0
$\begingroup$

I'm doing this integral that showed up in a book in quantum physics and I just want to check if what I did was correct.

So this is the integral, which looks like it's a Fourier transform of something:

$$\int_{-\infty}^{\infty} \frac{e^{ikx}}{x^2 + a^2} \, dx$$

So using methods from complex analysis I separated the function into the form $\frac{e^{ikx}}{(k-ia)(k+ia)}$ which would immediately show that I would have poles at $k = ia$ and $k = -ia$. Calculating for residues I'd get:

$$\text{At the simple pole $c$:}$$ $$\begin{align}\text{Res$(f,c)$} & = lim_{k \to c}f(z); \quad f(k) = \frac{a(k)}{b(k)} \\ & = \frac{a(c)}{b'(c)} \\ & = \frac{e^{-ax}}{2ia}\,; \frac{e^{ax}}{-2ia}\end{align}$$

Then using the residue theorem: $$\oint_\gamma f(z) \, dz = 2 \pi i \, \text{Res$(f,c)$}$$

Assuming $x > 0$ and by using the pole $k = ia$ I'd get: $$\oint f(z) \, dz = 2\pi i \, \frac{e^{-ax}}{2ia} = \frac{\pi}{a} e^{-ax}$$ Then if $x < 0$ and the pole $k = -ia$: $$\oint f(z) \, dz = 2 \pi i \frac{e^{ax}}{-2ax} = -\frac{\pi}{a} e^{ax}$$

I'm having second thoughts on my answer since I'm asked to normalize the resulting function and each of these functions does not converge on $(-\infty,\infty)$ therefore I might have the resulting function wrong.

$\endgroup$
  • $\begingroup$ First of all, the residue theorem includes a sign coming from how $\gamma$ circles your poles. Choose a suitable contour to close the integral over $\mathbb{R}$ into a suitable closed path in $\mathbb{C}$ for which the integral converges, that is pick the upper or lower (infinite) semicircle depending on the sign of $x$. Secondly, your (currently wrong) result behaves nicely. If $x<0$ and $a > 0$, $e^{ax}=e^{-a\vert x \vert}$ does not blow up. $\endgroup$ – Dominik Feb 1 '16 at 9:04
  • $\begingroup$ @Dominik So just to be clear, you're saying that the result from this contour integral would not blow up if I integrate it from $-\infty$ to $\infty$ and that the only thing wrong here is that I did not pick a contour to use? $\endgroup$ – mopy Feb 1 '16 at 9:53
  • $\begingroup$ You did not integrate along a closed path in what you wrote down, which you need to do in order to apply the residue theorem. You can do so by picking an infinite semicircle in the upper or lower plane, but be careful to pick the right one, for the path integral along the semicircle does only converge for certain k. $\endgroup$ – Dominik Feb 1 '16 at 10:04
  • $\begingroup$ @Dominik But I did, using the infinite semicircle you said, that's why there were different results from where $x > 0$ and $x < 0$ since the closed path would not enclose the other pole if I used the other one. $\endgroup$ – mopy Feb 1 '16 at 10:47
  • $\begingroup$ Sorry, I did not really see how you did get to the stated results. Note that the path that encircles the pole in $-ia$ does so in a mathematically negative orientation (clockwise), which has to be taken into account in the residue theorem (gives you a sign). $\endgroup$ – Dominik Feb 1 '16 at 10:53
1
$\begingroup$

Half Hint

Let $\gamma$ be the path along the real axis then circling back counter-clockwise through the upper half-plane, letting the circle get infinitely big.

$$ \begin{align} \left(\int_{-\infty}^\infty\frac{e^{ikx}}{x^2+a^2}\mathrm{d}x\right) &=\int_{\gamma}\frac{e^{ikx}}{x^2+a^2}\mathrm{d}x\\\\ &=2\pi i\,\mathrm{Res}\left(\frac{e^{ikx}}{x^2+a^2},ia\right)\\\\ &=2\pi i\,\lim_{z\to ia}\frac{e^{ikz}}{z+ia}\\\\ &=2\pi i\,\frac{e^{ik(ia)}}{2ia}\\\\ &=\frac{\pi}{a}e^{-ka} \end{align} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ what sorcery is this? residue? $\endgroup$ – user4951 Feb 1 '16 at 11:58
  • $\begingroup$ This only holds for $k \geq 0$. $\endgroup$ – Dominik Feb 1 '16 at 12:02
  • $\begingroup$ @JimThio Aye, the most pure application of Residue Calculus! $\endgroup$ – Mycroft Feb 1 '16 at 12:02
  • $\begingroup$ @Dominik This is why it's a "half hint" xD $\endgroup$ – Mycroft Feb 1 '16 at 12:04
  • 1
    $\begingroup$ When would you use the lower half-plane and why? $\endgroup$ – Ron Gordon Feb 1 '16 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.