8
$\begingroup$

Does anyone have any advice on how to evaluate the following integral?

$$\int_0^{\infty}\frac{2+7\mathrm{cos}(x^\pi-e)-7\mathrm{sin}(1+x^8)}{1+x^2} \mathrm{d}x$$

It looks like it converges, but I have no idea where to even begin evaluating it. Any tips would be appreciated, thanks.

$\endgroup$
  • 9
    $\begingroup$ In one word: WHY? $\endgroup$ – nbubis Jul 22 '13 at 1:56
  • 4
    $\begingroup$ What evidence do you have that this is evaluatable in any form but approximately? $\endgroup$ – Mariano Suárez-Álvarez Jul 22 '13 at 1:58
  • $\begingroup$ My best guess would be to use complex variable methods. Have you looked into that? $\endgroup$ – Suugaku Jul 22 '13 at 2:04
  • $\begingroup$ Using trig identities, your integral reduces to determining integrals of the form $\int_0^\infty \cos(x^p)/(1+x^2) dx$ and $\int_0^\infty \sin(x^p)/(1+x^2) dx$. These seem like reasonable integrals to try to compute, perhaps someone has done them before. $\endgroup$ – abnry Jul 22 '13 at 2:17
  • $\begingroup$ I started solving this insane integral. To solve the partial trigonometric integrals, I convert the common denominator into a geometric series, and the partial integrals can be solved. The results can be only be expressed in a Hypergeometric Function. $\endgroup$ – Arucard Nov 1 '13 at 0:06
3
$\begingroup$

Since the integrand is bounded above and below by: $$ \frac{16}{x^2+1} > f(x) > \frac{-14}{x^2+1}$$ Both which converge (t0 $8\pi$ and $-7\pi$ respectively), the integral converges to some value in between. Other than that, I'd bet against a closed form solution.

$\endgroup$
2
$\begingroup$

Not a fully answer, but hopefully useful information: for the cosine part, I noticed that $$\begin{aligned}\frac{\partial^2}{\partial x^2}I(a)&=\int_0^\infty\frac{\partial^2}{\partial x^2}\frac{\cos\left(x^\pi-a\right)}{1+x^2}\,dx\\ & = -\int_0^\infty \frac{\cos\left(x^\pi-a\right)}{1+x^2}\,dx\\ &=-I(a)\\\end{aligned}$$ and, then, our integral $I(e)$ $$I(e)=K\sin(e)$$ for some constant $K.$ One way to find $K$ is evaluating $I(\pi/2)$ or another convenient value. Numerical evidence suggests $K=-{\cos(1)+4\over \pi+1/3}$, but I don't know whether Wolfram is pointing a rational value or giving up calculations. I'd like someone to test it in a more powerful environment.

$\endgroup$

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.