I want to solve an integral of the type:

$\int_c^d \frac{1}{x^s} \int_{x}^{\infty} t^{r-1} e^{-t} dt dx = \int_c^d \frac{1}{x^s} \Gamma(r, x) dx$

So, the inner integral is an incomplete Gamma function.

How do I best solve this? I tried to integrate the inner integral first, but I don't know, what is the integral of the incomplete Gamma function? Is there a closed form solution?

Is it better to use integrations by parts on the whole integral with $v'(x) = \frac{1}{x^s}$ and $u(x) = \int_{x}^{\infty} t^{r-1} e^{-t} dt$ ? In this case I'm a little struggling because I'm not sure about how to apply the fundamental theorem to the integral with the variable lower limit. My calculus skills are pretty rusty at the moment ;)

Thanks for the help!

Edit: Solution using integration by parts: \begin{align} \int_c^d \frac{1}{x^s} \int_{x}^{\infty} t^{r-1} e^{-t} dt dx &= \int_c^d \frac{1}{x^s} \Gamma(r, x) dx \\ & = \frac{x^{1-s}}{1-s} \Gamma(r,x) \Big|_c^d - \int_c^d \frac{x^{1-s}}{1-s} (-exp(-x)) x^{r-1} dx \\ & = \frac{x^{1-s}}{1-s} \Gamma(r,x) \Big|_c^d + \frac{1}{1-s} \int_c^d x^{r-s} exp(-x)dx \\ & = \frac{b^{1-s}}{1-s} \Gamma(r,b) - \frac{a^{1-s}}{1-s} \Gamma(r,a) + \frac{1}{1-s}\left(\Gamma(r-s+1, a) - \Gamma(r-s+1, b)\right) \end{align}

  • $\begingroup$ Thanks for the reply. I edited the solution. Could someone clarify why it holds: $ \frac{d}{dx} \Gamma(r,x) = - exp(-x)x^{r-1} $? $\endgroup$ – jenna Oct 27 '15 at 10:45

Integration by parts just gives: $$ \int \frac{\Gamma(r,x)}{x^s}\,dx = \frac{x^{1-s}}{1-s}\,\Gamma(r,x)-\frac{1}{1-s}\,\Gamma(r-s+1,x).$$


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