I am reading an essay that says that it is true that

\begin{equation} \frac{1}{2\pi i} \int^\infty_{-\infty} \frac{e^{ixy}}{y - i} \, dy = \begin{cases} e^{-x} & \text{for x }>0 \\ 0 & \text{for x} < 0 \end{cases} \end{equation}

I would like to know how to compute this integral, can anyone help please ? Many thanks!

  • 1
    $\begingroup$ Did you transcribe the formula correctly or are there some missing $2\pi$ factors? Because $e^{-x}\mathbf 1_{(0,\infty}$ and $\frac{1}{1+iy}$ are a Fourier transform pair. $\endgroup$ – Dilip Sarwate Jan 11 '12 at 22:16
  • $\begingroup$ of course you are right! i missed the factor, thanks for pointing this out! $\endgroup$ – harlekin Jan 11 '12 at 22:23

You need knowledge of complex analysis. First, you should locate singularities for the function you are integrating. Since ${e^{ixy}}$ is entire, only singularity is at $y = i$. Second, you need Residue theorem http://en.wikipedia.org/wiki/Residue_theorem. The third step is closing the contour of integration, as is described here http://en.wikipedia.org/wiki/Residue_theorem#Example. The fourth step is showing that integral by arc is going to 0 as we limit the radius of contour to infinity.

For your problem, it would go like this:

$$\displaystyle \frac{1} {{2\pi i}}\int_{ - a}^a {\frac{{{e^{ixy}}}} {{y - i}}dy} + \frac{1} {{2\pi i}}\int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} = \frac{1} {{2\pi i}}\int_C {\frac{{{e^{ixy}}}} {{y - i}}dy} = {\text{res}}\left( {\frac{{{e^{ixy}}}} {{y - i}},i} \right) = {e^{ - x}} \text{ for }a > 1$$

Now, we will need Jordan's lemma http://en.wikipedia.org/wiki/Jordan%27s_lemma to obtain $$\displaystyle \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} } \right| \leqslant \frac{\pi } {x}\frac{1} {{a - 1}}$$ (point on imaginary line is closest to $i$ so the maximum of $\displaystyle \frac{1} {{\left| {y - i} \right|}}$ is obtained in that point) from which it follows that

$$\displaystyle \int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} \leqslant \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} } \right| \leqslant \frac{\pi } {x}\frac{1} {{a - 1}}$$ and

$$\displaystyle \mathop {\lim }\limits_{a \to + \infty } \frac{\pi } {x}\frac{1} {{a - 1}} = 0 \Rightarrow \mathop {\lim }\limits_{a \to + \infty } \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} } \right| = 0 \Rightarrow \mathop {\lim }\limits_{a \to + \infty } \int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} = 0$$

So, $$\displaystyle {e^{ - x}} = \mathop {\lim }\limits_{a \to + \infty } \left( {\int_{ - a}^a {\frac{{{e^{ixy}}}} {{y - i}}dy} + \int_{{\text{arc}}} {\frac{{{e^{ixy}}}} {{y - i}}dy} } \right) = \int_{ - \infty }^\infty {\frac{{{e^{ixy}}}} {{y - i}}dy} $$

For $x < 0$ everything is the same, except that part of contour on a real line has opposite orientation, so we close the contour with an arc on a lower half-plane. Since our function doesn't have any singularities in the lower half plane, sum of residues in that area is 0, and so is the value of the integral in that case.

  • $\begingroup$ that's great, many thanks! the only thing that I don't yet understand is why the integral evaluates to $0$ if $x < 0$, could you maybe give a hint? $\endgroup$ – harlekin Jan 11 '12 at 22:43
  • $\begingroup$ aahh of course, now I see how the argument works for the case $x < 0$, your answer really helped a lot many thanks! $\endgroup$ – harlekin Jan 11 '12 at 23:11

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