I have to find the Fourier transform of the function $$f(x) = \frac{1}{\sqrt{2\pi}(x^2 + a^2)}$$ where $a > 0$.
I start out by checking if the function is differentiable. Since it is defined everywhere, and also its derivative is defined everywhere there are no problems.
Then I check if $ \int_{-\infty}^{\infty}|f(x)|dx < \infty.$ To see this we integrate:
$$\int_{-\infty}^{\infty}|f(x)|dx = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}(x^2 + a^2)}dx = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} \frac{1}{(x^2 + a^2)}dx = \sqrt{\frac{2}{\pi}}\arctan\left(\frac{x}{a}\right)\frac{1}{a} \Big|_0^\infty = \frac{1}{a}\sqrt{\frac{2}{\pi}}\left(\frac{\pi}{2} – 0\right) < \infty $$
and so we can compute the Fourier transformation using
$$\int_{-\infty}^{\infty}f(x)e^{iyx} dx = \int_{-\infty}^{\infty} \frac{e^{iyx}}{\sqrt{2\pi}(x^2 + a^2)}dx$$
I hope I didn't make a mistake in the first integral. For the second is there a trick to compute it because it seems so hard to calculate?
Best Answer
Hint
Use residu theorem on $$[-R,R]\cup\{Re^{i\theta }\mid \theta \in [0,\pi]\}.$$