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I'm working through Big Rudin's (Real and Complex Analysis) Fourier Transform chapter, and the following complex integral is part of a discussion on the Inverse Transform that Rudin mentions briefly without resolving, and I'm very curious how this is done.

$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt$

What I remember from my Complex Analysis days, is to take the limit of the integral evaluated around a contour from -R to R, and then around $C_R$, the semicircle connecting -R and R. Since this is closed I can use the Cauchy Residue Theorem, and by Jordan's Lemma hopefully the integral around $C_R$ will vanish, so I'm left with

$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt=2\pi*\{\text{sum of residues}\}$

However, this integral has no residues, so simply taking the limit of this integral from -R to R should work, but my rusty integration techniques aren't helping and Mathematica is not giving me any results.

Any help in solving this is much appreciated!

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  • $\begingroup$ Split the integral between $[-\infty,0[$ and $[0,+\infty[$. $\endgroup$
    – C. Falcon
    Commented Dec 1, 2015 at 17:56
  • $\begingroup$ Your function is not holomorphic so why are you talking about residues... this is just a real analysis exercise. $\endgroup$
    – mercio
    Commented Dec 1, 2015 at 18:06
  • $\begingroup$ @mercio Thx for your comment, I'm editing it now that I see it better. I began asking this as a complex question because it is part of Rudin's Complex analysis section in a chapter on the Fourier Transform, and this integral is inside another integral that is complex, so I got carried away $\endgroup$
    – Mike
    Commented Dec 1, 2015 at 18:09

2 Answers 2

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Let $F(x)$ be the function defined by the Fourier Integral

$$F(x)=\int_{-\infty}^\infty e^{-|\lambda t|}e^{itx}\,dt$$

Since the integral converges for $\lambda \ne 0$, we can write the integral as

$$F(x)=\lim_{R\to \infty}\int_{-R}^R e^{-|\lambda t|}e^{itx}\,dt$$

One might be tempted to evaluate this integral be closing the real line contour with a semi-circle, centered at the origin and with radius $R$, in the upper-half or lower-half plane and then using the Residue Theorem. However, this approach is fatally flawed since $e^{-|\lambda||z|}$ is not an Analytic Function. Therefore, we proceed along a different way forward.

First, we use Euler's Formula and exploit even and odd components of the integrand to reveal

$$F(x)=2\int_0^\infty e^{-|\lambda|t}\cos (xt)\,dt$$

Then, Integrating by Parts twice yields

$$\begin{align} F(x)&=\lim_{R\to \infty}\left.\left(\frac{e^{-|\lambda|t}\left(-|\lambda|\cos (xt)+x\sin(x t)\right)}{x^2+|\lambda|^2}\right)\right|_{t=0}^{t=R}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{|\lambda|}{x^2+|\lambda|^2}} \end{align}$$

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  • $\begingroup$ thanks a lot, this is really insightful. I've never used Euler's Identity I'll read up on it! $\endgroup$
    – Mike
    Commented Dec 1, 2015 at 18:31
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    $\begingroup$ ...which yields a delta distributionin the limit $\delta \rightarrow 0$. This is one (not totally rigorous) way to prove that the FT of 1 is the delta distribution. $\endgroup$
    – tired
    Commented Dec 1, 2015 at 18:40
  • $\begingroup$ @tired Excellent Point. The result is one regularized form for the Dirac Delta. Does Rudin's Book discuss tempered distributions? $\endgroup$
    – Mark Viola
    Commented Dec 1, 2015 at 18:47
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    $\begingroup$ @Dr.MV You have somewhere lost a factor $2$, the correct result is $\frac{2\lvert \lambda\rvert}{x^2 + \lvert\lambda\rvert^2}$. $\endgroup$ Commented Dec 1, 2015 at 19:05
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    $\begingroup$ @Craig And hence one obtains $h_{\lambda}(x) = \sqrt{\frac{2}{\pi}}\cdot \frac{\lvert \lambda\rvert}{x^2 + \lvert\lambda\rvert^2}$, which nicely yields $$\int_{\mathbb{R}} h_{\lambda}(x)\, dm(x) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} h_{\lambda}(x)\,dx = 1.$$ $\endgroup$ Commented Dec 1, 2015 at 19:08
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Use is made of even function and Laplace transform: \begin{align} F(x)&=\int_{-\infty}^\infty e^{-|\lambda t|}e^{itx}\,dt\\ &\overset{s=|\lambda|}=2\int_{0}^\infty e^{-s t}\cos(tx)\,dt\\ &=\frac{2s}{s^2+x^2}. \end{align}

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