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