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.
On
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.
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.