Problem Find the Fourier transformation of $u(x) = \frac{1}{1+x^2}$

I want $\int_\mathbb R e^{-itx} \frac{1}{1+x^2} dx$. Let $f(z) = e^{-itz} \frac{1}{1+z^2}$, $z \in \mathbb C$, let's integrate this over the semi-circunference of radius $R$ and the line $[-R, R]$.

First, the integral on the circumference vanishes (with the change $z = Re^{i\theta}$)

$$\left|\int_{\gamma_R} \frac{e^{-izt}}{z^2+1}dz\right| = \left|\int_0^\pi \frac{e^{-itRe^{i\theta}}}{R^2e^{2i\theta}+1}Rie^{i\theta}d\theta\right| \le \int_0^\pi \frac{R}{|R^2e^{2i\theta}+1|}d\theta \le \int_0^\pi\frac1R d\theta= \frac\pi R$$ That goes to $0$ as $R \to \infty$.

The integral on the whole path is $2\pi i \text{ Res }(f, i)$, and since the integral on the circumference vanishes we have

$$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = 2 \pi i \text{ Res }(f, i)$$

I know that

If $z_0 \neq \infty$ is a pole of order $m$ for $f$, then set $g(z) = (z-z_0)^mf(z)$ and $\displaystyle \text{ Res }(f, z_0) = \frac{g^{(m-1)}(z_0)}{(m-1)!}$

Since $z_0 = i$ is a pole of order $1$, I set $\displaystyle g(z) = \frac{e^{-izt}}{z+i}$ and the residue should be $\displaystyle g(i) = \frac{e^t}{2i}$

And the integral then is $$\int_\mathbb R \frac{e^{-ixt}}{1+x^2} = \pi e^t$$

I know this is wrong (the fourier transform isn't even bounded) but I don't know where the error is. I am pretty sure the residue is wrong, but why?

  • $\begingroup$ The residue is right. The integral over the semicircle only tends to $0$ if $t$ satisfies a certain condition. If you know what the result should be, you probably can identify the condition, and fix your mistake/treat the remaining case. $\endgroup$ – Daniel Fischer Nov 18 '14 at 20:27

The mistake is your answer to the question

What is $$\bigl\lvert e^{-izt}\bigr\rvert\,?$$

Answer that correctly, and you will see what goes on.

We have $\lvert e^{-izt}\rvert = e^{t\operatorname{Im} z}$, so the exponential factor is bounded only if $t\operatorname{Im} z \leqslant 0$.

Thus we can only use the semicircle of radius $R$ in the upper half-plane to close the contour and evaluate the integral over the real line by the residue theorem if $t \leqslant 0$. In that case, it indeed evaluates to $\pi e^t$.

For $t > 0$, we can argue with the parity, $\frac{1}{1+x^2}$ is an even function, so

$$\int_{-\infty}^\infty \frac{\sin (xt)}{1+x^2}\,dx = 0$$

for all $t\in \mathbb{R}$ and therefore

$$\int_{-\infty}^\infty \frac{e^{-ixt}}{1+x^2}\,dx = \int_{-\infty}^\infty \frac{e^{ixt}}{1+x^2}\,dx,$$

so the integral is $\pi e^{-t}$ for $t > 0$, altogether $\pi e^{-\lvert t\rvert}$. Alternatively, one can close the contour with a semicircle in the lower half-plane for $t > 0$, and the residue theorem then yields

$$\int_{-\infty}^\infty \frac{e^{-ixt}}{1+x^2}\,dx = -2\pi i \operatorname{Res}\left(\frac{e^{-izt}}{1+z^2}; -i\right) = -2\pi i \frac{e^{-t}}{-2i} = \pi e^{-t}.$$

  • $\begingroup$ Aah, I see. $|e^{-itRe^{i\theta}}| \neq 1$, it is instead $e^{tR\sin\theta}$. So if $t<0$ the inequality makes sense as the last term goes to $0$. If $t>0$ the last term diverges, so I should treat this case differently. If what I've just said makes any sense, can you offer an hint on how to do that? $\endgroup$ – Ant Nov 18 '14 at 20:38
  • $\begingroup$ It makes sense. More than that, it is correct. The two standard ways to deal with $t > 0$ are a) parity, $\frac{1}{1+x^2}$ is even, hence something useful follows about the Fourier transform, or b) for $t > 0$ you use a semicircle in the lower half-plane. Then note that since the contour is traversed in the negative sense, the factor is $-2\pi i$ and not $2\pi i$. $\endgroup$ – Daniel Fischer Nov 18 '14 at 20:42
  • $\begingroup$ Great. Using the semicircle on the lower half plane (and setting $z = -Re^{i\theta}$) I get once again that the integral on the semi circle vanishes; the residue (calculated in $-i$ this time) is $\frac{e^{-t}}{-2i}$, and putting together the result from $t>0$ and $t<0$ I would say that this Fourier transform equals $\pi e^{-|t|}$ (wolfram reports $\sqrt{\frac{\pi}{2}} e^{-|t|}$ but it uses a different definition.. Actually dividing my answer by $\sqrt{2\pi}$ fixes it, so I guess it is correct!) :D $\endgroup$ – Ant Nov 18 '14 at 20:57
  • $\begingroup$ Yes, it's correct. Wolfram uses $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-ixt}\,dx$$ for the Fourier transform (because that normalisation makes it an isometry in $L^2(\mathbb{R})$), so the results differ by a constant factor. $\endgroup$ – Daniel Fischer Nov 18 '14 at 21:03
  • $\begingroup$ Perfect. Thank you very much! You've been very clear! :-D $\endgroup$ – Ant Nov 18 '14 at 21:10

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