# Evaluate $\int_0^\infty \frac{1}{(x+1)(\pi^2+\ln(x)^2)}dx$

I've been given this integral, and I have no idea how to evaluate it. It has a really nice answer of $\frac{1}{2}$ but I have no idea. I tried converting it into a series. I tried differentiating under the integral. I can't find a way of using complex numbers. The only thing I can think of is complex analysis which I don't know. Is there a way to solve it without complex analysis? If not, I wouldn't mind that as a solution.

Here's the integral:

$$\int_0^\infty \frac{1}{(x+1)(\pi^2+\ln(x)^2)}dx$$

• Substitute $x=\frac1y$ and add what you get with the initial integral, then simplify and let $\ln x =t$ – Zacky Aug 22 '18 at 23:30
• I was thinking of substituting $u = \ln(x)$, but Zacky might be onto something. – 4-ier Aug 22 '18 at 23:31
• Mathematica yields $1/2$. – David G. Stork Aug 22 '18 at 23:31

## 1 Answer

$$\int_{0}^{+\infty}\frac{dx}{(x+1)(\pi^2+\log^2 x)}=\int_{0}^{1}\frac{dx}{(x+1)(\pi^2+\log^2 x)}+\int_{0}^{1}\frac{dx}{x(x+1)(\pi^2+\log^2 x)}$$ equals $$\int_{0}^{1}\frac{dx}{x(\pi^2+\log^2 x)}\stackrel{x\mapsto e^t}{=}\int_{-\infty}^{0}\frac{dt}{\pi^2+t^2}=\int_{0}^{+\infty}\frac{du}{\pi^2+u^2}=\left[\frac{\arctan(u/\pi)}{\pi}\right]_{0}^{+\infty}=\frac{1}{2}.$$

An overkill is to exploit the integral representation of Gregory coefficients.

• Nice mentioning Gregory coefficients. Have any idea about $$\int_{0}^{\infty}\frac{dx}{(x^2+1)(\pi^2+\log^2 x)}$$ Or possibly a generalization? – Zacky Aug 22 '18 at 23:39
• @Zacky: with the same trick as above it reduces to $\int_{0}^{+\infty}\frac{dx}{(\pi^2+x^2)\cosh(x)}$, which by the Fourier transform equals $\frac{1}{2}\int_{0}^{+\infty}\frac{ds}{e^{\pi s}\cosh(\pi s/2)}$ or $\frac{2}{\pi}-\frac{1}{2}$ by letting $s=\frac{2}{\pi}\log u$. – Jack D'Aurizio Aug 22 '18 at 23:55
• @Zacky: $$\int_{-\infty}^{+\infty}f(x)g(x)\,dx = \int_{-\infty}^{+\infty}(\mathscr{F} f)(s)(\mathscr{F}^{-1}g)(s)\,ds.$$ $\frac{1}{\cosh}$ is essentially a fixed point for $\mathscr{F}$ while the Cauchy distribution and the Laplace distribution are conjugated. – Jack D'Aurizio Aug 23 '18 at 0:03