I know this should be relatively simple, but I'm not getting the complete answer correct when I check with Wolframalpha. Here is my attempt.

Going straight from the definition, with $x,t,a \in \mathbb{R}$

$$\begin{align*} \mathcal{F}\left[\frac{1}{t^2+a^2}\right](x)&=\int_{-\infty}^{\infty}\frac{e^{-itx}}{t^2+a^2}dt\\ &=\int_{-\infty}^{\infty}\frac{e^{-itx}}{(t-ia)(t+ia)}dt \end{align*}$$

Now this is relatively easy with the Residue Theorem (or so I suppose - I think I might miss a step here). Taking the countour $\gamma_R$ to be the half-circle that lies in the upper-half of the complex plane and whose radius is $R$, we can show that the residue of the integrand at the pole at $ia$ (as you can clearly see above, $e^{-itx}$ is entire) multiplied by $2\pi i$ will give us the value of the improper integral above.

I have shown my calculations below. It seems that I am missing an absolute value sign in my answer, but I can't see where it comes into play, even when $x<0$. Could you point out my mistake(s)?


Replace all the $t$'s with $z$'s in the integral above.


Now we are in the complex domain, and this integral represents integrating the integrand over the real line, going from $-\infty$ to $\infty$. So we consider the semicircle contour $\gamma_R$.

If $R \rightarrow \infty$, integrating the part of the contour on the real line will give us our original integral. With that said, we apply the residue theorem. We know that integrating over the entire semicircle can be simplified by the residue theorem, and the only residue that we need to calculate is that at the pole $ia$ (which is inside the contour).

$$2 \pi i \textrm{Res} (f; ia) = \left. \frac{e^{-itx}}{t+ia}\right|_{t=ia} = \frac{\pi}{a}e^ax$$

But what I just got only applies to the entire contour $\gamma_R$, not the horizontal line in specific. I now show that it actually does apply to the horizontal line in specific (the real line), as the top part of this semicircle goes to $0$ as $R \rightarrow \infty$.

$$ \left| \int_{\textrm{top half}} \frac{e^{-izx}}{z^2+a^2}dz \right| \leq \pi R \left|\frac{1}{a^2-R^2}\right|, \, \rightarrow 0 \,\, \textrm{ as } R \rightarrow \infty $$

So the top-half doesn't matter when $R$ is sufficiently big.

This finally leaves us with


  • 3
    $\begingroup$ When you check that the integral of the top semi-circle goes to 0, don't forget that $z$ is a complex number! $e^{-izx}$ is not of modulus 1 but of modulus $e^{x Im(z)}$ so the contour you have to use depend on the sign of $x$ $\endgroup$
    – Mirajane
    Jan 12 '15 at 22:20
  • $\begingroup$ Thank you. I understand now. That was pretty quick. $\endgroup$ Jan 12 '15 at 22:24

Actually your conclusion that $e^{-izx}$ is small at the top of the semicircle only holds if $x$ is negative.

When $x$ is positive, the value of $e^{-izx}$ at the top of the semicircle where $z=Ri$ will be $e^{-i^2Rx}=e^{Rx}$ which is large. So for positive $x$ you need a semicircle in the lower half-plane.

For $x=0$ the exponential is $1$ everywhere, so you can use either contour because the $z^2+a^2$ denominator gets to dominate no matter which direction you go.

  • $\begingroup$ Oh! I see! Dear gosh what a fumble! $\endgroup$ Jan 12 '15 at 22:23

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