1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Substitute $x=\frac1y$ and add what you get with the initial integral, then simplify and let $\ln x =t$ $\endgroup$ – Zacky Aug 22 '18 at 23:30
  • $\begingroup$ I was thinking of substituting $u = \ln(x)$, but Zacky might be onto something. $\endgroup$ – 4-ier Aug 22 '18 at 23:31
  • $\begingroup$ Mathematica yields $1/2$. $\endgroup$ – David G. Stork Aug 22 '18 at 23:31
6
$\begingroup$

$$\int_{0}^{+\infty}\frac{dx}{(x+1)(\pi^2+\log^2 x)}=\int_{0}^{1}\frac{dx}{(x+1)(\pi^2+\log^2 x)}+\int_{0}^{1}\frac{dx}{x(x+1)(\pi^2+\log^2 x)}$$ 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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Zacky Aug 22 '18 at 23:39
  • 1
    $\begingroup$ @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 $s=\frac{2}{\pi}\log u$. $\endgroup$ – Jack D'Aurizio Aug 22 '18 at 23:55
  • 1
    $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Aug 23 '18 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.