Let the Fourier transform of a function and its inverse be given as:
$$\mathcal{F}[f](\lambda) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{-i\lambda t} dt$$
$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \mathcal{F}[f](\lambda)e^{i\lambda t} d\lambda$$
I wish to compute the following integral for $a,b>0$:
$$I = \int_{-\infty}^{\infty} \frac{1}{(a^2+x^2)(b^2+x^2)} dx$$
I know that the fourier transform of the function $\psi_a:\mathbb{R}\rightarrow \mathbb{R}\quad \psi_a(t) = \frac{1}{2\pi}\frac{2a}{a^2+t^2}$ is $\mathcal{F}[\psi_a](\lambda) = \frac{1}{\sqrt{2\pi}}e^{-a|\lambda|}$. I have also shown using the convolution theorem that $\psi_{a+b} = \psi_a * \psi_b$, where $*$ denotes convolution.
The integral $I$ looks similar to $\psi_a\psi_b$ without the constants in the denominator. Can this somehow be used to solve the integral?