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Find the Fourier transform of the following function: $$ \frac{x}{x^2+a^2} $$

Attempt: We want to solve $$ \hat{F}(k) = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2}.$$ To solve for this real integral, I consider a semicircle contour. The origin is $z = 0$, and the half-circle will enclose the upper half plane. Call the total contour $C$ and the arc of the circle $C_R$. Then $$\oint_C \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2} + \oint_{C_R} \frac{z e^{-ikz}}{z^2+a^2}. $$ I claim that the second integral on the righthand side goes to zero for large $R$. This is because

If on a circular arc $C_R$ of radius R and center $z=0$ , $zf(z)\rightarrow 0$ uniformly as R → ∞, then $$ \oint_{C_R} f(z) = 0 $$

for us $$ f(z) =\frac{z e^{-ikz}}{z^2+a^2} \Rightarrow zf(x) = \frac{z^2 e^{-ikz}}{z^2+a^2} $$ which should converge ($z^2/z^2$ approach 1, but the decaying exponential should drop the expression down to zero for large $R$). Thus $$\oint_C \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2}$$ which implies $$ \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2} = 2π i \,\,\, \text{Res}(f;ia) = π i e^{ak} = \text{The WRONG answer}$$

can someone help? I feel like my issue is with the $C_R$ integral…

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    $\begingroup$ If $k>0$ then the contour must be closed in the lower half plane, while for $k<0$ it must be closed in the upper half plane. $\endgroup$ Apr 24, 2018 at 21:41
  • $\begingroup$ Isn't $k$ assume to be $>0$ in Fourier transforms? $\endgroup$
    – talrefae
    Apr 24, 2018 at 21:46
  • $\begingroup$ Oh wait I think I see where youre going with this. So if $k<0$, can I use Jordan's lemma to show that $\oint_{C_R} \rightarrow 0$? I'd also need to calculate the residue of $-ia$ rather than $ia$… $\endgroup$
    – talrefae
    Apr 24, 2018 at 23:01

1 Answer 1

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I got $π𝑖𝑒^{-𝑎𝑘}$. We only integrate in the upper half plane, the only pole there is $+ia$. Using the formula, gives you...

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