H-function for the following integral

Question

I stumbled upon the integral $\int\limits_0^{+\infty} u^\nu\exp(-au-bu^\rho)du$, $\Re(a)>0,\,\,\Re(b)>0,\,\,\rho>0$. I cannot find any way to represent it using the Fox-H function. Any hints? PS: This is the Krätzel function, right? The one known as reaction rate integral.

Answer 1

Expand the factor $exp(-bu^\rho)$ into a Taylor series. You get:

$\int_0^\infty u^\nu exp(-au-bu^\rho)du=\sum_{i=0}^\infty \frac{1}{i!} (-b)^i \int_0^\infty u^{\nu+i \rho} exp(-au)du$. With the Substitution $v=au$ you get the following series of Gamma functions:

$\int_0^\infty u^\nu exp(-au-bu^\rho)du=\sum_{k=0}^\infty \frac{1}{k!} (-b)^k (\frac{1}{a})^{\nu + k \rho + 1} \Gamma(\nu + k \rho + 1)$.

It is possible to express this sum of Gamma functions by contour integrals. For example, may be $f(s)$ an arbitrary function, then you would have the following contour integral:

$\frac{1}{2 \pi i}\oint \Gamma(qs+r) f(s) ds = \sum_{k=1}^\infty Res(\Gamma(qs+r),\frac{-k-r}{q}) f(\frac{-k-r}{q})$.

Here you see what are the poles of the Gamma function. It holds that $Res(\Gamma(qs+r),\frac{-k-r}{q})$ is proportional to the factor $\frac{(-1)^k}{k!}$.

Now the H-function can be used. The coefficients $(-b)^k (\frac{1}{a})^{\nu + k \rho + 1}$ arise when a suitable $z$ is Chosen for the factor $z^{-s}$.