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Find the Fourier transform of $$f(t)=\frac1{1+t^2}$$ using contour integration that $$F\{f(t)\}=\int^\infty_{-\infty}\frac1{1+t^2}e^{2\pi ft}dt$$ How can I do this?

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  • $\begingroup$ Shouldn't the exponent be $2\pi ift$ instead of $2\pi ft$? $\endgroup$
    – Ruslan
    Dec 2, 2014 at 10:52
  • $\begingroup$ @Ruslan Yes, it should be $2\pi ift$ $\endgroup$ Dec 2, 2014 at 13:34

1 Answer 1

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First, I don't think you should have a function of $t$ given by $f(t)$ and then to use $f$ as your transformation variable. That is just going to cause problems.

Anyway, let

$$ g(t) = e^{-a\lvert t \rvert} $$

(you will see why we pick this function soon).

Then

$$\begin{align} \mathscr{F} (g(t)) & = \int_{-\infty}^{\infty} e^{-a \lvert t \rvert} e^{-2\pi ift} dt \\ & = \int_{-\infty}^{0} e^{at} e^{-2\pi ift} dt + \int_{0}^{\infty} e^{-at} e^{-2\pi ift} dt \\ & = \frac{e^{t(a-2\pi if)}}{a-2\pi if} \Bigg \rvert_{-\infty}^{0} + \frac{e^{-t(a+2\pi if)}}{-(a+2\pi if)} \Bigg \rvert_{0}^{\infty} \\ & = \frac{2a}{a^{2}+4\pi^{2}f^{2}} \end{align}$$

Using the inversion theorem

$$ \implies e^{-a\lvert t \rvert} = \int_{-\infty}^{\infty} \frac{2a}{a^{2}+4\pi^{2}f^{2}} e^{2\pi ift} df $$

Setting $a = 2 \pi$ puts the denominator of the integrand in the form we want

$$ \implies \pi e^{-2\pi \lvert t \rvert} = \int_{-\infty}^{\infty} \frac{1}{1+f^{2}} e^{2\pi ift} df $$

Let $t \mapsto -t$

$$\begin{align} \implies \pi e^{-2\pi \lvert -t \rvert} & = \pi e^{-2\pi \lvert t \rvert} \\ & = \int_{-\infty}^{\infty} \frac{1}{1+f^{2}} e^{-2\pi ift} df \end{align} $$

Therefore

$$ \mathscr{F}\Bigg(\frac{1}{1+t^{2}}\Bigg) = \pi e^{-2\pi \lvert f \rvert} $$

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