# What's the integration of a Gamma going from 0 to 1?

I am trying to understand how to integrate this:

$$\int_{0}^{1} x^{\alpha -1}e^{-x}dx$$

I came to a point where integration by parts is too long and maplesoft is giving me this:

exp(-1/2)*WhittakerM((1/2)*alpha, (1/2)alpha+1/2, 1)/(alpha(1+alpha))+exp(-1/2)*WhittakerM((1/2)*alpha+1, (1/2)*alpha+1/2, 1)/alpha

Is there a simple way to get the answer to this primitive ?

Sorry if I'm not specific enough Thx for the answers !

• The integral can be expressed in terms of the incomplete Gamma function, but I don't know if anything more can be said in general. – carmichael561 Nov 15 '16 at 4:47

Using the integral defintion of the confluent hypergeometric function $${}_{1}\mathrm{F}_{1}(a,b;x) = \frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)} \int\limits_{0}^{1} \mathrm{e}^{xt} t^{a-1} (1-t)^{b-a-1} dt$$ for $\mathrm{Re}\,b \gt \mathrm{Re}\,a \gt 0$

We have \begin{align} I &= \int\limits_{0}^{1} x^{\alpha - 1} \mathrm{e}^{-x} dx \\ &= \frac{\Gamma(\alpha)\Gamma(1)}{\Gamma(\alpha+1)} {}_{1}\mathrm{F}_{1}(\alpha, \alpha + 1; -1) \\ &= \frac{1}{\Gamma(\alpha+1)} {}_{1}\mathrm{F}_{1}(\alpha, \alpha + 1; -1) \end{align}

Note that the Maple answer comes from the relationship between the Whittaker M function and the confluent hypergeometric function $$\mathrm{M}_{p,q}(z) = z^{q+1/2} \mathrm{e}^{-z/2} {}_{1}\mathrm{F}_{1}\left(\frac{1}{2} -p+q, 2q+1;z \right)$$ for $|\mathrm{arg}z| \lt \pi$

Just as carmichael561 commented, almost by definition, $$I=\int x^{\alpha -1}e^{-x}dx=-\Gamma (\alpha ,x)$$ where appears the incomplete Gamma function.

So, $$J=\int_{0}^{1} x^{\alpha -1}e^{-x}dx=\Gamma (\alpha )-\Gamma (\alpha ,1)$$ provided $\Re(\alpha )>0$.