Mellin transform method

Remarks

Step 1 rests on the identity $\mathcal{M}\left[\int_0^\infty k(x/t) \varphi(t) \, dt/t\right](s) = \mathcal{M}k(s) \, \mathcal{M}\varphi(s)$, which follows from the substitution $x = t u$.

Example: the first-kind equation $\int_0^\infty e^{-x/t} \varphi(t) \, \frac{dt}{t} = f(x)$ with $f(x) = \int_0^\infty e^{-x/t - t} \, \frac{dt}{t}$. Here $k(u) = e^{-u}$ and $\mathcal{M}k(s) = \Gamma(s)$ for $\operatorname{Re} s > 0$. By step 1 applied to $e^{-t}$, $\mathcal{M}f(s) = \Gamma(s)^2$, so step 2 gives $\mathcal{M}\varphi(s) = \Gamma(s)$ and step 3 gives $\varphi(t) = e^{-t}$.

A kernel $h(x, t)$ that is homogeneous of degree $-1$, that is $h(cx, ct) = c^{-1} h(x, t)$ for all $c > 0$, can be written $h(x, t) = \frac{1}{t} h\!\left(\frac{x}{t}, 1\right)$, so the method applies to $\int_0^\infty h(x, t) \varphi(t) \, dt$. For $h(x, t) = \frac{1}{x + t}$ one gets $k(u) = \frac{1}{1 + u}$ and $\mathcal{M}k(s) = \frac{\pi}{\sin \pi s}$ for $0 < \operatorname{Re} s < 1$.

Non-example: the kernel $e^{-(x - t)^2}$ on $(0, \infty)$ is not a function of $x / t$ times $\frac{1}{t}$, so the Mellin transform does not turn the equation into a product.

The substitution $x = e^{u}$, $t = e^{v}$ turns the equation into a convolution integral equation on $\R$, and the method into the Fourier transform method.