Inverse of an integral transform

Question

Suppose that in a certain domain of analyticity we're given a function $A(s)$ in terms of the integral : $$A(s)=\int_{0}^{\infty}\frac{a(t)}{t(t^{2}+s^{2})}dt$$ How can we recover $a(t)$?

Answer 1

If we set $b(t)=\frac{a(t)}{t^2}$, since: $$\frac{t}{t^2+s^2}=\int_{0}^{+\infty}\cos(s x)\,e^{-tx}dx,$$ we have: $$A(s) = \int_{0}^{+\infty}\int_{0}^{+\infty}b(t)\cos(sx)\,e^{-tx}\,dx\,dt,$$ $$A(s) = \int_{0}^{+\infty}B(x)\,\cos(sx)\,dx,\tag{1}$$ where $B$ is the Laplace transform of $b$. $(1)$ gives that $A$ is the Fourier cosine transform of B, so $a(t)$ is just $t^2$ times the inverse Laplace transform of the inverse Fourier cosine transform of $A$.