The following integral appeared on the $8$th Open Mathematical Olympiad of the Belarusian-Russian University. $$I=\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$$ I used power series: $$x-\sin x = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}\rightarrow I=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n+1)!} \int_0^\infty \frac{x^{2n-2}}{x^2+4}dx$$ Taking the inner integral and substituting $\displaystyle{x^2=4t \rightarrow dx=\frac{dt}{\sqrt{t}}}$ gives: $$\int_0^\infty \frac{x^{2n-2}}{x^2+4}dx=4^{n-2}\int_0^\infty \frac{t^{n-1-\frac12}}{t+1}dt$$ $$=4^{n-2} B\left(n-\frac12, 1-n+\frac12\right)=4^{n-2}\Gamma\left(n-\frac12\right)\Gamma\left(1+\frac12-n\right)$$ And using Euler's reflection formula: $$\Gamma\left(n-\frac12\right)\Gamma\left(1+\frac12-n\right)=\pi \csc\left({n\pi-\frac{\pi}{2}}\right)=-\pi\sec(n\pi)=(-1)^{n+1}\pi$$ $$I=\pi\sum_{n=1}^\infty \frac{4^{n-2}}{(2n+1)!}=\frac{\pi}{32} \sum_{n=1}^\infty \frac{2^{2n+1}}{(2n+1)!}=\frac{\pi}{32}(\sinh 2 -1)$$ I did not found the official solution, but the answer given $\displaystyle{\frac{\pi}{32}\left(\frac{e^2-1}{e^2}\right)},\,$ doesn't match. Can you help me find my mistake? And maybe share some different methods to solve this integral?

  • $\begingroup$ Can residue calculus be used? $\endgroup$ – MisterRiemann Oct 28 '18 at 10:59
  • $\begingroup$ I'm pretty bad at using contour integration, but sure if you can do it feel free to share, I will try to understand. $\endgroup$ – Zacky Oct 28 '18 at 11:00
  • 1
    $\begingroup$ Residues seems like the most simple way, especially looking at the answer $\endgroup$ – Yuriy S Oct 28 '18 at 11:11
  • 2
    $\begingroup$ The Laplace transform is another viable way. $\mathcal{L}(x-\sin x)=\frac{1}{s^2}-\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^3(x^2+4)}\right)=\frac{s^2-\sin^2(s)}{8}$ reduce the original problem to the evaluation of $\int_{\mathbb{R}}\frac{\sin^2(s)\,ds}{s^2(1+s^2)}$. $\endgroup$ – Jack D'Aurizio Oct 28 '18 at 12:08
  • 1
    $\begingroup$ Wow, this is also nice! And since $$\int_0^\infty \frac{\sin^2 x}{x^2}dx=\frac{\pi}{2}$$ This cancels out with $$\int_0^\infty \frac{x^2}{x^2(x^2+1)}dx$$ And the original integral equals to $$\frac{1}{8}\int_0^\infty \frac{\sin^2 x}{1+x^2}dx$$ And now the easiest way to evaluate it I think is to rewrite $2\sin^2 x = 1-\cos(2x).\,$Thank you! $\endgroup$ – Zacky Oct 28 '18 at 12:16

Let $J(a)=\displaystyle\int_{0}^{\infty}\frac{ax-\sin ax}{x^3(x^2+1)}dx$. Differentiating by $a$ three times (which is admissible under the integral sign, because the resulting integrals converge uniformly for $a$ in any finite interval), we get $$J'''(a)=\displaystyle\int_{0}^{\infty}\frac{\cos ax}{x^2+1}dx=\frac{\pi}{2}e^{-|a|}.$$ With $J(0)=J'(0)=J''(0)=0$ this gives $J(a)=\dfrac{\pi}{2}\Big(1-|a|+\dfrac{a^2}{2}-e^{-|a|}\Big)\operatorname{sgn}(a)$.

The answer is $\dfrac{J(2)}{16}=\dfrac{\pi}{32}(1-e^{-2})$ as expected.

  • $\begingroup$ (I'm giving this as an alternative to the contour integration method.) $\endgroup$ – metamorphy Oct 28 '18 at 11:30
  • $\begingroup$ This is beautiful, thank you! $\endgroup$ – Zacky Oct 28 '18 at 11:35
  • 1
    $\begingroup$ You still need to prove the integral for the third derivative, which is most easily done by contour integration. But +1 nonetheless $\endgroup$ – Yuriy S Oct 28 '18 at 11:48

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.