# Fourier transform problem using complex analysis?

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…

• 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. Apr 24, 2018 at 21:41
• Isn't $k$ assume to be $>0$ in Fourier transforms? Apr 24, 2018 at 21:46
• 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 $ia$… Apr 24, 2018 at 23:01

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