$$I=\int_{-\infty}^{+\infty}f(x)\,dx=\int_{-\infty}^{+\infty} \frac{\cos(x)}{\exp(x^2)}\,dx$$
I tried using the Limit comparison for this one:
$$f(x)\leq g(x) = e^{-x^2}$$
Now I can take the integral of $g(x)$. But my problem are the limits $-\infty$ and $+\infty$. How am I supposed to solve this in order to show the convergence? If possible I want to calculate the real value for my integral, i.e. what it converges to.
On
Hint: Let $x=t^2$ and use Euler's formula in conjunction with your knowledge of Gaussian integrals, in order to arrive at the desired result, which has already been posted in the comment section. This will involve completing the square in the exponent, and then making a second simple substitution.
The integral is absolutely convergent by your bound, and:
$$ I = \int_{0}^{+\infty}\frac{\cos(\sqrt{x})}{\sqrt{x}}\,e^{-x}\,dx.\tag{1}$$ By expanding $$ \frac{\cos(\sqrt{x})}{\sqrt{x}}=\sum_{n=0}^{+\infty}\frac{(-1)^n x^{n-\frac{1}{2}}}{(2n)!}\tag{2}$$ and integrating termwise through: $$ \int_{0}^{+\infty} x^{n-\frac{1}{2}}e^{-x}\,dx = \Gamma\left(n+\frac{1}{2}\right)\tag{3}$$ we get, through the Legendre duplication formula: $$ \color{red}{I} = \sum_{n\geq 0}\frac{(-1)^n\, \Gamma\left(n+\frac{1}{2}\right)}{\Gamma(2n+1)}=\Gamma\left(\frac{1}{2}\right)\sum_{n\geq 0}\frac{(-1)^n }{4^n\,n!}=\color{red}{\sqrt{\pi}\,e^{-1/4}}\tag{5}$$ and along the same lines we may also prove: $$ \mathcal{L}\left(\frac{\cos\sqrt{x}}{\sqrt{x}}\right) = \sqrt{\frac{\pi}{s}} e^{-\frac{1}{4s}}.\tag{6}$$