Integral of the day XXVI

Integral involving the Dirichlet eta function $\eta(v)$

Today we show the proof of this integral proposed by @Ali39342137: $$\int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)$$ For the proof we use the expansion in partial fractions of the hyperbolic cosecant function and the derivative of the Dirichlet eta function.

Proof:

Recall that the hyperbolic cosecant can be expanded as partial fractions: $$\operatorname{csch} x = \frac{1}{\sinh x} = \frac{1}{x}-\frac{2x}{\pi^2+x^2}+ \frac{2x}{4\pi^2+x^2}-\frac{2x}{9\pi^2+x^2} + \cdots = \frac{1}{x} + 2\sum_{j=1}^{\infty} \frac{(-1)^jx}{j^2\pi^2+x^2}$$ Hence, $$\begin{align*} \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx =& \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{x} - 2\sum_{j=1}^{\infty} \frac{(-1)^jx}{j^2\pi^2+x^2}\right)\frac{\ln(x)}{x} dx \\ =& -2\int_{0}^{\infty} \sum_{j=1}^{\infty} \frac{(-1)^j\ln(x)}{j^2\pi^2+x^2} dx \\ =& -2\sum_{j=1}^{\infty}\int_{0}^{\infty} (-1)^j\frac{\ln(x)}{j^2\pi^2+x^2} dx \\ =& -\frac{2}{\pi^2}\sum_{j=1}^{\infty}\frac{(-1)^j}{j^2} \int_{0}^{\infty} \frac{\ln(x)}{1+\left(\frac{x}{j\pi}\right)^2} dx \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \int_{0}^{\infty} \frac{\ln(j\pi w)}{1+w^2} dx \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \int_{0}^{\infty} \frac{\ln(j\pi)+\ln(w) }{1+w^2} dw \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\ln(j\pi)\int_{0}^{\infty} \frac{1}{1+w^2} dw + \int_{0}^{\infty} \frac{\ln(w) }{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \int_{0}^{\infty} \frac{\ln(w) }{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \int_{0}^{\infty} \frac{\frac{d}{dt}\Big|_{t=0+} w^t}{1+w^2} dw\right] \\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} \frac{ w^t}{1+w^2} dw\right] \\ \end{align*}$$ Recall the integral representation of the secant function: $$\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2x}{\pi}}}{y^2+1} dy \quad \left|x\right|\lt \frac{\pi}{2}$$ If we put $\displaystyle x = \frac{t\pi}{2}$ $$\sec\left( \frac{t\pi}{2}\right) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{t}}{y^2+1} dy$$ Hence $$\int_{0}^{\infty} \frac{y^{t}}{y^2+1} dy = \frac{\pi}{2}\sec\left( \frac{t\pi}{2}\right)$$ Therefore $$\begin{align*} I =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+}\int_{0}^{\infty} \frac{ w^t}{1+w^2} dw\right]\\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \frac{d}{dt}\Big|_{t=0+} \frac{\pi}{2}\sec\left( \frac{t\pi}{2}\right)\right]\\ =& -\frac{2}{\pi}\sum_{j=1}^{\infty}\frac{(-1)^j}{j} \left[\frac{\pi}{2} \ln(j\pi) + \underbrace{\lim_{t \to 0+} \frac{\pi^2}{4}\sec\left( \frac{t\pi}{2}\right)\tan\left( \frac{t\pi}{2}\right)}_{=0}\right]\\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j\pi)}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\left[\ln(j)+ \ln(\pi)\right]}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j} \\ =& \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\ln(2) \\ \end{align*}$$ Recall the definition of the Dirichlet eta funcion: $$\eta(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j+1}}{j^v}$$ Taking the derivative $$\eta'(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j^v}$$ Therefore $$\lim_{v \to 1} \eta'(v) = \sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j}$$ Now recall the relationship: $$\eta(v) = (1-2^{1-V}) \zeta(v)$$ Hence $$\eta'(v) = 2^{-v}\left[(2^v-2)\zeta'(v) + \ln(4)\zeta(v) \right]$$ Taking the limit as $v \to 1$ can be proven that: $$\lim_{v \to 1} \eta'(v) = \gamma\ln(2) -\frac{1}{2}\ln^2(2)$$ Therefore $$\sum_{j=1}^{\infty} \frac{(-1)^{j}\ln(j)}{j} = \gamma\ln(2) -\frac{1}{2}\ln^2(2)$$ Then we have $$I = \sum_{j=1}^{\infty}\frac{(-1)^{j+1}\ln(j)}{j} + \ln(\pi)\ln(2) = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)$$ Therefore we can conclude $$\boxed{ \int_{0}^{\infty} \left(\frac{1}{x} - \frac{1}{\sinh x}\right)\frac{\ln(x)}{x} dx = -\gamma\ln(2) +\frac{1}{2}\ln^2(2) + \ln(\pi)\ln(2)}$$