How can I demonstrate the following equality: $\int_0^\infty \frac{\cos(yx)}{y^{2}+\lambda^{2}}\, dy = \frac{\pi}{2\lambda}e^{-\lambda x}$

Question

I tried to solve with Fourier Transform. It is an exercise that appears in the context of Fourier Transforms.

Answer 1

Let $f(x) = \mathrm{e}^{-\lambda|x|}$ with $\lambda>0$. Warning, I add an absolute value and I assume $\lambda$ positive.

The Fourier transform is $F(y) = \int_{-\infty}^{+\infty}f(x)\mathrm{e}^{-\mathrm{i}xy}\mathrm{d}x = \frac{2\lambda}{y^2+\lambda^2}$.

Thus, $f(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}F(y)\mathrm{e}^{+\mathrm{i}xy}\mathrm{d}y = \frac{1}{\pi}\int_{0}^{+\infty}F(y)\cos(xy)\mathrm{d}y$ because $F$ is even.

$$ \frac{\pi}{2\lambda}\mathrm{e}^{-\lambda|x|} = \int_{0}^{+\infty}\frac{\cos(xy)}{y^2+\lambda^2}\mathrm{d}y $$