This integral was also posted recently on other sites like AoPS, but since it didnt get an answer I thought no one minds if I post this here too. $$I=\int_{0}^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}$$ I have no successful attempt that lead actually to something relevant, but I would love to see a closed form.


Of course it has a nice closed form. Since $$ \frac{1}{5-4\cos(2x)} = \frac{1}{3}+\frac{2}{3}\sum_{n\geq 1}\frac{\cos(2nx)}{2^n}\tag{1} $$ and $$ \int_{0}^{+\infty}\frac{\cos(2nx)}{1+x^2}\,dx = \frac{\pi}{2}e^{-2n}\tag{2}$$ we have $$ \int_{0}^{+\infty}\frac{dx}{(1+x^2)(5-4\cos(2x))}=\frac{\pi}{6}+\frac{2}{3}\sum_{n\geq 1}\frac{\pi}{2^{n+1}}e^{-2n}=\color{red}{\frac{\pi}{6}\cdot\frac{2e^2+1}{2e^2-1}}.\tag{3} $$ The same technique can be used for computing $ \int_{0}^{+\infty}\frac{dx}{(1+x^2)^m (5-4\cos(2x))^n}$, too.

| cite | improve this answer | |
  • 1
    $\begingroup$ You are correct about the possible generalization and the last series is just a geometric series. $\endgroup$ – Jack D'Aurizio Apr 20 '18 at 8:12

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.