Solving integral that contain exponential function and lower incomplete gamma function

Question

I have the following integral;

$$y=\int_0^\infty\frac{e^{-xf}}{m+x}\gamma(a,hx)~dx$$ where $f,m,h\in\mathbb{R}^+$ , $a\in\mathbb{N}$ , $\gamma\left(a,h x\right)$ is the lower incomplete gamma function

Can anyone help me how to solve it?

Thank you very much

Answer 1

Finding a closed form for this integral seems difficult.

Let us try to obtain an analytic expression.

We may start to write $$ \begin{align} y & =\int_{0}^{m}\frac{e^{- x f}}{m+x}\gamma\left(a,h x\right) dx +\int_{m}^{+\infty}\frac{e^{- x f}}{m+x}\gamma\left(a,h x\right) dx \\\\ & =\sum_{k=0}^{\infty} (-1)^k\left(\frac{1}{m^k}\!\!\int_{0}^{m}\!\!x^ke^{-f x}\gamma\left(a,h x\right) dx +m^k\!\!\int_{m}^{+\infty}\!\!x^{-k}e^{-f x}\gamma\left(a,h x\right) dx\right) \end{align} $$ and then inserting the series expansion you gave in your comment for $\gamma\left(a,h x\right)$.

We will end up with a double series.

Answer 2

Hint:

$\int_0^\infty\dfrac{e^{-xf}}{m+x}\gamma(a,hx)~dx$

$=\int_0^\infty\dfrac{e^{-fx}}{x+m}\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}}{n!(n+a)}dx$

$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}x^{n+a}e^{-fx}}{n!(n+a)(x+m)}dx$

$=\int_m^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}(x-m)^{n+a}e^{-f(x-m)}}{n!(n+a)x}d(x-m)$

$=\int_m^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nh^{n+a}e^{fm}(x-m)^{n+a}e^{-fx}}{n!(n+a)x}dx$

$=\int_m^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{n+a}\dfrac{(-1)^nh^{n+a}e^{fm}C_p^{n+a}(-1)^{n+a-p}m^{n+a-p}x^pe^{-fx}}{n!(n+a)x}dx$

$=\int_m^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^{n+a}\dfrac{(-1)^{p+a}(n+a-1)!h^{n+a}m^{n+a-p}e^{fm}x^{p-1}e^{-fx}}{n!p!(n+a-p)!}dx$