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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.

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  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Aug 28 at 12:45
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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.

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  • $\begingroup$ I was afraid to ask the math community because of this reason (they will just mention a lemma or theorem which I have no idea about and answers my question in one line). :) Jokes apart, trying to understand the application of the Riemann Lebesgue theorem in this case, how do you ensure the Lebesgue integral of |f| is finite in the range of integration just by looking at it? Is it because the function |f| is bounded everywhere and goes to zero at $\pm\infty$? $\endgroup$ – abhijit975 Aug 28 at 15:21
  • $\begingroup$ Removing $e^{2ika}$, the absolute value of the integral is bounded by constant $\times e^{-\beta k^2}$ with $\beta >0$. This function is evidently integrable (it is a Gaussian function), so it is integrable your function as well. It is simply unrealistic in physics to assume that all integral you will encounter in your career can be explicitly computed! $\endgroup$ – Valter Moretti Aug 28 at 15:49
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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.

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