Integral involving Gamma Function

Question

I am solving the following integral:

$$ \int_{-1}^{K} u^B e^{-u} du $$

The solution of the integral is a lower incomplete Gamma Function if -1 is replaced with 0. Can anybody help me in solving the integral so that the solution involves an incomplete gamma function (either upper or lower)?

Regards

Answer 1

I believe this will work.

$$I=\int_{-1}^K u^B e^{-u}du=\int_{-1}^{\infty}u^B e^{-u}du-\int_K^{\infty}u^B e^{-u}du $$

Then,

$$\Gamma(z,x)=\int_x^{\infty}u^{z-1} e^{-u}du $$

So our integrals are...

$$ I=\Gamma\left(B+1,-1\right)-\Gamma\left(B+1,K\right) $$

The incomplete gamma function for with a negative bound seems to behave ok. I did a little research and found the "generalized incomplete gamma function." Which is,

$$\Gamma(z,x_1,x_2) =\int_{x_1}^{x_2}u^{z-1} e^{-u}du $$

This is defined for complex $a$, $x_1$, and $x_2$. Full details on branch cuts and such can be found here: http://functions.wolfram.com/GammaBetaErf/Gamma3/04/05/03/