I have this massive integral that I would just put into an integral calculator normally (not trying to do this out manually because of the complexity), but the existence of the gamma function within the integrand makes it impossible for me to enter on the integral calculator sites I know. This is the integral:
$$\int_0^∞ e^{tx} \cdot \frac{ba}{Γ(x)} \cdot x^{a-1} \cdot e^{-bx} dx$$
$t$, $a$, and $b$ are constants.
What is the solution to this integral?
From the reflection formula we have $$\frac{1}{\Gamma(x)} = \frac{1}{-2i\pi} \int_C (-z )^{-x} e^{-z}dz$$ So you want $$\int_0^\infty \frac{x^{a-1} e^{-bx}}{\Gamma(x)}dx=\frac{1}{-2i\pi}\int_C e^{-z}\int_0^\infty (-z)^{-x} x^{a-1} e^{-bx}dx dz \\= \frac{\Gamma(a)}{-2i\pi}\int_C e^{-z} (b+\log(-z))^{-a} dz $$ For $a,b$ real it is $$= \frac{\Gamma(a)}{-\pi}\Im(\int_0^\infty e^{-y} (b+\log y+i\pi)^{-a} dy)$$