}^{+\infty}\frac{a\log(1+t^2)}{e^{a}-1}adt= \log(a^2)\int\limits_{0}^{+\infty}\frac{u}{e^{u}-1}du+ a^2\int\limits_{0}^{+\infty}\frac{t\log(1+t^2)}{e^{a}-1}dt $$ La primera integral en la última expresión es bueno saber, y su cálculo se puede encontrar aquí. Pero en cuanto a la segunda... aquí lo Mathematica 8 da $$ -\frac{1}{12 a^2}(3^2-6 a^2 \log (a)+6^2 \log (\pi )+a^2 \log (64)+24 \pi \text{log$\Gamma $}\left(\frac{a}{2 \pi }\right)+12 \pi a+4 \pi ^2 \log (a)-48 \pi ^2 \psi ^{(-2)}\left(\frac{a}{2 \pi }\right)-48 \pi ^2 \log (a)+4 \pi ^2-4 \pi ^2 \log (\pi )-\pi ^2 \log (16)) $$ Así, el resultado es $$ \frac{1}{12} (-3^2+6^2 \log (a)-6 a^2 \log (\pi )-a^2 \log (64)-24 \pi \text{log$\Gamma $}\left(\frac{a}{2 \pi }\right)-12 \pi+48 \pi ^2 \psi ^{(-2)}\left(\frac{a}{2 \pi }\right)+48 \pi ^2 \log (a)-4 \pi ^2+4 \pi ^2 \log (\pi )+\pi ^2 \log (16)) $$

Answer 2

Sólo he podido encontrar la fórmula siguiente: $a = 2\pi\alpha$,

$$\begin{align} \int{0}^{\infty} \frac{u \log(4\pi^2\alpha^2+u^2)}{e^u - 1} \; du &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \ &\quad + \pi^2 \alpha \left(2\alpha \log\alpha - \alpha + 2 - 4 \log\Gamma(\alpha+1)\right) \ &\quad + 4\pi^2 \int{0}^{\alpha} \log\Gamma(u+1)\;du. \end{align}$$ Indeed, let $$I(a) = \int{0}^{\infty} \frac{u \log(a^2+u^2)}{e^u - 1} \; du.$$ Then we have $$\begin{align*}I(0) & = 2 \int{0}^{\infty} \frac{u \log u}{e^u - 1} \; du = 2 \left. \frac{d}{ds}\zeta(s)\Gamma(s) \right|_{s=2} \ & = 2(\zeta'(2) + \zeta(2)\psi0(2)) = 2\left(\zeta'(2) + \frac{\pi^2}{6}(1-\gamma)\right). \end{align}$$ Also, $$\begin{align} I'(a) &= \int{0}^{\infty} \frac{2au}{u^2+a^2} \frac{du}{e^u-1} \ &= 2a \int{0}^{\infty} \sum{n=1}^{\infty} e^{-nu} \left(\int{0}^{\infty} \sin(ux)e^{-ax}\;dx\right)\;du \ &= 2a \int{0}^{\infty} \sum{n=1}^{\infty} \left(\int{0}^{\infty} \sin(xu)e^{-nu}\;du\right)e^{-ax}\;dx \ &= 2a \int{0}^{\infty} \left(\sum{n=1}^{\infty} \frac{x}{x^2+n^2}\right)e^{-ax}\;dx \ &= a \int{0}^{\infty} \left(\pi\coth(\pi x) - \frac{1}{x}\right)e^{-ax}\;dx \end{align}$$ Proceeding, $$\begin{align} I'(a) &= a \left[ \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \right]{0}^{\infty} + a^2 \int{0}^{\infty} \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \; dx \ &= -a\log\pi + a^2 \int{0}^{\infty} \left(\pi x + \log\left(1 - e^{-2\pi x}\right)-\log2 + \log a -\log ax \right)e^{-ax} \; dx \ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right)+a^2 \int{0}^{\infty} e^{-ax}\log\left(1 - e^{-2\pi x}\right) \; dx \ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a^2 \int{0}^{\infty} \sum{n=1}^{\infty} \frac{1}{n} e^{-(a+2n\pi)x} \; dx \ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \sum{n=1}^{\infty} \frac{\frac{a}{2\pi}}{n\left(n + \frac{a}{2\pi}\right)} \ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \left(\gamma + \psi_0 \left( \frac{a}{2\pi} +1 \right)\right) \ &= \pi + a\left(\log\left(\frac{a}{2\pi}\right) - \psi0 \left( \frac{a}{2\pi} +1 \right)\right). \end{align}$$ Thus integrating, $$\begin{align} I(a) &= I(0) + \int{0}^{a} I'(t) \; dt \ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + \pi a + \int_{0}^{a} t \left(\log\left(\frac{t}{2\pi}\right) - \psi0 \left( \frac{t}{2\pi} +1 \right)\right) \; dt \ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + 2 \pi^2 \alpha + 4\pi^2 \int{0}^{\alpha} u \left(\log u - \psi0 (u+1)\right) \; du \ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \ &\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \int{0}^{\alpha} u \psi0 (u+1) \; du \ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \ &\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \left[ u \log\Gamma(u+1) \right]{0}^{\alpha} \ &\quad + 4\pi^2 \int{0}^{\alpha} \log\Gamma(u+1) \; du, \end{align}$$ which gives the desired result. You may verify this formula numerically with the following Mathematica code a = 8; NIntegrate[(u Log[4 Pi^2 a^2 + u^2])/(Exp[u] - 1), {u, 0, Infinity}] 2 Zeta'[2] + Pi^2/3 (1 - EulerGamma) + Pi^2 a (2 a Log[a] - a + 2 - 4 LogGamma[1 + a]) + 4 Pi^2 NIntegrate[LogGamma[1 + u], {u, 0, a}] Clear[a]; - - - - - - In fact, we can give a neater form as follows: $$\begin{align} \int{0}^{\infty} \frac{u \log(a^2+u^2)}{e^{2\pi u} - 1} \; du &= \zeta'(-1) + \frac{a}{2}\left(1 - \frac{a}{2}\right) + \frac{a^2}{2}\log a - a \log\Gamma(a+1)\ &\quad + \int_{0}^{a} \log\Gamma(t+1) \; dt. \end{align*} $$