\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}

In today´s post We will compute these wonderfull integrals:

$\int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx=\frac{\ln 2}{3} - \log \pi - \frac{1}{2} + 6 \ln A$

$\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}$

$\int_0^\infty\frac{x \ln(1+x^2)}{e^{\pi x}-1}\,dx =-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34$ (1)

$\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi$ (2)

$\int_0^\infty\frac{x \ln(1+x^2)}{e^{4\pi x}-1}\,dx=\frac{\ln \pi}{4}-\frac{36}{48}+\frac{35} {48} \ln2+\frac14 \zeta^{\prime}(-1)$ (3)

Lemma 1

$\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2}{e^{2 \pi x}-1}$

Proof:

$\begin{aligned} &\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2}{e^{2 \pi x}-1}\\ &\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2e^{-\pi x}}{e^{ \pi x}-e^{-\pi x}}\\ &\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{e^{-\pi x}}{\sinh \pi x}\\ &\frac{e^{-\pi x}+1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}\\ &\frac{1}{\sinh \pi x}=\frac{2}{(e^{\pi x}-1)(e^{-\pi x}+1)}\\ &\frac{1}{\sinh \pi x}=\frac{1}{\sinh \pi x} \qquad \blacksquare\\ \end{aligned}$

Therefore, from Lemma 1 and from (1) and (2) we obtain

$\begin{aligned} \int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx&=2\int_0^\infty\frac{x \ln(1+x^2)}{e^{\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx \\&=2\left(-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4}-\zeta^{\prime}(-1)-\ln \sqrt{2 \pi}+\frac{3}{4}\right)\\ &=2\left(-3 \zeta^{\prime}(-1)+\frac{2}{3}\ln 2-\frac12 \ln 2-\frac12 \ln \pi\right)\\ &=-6 \zeta^{\prime}(-1)+\frac{4}{3}\ln 2- \ln 2- \ln \pi\\ &=-6 \zeta^{\prime}(-1)+\frac{\ln 2}{3}- \ln \pi\\ &=-6 \left(\frac{1}{12}-\ln A \right)+\frac{\ln 2}{3}- \ln \pi\\ &=-\frac{\ln e}{2}+\ln A^6 +\frac{\ln 2}{3}- \ln \pi\\ &=\ln \, \frac{\sqrt[3]{2}\, A^6}{\pi \sqrt{e}} \qquad \blacksquare \end{aligned}$

Lemma 2

$\frac{1}{e^{2\pi x}+1}=\frac{1}{e^{2\pi x}-1}-\frac{2}{e^{4\pi x}-1}$

Therefore from Lemma 2 and (2) and (3) we obtain

$\begin{aligned} \int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx&=\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(1+x^2)}{e^{4\pi x}-1}\,dx\\ &=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi-2\left( \frac{\ln \pi}{4}-\frac{36}{48}+\frac{35}{48} \ln2+\frac14 \zeta^{\prime}(-1)\right)\\ &=\frac12\zeta^{\prime}(-1)-\frac{23}{24}\ln 2+\frac{18}{24}\\ &=\frac12\left(\frac{1}{12}-\ln A \right)-\frac{23}{24}\ln 2+\frac{18}{24}\\ &=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2} \qquad \blacksquare \end{aligned}$

Search This Blog

MY MATEX NOTES