I have been asked to derive the Fourier Transform for $$f(x)=\frac{1}{x^2+a^2}$$ where $a>0$.
I know the Fourier Transform is equal to $$\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{-ikx}}{x^2+a^2}\;dx$$
By using a semicircular contour between $-L$ and $L$ on the x axis with a curve in the positive $x$ plane, I have found through the Residue Theorem that this integral is equal to $$\sqrt{\frac{\pi}{2}}\frac{e^{ka}}{a}$$
The answer given to me says however that it is in fact equal to $$\sqrt{\frac{\pi}{2}}\frac{e^{-|k|a}}{a}$$
Can someone please explain to me where I've gone wrong so as to miss the negative sign and also the modulus?
We define $$\hat{f}(k)=\mathcal F\{f(x)\}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i k x}\,\mathrm dx $$ and $$ f(x)=\frac{1}{x^2+a^2} $$ It's easy to find that $$\mathcal F\{\mathrm e^{-a|x|}\}=\sqrt{\frac{2}{\pi}}\frac{a}{k^2+a^2}$$ and using the duality property $\mathcal F\{\hat f(x)\}=f(-k)$, we have
$$\mathcal F\left\{\frac{1}{x^2+a^2}\right\}=\sqrt{\frac{\pi}{2}}\frac{\mathrm e^{-a|k|}}{a}$$