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

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.

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In one word: WHY? –  nbubis Jul 22 '13 at 1:56
What evidence do you have that this is evaluatable in any form but approximately? –  Mariano Suárez-Alvarez Jul 22 '13 at 1:58
My best guess would be to use complex variable methods. Have you looked into that? –  Suugaku Jul 22 '13 at 2:04
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. –  nayrb Jul 22 '13 at 2:17
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. –  Arucard Nov 1 '13 at 0:06
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.
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.