Integral evaluation I am reading this paper where I have encountered the following integral:
$$ I_2 = \lim_{a\rightarrow \infty}\int_0^\infty e^{-\beta k^2} \frac{\cos(2ka)}{\kappa^2 + k^2}dk.$$
where $\beta>0$ is the inverse temperature, and $\kappa^2$ is also a positive number. The limit $a\rightarrow\infty$ is to take care of an ultraviolet divergence that was done in a previous step. Now the paper states that the above integral is zero. The problem is Mathematica cannot solve it and I am not sure what I did is correct. I proceeded as following:
$$ I_2 = \frac{1}{2}\lim_{a\rightarrow \infty}\int_{-\infty}^\infty e^{-\beta k^2} \frac{e^{2ika}}{\kappa^2 + k^2}dk $$
which follows from the integrand being an even function. Now if we use Jordon's Lemma, and close the contour in the upper-half plane ($a>0$), we get 
$$I_2 =  \frac{\pi}{2\kappa}\lim_{a\rightarrow \infty} e^{-2\kappa a}e^{\beta \kappa^2} =0$$ 
My question is whether what I did is correct and if one could apply Jordon's Lemma in this case.
I should probably ask this in the MathStackExchange but since I encountered the integration from a physics point of view and being a physics student myself I think it would be more legible to me if someone from the physics community answers it. 
 A: The fact that $I_2$, in the second form, vanishes is nothing but a direct application of  the so-called Riemann-Lebesgue lemma. No computations are necessary.
A: This is just complex analysis, in or out of context, so math.SE would've been a better choice, but let me try first:
Following from Wikipedia, the only condition to use the Jordan's lemma is if your function is continuous on a semicircular contour ($z=re^{i\theta}:r\in\mathbb{R}^+,\theta\in[0,\pi]$). Since there are 2 poles at $k = \pm i\kappa$ we also need $r\ne|\kappa|$. We want to see what happens as $r\to\infty$ so the last statement does not constitute a problem in the limit, and we can use Jordan's lemma without a problem.
But, Jordan's lemma uses maximum value of a function on a contour. And frankly I did not check the math, but you need to separate your function $f(z)$ (all $k$'s transform into $z$'s when you are dealing with contours):
$$f(z) = g(z)\cdot e^{iaz}$$
$$M_r := \max_{\theta\in[0,\pi]}\left|g(re^{i\theta})\right|$$
$$f(z) = \frac{1}{2}\ e^{-\beta z^2} \frac{e^{2iza}}{\kappa^2 + z^2} = \frac{e^{-\beta z^2}}{2\,(\kappa^2 + z^2)}\cdot e^{2iza}$$
$$g(z) := \frac{e^{-\beta z^2}}{2\,(\kappa^2 + z^2)} \therefore\ g(re^{i\theta}) = \frac{e^{-\beta (re^{i\theta})^2}}{2\,(\kappa^2 + (re^{i\theta})^2)}$$
$$|g(re^{i\theta})| = \sqrt{\frac{e^{-\beta r^2e^{i2\theta}}}{2\,(\kappa^2 + r^2e^{i2\theta})}\cdot{\frac{e^{-\beta r^2e^{-i2\theta}}}{2\,(\kappa^2 + r^2e^{-i2\theta})}}}$$
$$=\sqrt\frac{e^{-2\beta r^2 \cos{2\theta}}/4}{\kappa^4+r^4+2\kappa^2r^2\cos(2\theta)}$$
If you noticed, $\displaystyle{\lim_{r\to\infty}} |g(re^{i\theta})| = 0$ for all $\theta$ values, from which follows that the integral itself is zero.
