VARIATION OF BINET´S INTEGRAL

In this post we obtain the following integrals

$\begin{aligned} &\int_{0}^{\infty}\frac{\arctan(x)}{e^{4\pi x}-1}dx=\frac12-\frac{3 \ln 2}{8}-\frac{\ln 2 \pi}{8}\\ &\\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{2\pi x}-1}dx=\frac12-\frac{\ln 2 \pi}{4}\\ &\\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{\pi x}-1}dx=\frac12-\frac{\ln 2}{2}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{\pi x/2}-1}dx=\frac12-\ln \pi-2 \ln 2+2 \ln \Gamma\left(\frac14 \right) \\ & \\ & \int_{0}^{\infty}\frac{\arctan(x)}{\sinh 2 \pi x}dx=\frac{\ln 2}{2}-\frac{\ln \pi}{4}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{\sinh \pi x}dx=\frac{\ln \pi}{2}-\frac{\ln 2}{2}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{\sinh \frac{\pi x}{2}}dx=4 \ln \Gamma\left(\frac14\right)-2 \ln \pi-3 \ln 2\\ &\\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{ 2\pi x}+1}dx=\frac{3}{4}\ln 2-\frac12\\ &\\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{ \pi x}+1}dx=\frac{\ln \pi}{2}-\frac{1}{2}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{ \pi x/2}+1}dx=2 \ln \Gamma\left(\frac14\right)-\frac12-\ln 2\pi \end{aligned}$

We have previously proved the following result

$2\int_{0}^{\infty}\frac{\arctan\left(\frac{x}{z}\right)}{e^{2\pi x}-1}dx=\log(\Gamma(z))-z\log(z)+z+\frac{1}{2}\log(z)-\frac12\ln(2 \pi)$ (1)

Letting $x \to zx$ we obtain

$\int_{0}^{\infty}\frac{\arctan(x)}{e^{2\pi z x}-1}dx=\frac{\log(\Gamma(z))}{2z}-\frac{\log(z)}{2}+\frac12+\frac{1}{4z}\log(z)-\frac{1}{4z}\ln(2 \pi)$ (2)

From (2), we can easily obtain the following relations

Lemma 1

$\frac{1}{\sinh \pi x}=\frac{2}{e^{\pi x}-1}-\frac{2}{e^{ 2 \pi x}-1}$ (4)

Proof:

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

From (4) and (3) we obtain

$\begin{aligned} \int_{0}^{\infty}\frac{\arctan(x)}{\sinh 2 \pi x}dx&=2\int_{0}^{\infty}\frac{\arctan(x)}{e^{2\pi x}-1}dx-2\int_{0}^{\infty}\frac{\arctan(x)}{e^{4\pi x}-1}dx\\ &=2\left(\frac12-\frac{\ln 2 \pi}{4}\right)-2\left(\frac12-\frac{3 \ln 2}{8}-\frac{\ln 2 \pi}{8}\right)\\ &=\frac{\ln 2}{2}-\frac{\ln \pi}{4} \qquad \blacksquare\\ \end{aligned}$

In the same fashion we obtain

$\begin{aligned} &\int_{0}^{\infty}\frac{\arctan(x)}{\sinh \pi x}dx=\frac{\ln \pi}{2}-\frac{\ln 2}{2}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{\sinh \frac{\pi x}{2}}dx=4 \ln \Gamma\left(\frac14\right)-2 \ln \pi-3 \ln 2 \end{aligned}$

Lemma 2:

$\frac{1}{e^x+1}=\frac{1}{e^{\pi x}-1}-\frac{2}{e^{ 2 \pi x}-1}$ (5)

From Lemma 2 and (3) we obtain:

$\begin{aligned} \int_{0}^{\infty}\frac{\arctan(x)}{e^{2 \pi x}+1}dx&=\int_{0}^{\infty}\frac{\arctan(x)}{e^{2\pi x}-1}dx-2\int_{0}^{\infty}\frac{\arctan(x)}{e^{4\pi x}-1}dx\\ &=\left(\frac12-\frac{\ln 2 \pi}{4}\right)-2\left(\frac12-\frac{3 \ln 2}{8}-\frac{\ln 2 \pi}{8}\right)\\ &=\frac{3}{4}\ln 2-\frac12 \qquad \blacksquare\\ \end{aligned}$

Similarly

$\begin{aligned} &\int_{0}^{\infty}\frac{\arctan(x)}{e^{ \pi x}+1}dx=\frac{\ln \pi}{2}-\frac{1}{2}\\ & \\ &\int_{0}^{\infty}\frac{\arctan(x)}{e^{ \pi x/2}+1}dx=2 \ln \Gamma\left(\frac14\right)-\frac12-\ln 2\pi \end{aligned}$

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