Solving integral equations by $L$ and $L^{-1}$ operators

Abstract:

\begin{equation}\tag {$1$}g(u) = \int _0^\infty {k(ux)f(x)dx = \frac {1}{{2\pi i}}\int _C^{} {K(s)F(1 - s){u^{ - s}}ds,} } \end{equation} where $g(u)$ and $k(u)$ are known and $f(x)$ is to be found. $K(s)$ is the Mellin transform of $k(x)$ and $F(s)$ of $f(x)$; hence the second equality. L and ${L^{ - 1}}$ denote the Laplace transform and its inverse. If \begin{equation}\tag {$2$}{{K(s) = \prod \limits _{i = 1}^n {\Gamma ({\alpha _i}s + {\beta _i})} } {\prod \limits _{j = 1}^m {\Gamma ({\alpha _j}s + {\beta _j})} }}\end{equation} then I show that a suitable combination of L and ${L^{ - 1}}$ operators, applied to (1), can eliminate $K(s)$ from the second integrand. This leaves $F(1 - s)$ standing free and the Mellin transform then obtains $f(x)$ from $F(1 - s)$. This solution needs tables of Laplace transforms only. When (2) does not hold, an L and ${L^{ - 1}}$ combination may turn (1) into an integral equation whose solution is already known.