Gamma function, finding proper substitution for integral

Question

Let us attempt to calculate $$\int_a^\infty e^{-x/2}x^{y-1}\mbox{d} x, a\geq 0 $$ First, we need the limits of integration, therefore let $u : = \frac{x-a}{2}$ i.e $x = 2u+a$ and $\mbox{d}x = 2\mbox{d}u$. $$2e^{-a/2}\int_0^\infty e^{-u}(2u+a)^{y-1}\mbox{d}u $$

Which is ALMOST what I want, save for the $(2u+a)$. The objective is to make use of $$\int_0^\infty e^{-t}t^{y-1}\mbox{dt} =\Gamma (y),\ y>1 $$

Is there a better substitution to make that gives us necessary limits and solves our problem with the $(2u+a)$ factor?

Answer 1

Let $z=x/2$ \begin{align} \int\limits_{a}^{\infty} \mathrm{e}^{-x/2} x^{y-1}\mathrm{d} x &= 2^{y} \int\limits_{a/2}^{\infty}\mathrm{e}^{-z} z^{y-1}\mathrm{d} z \\ &= 2^{y} \Gamma\left(y,\frac{a}{2}\right) \end{align} where $$\Gamma(a,x) = \int\limits_{x}^{\infty}\mathrm{e}^{-t} t^{a-1}\mathrm{d} t$$ is the upper incomplete gamma function.

Answer 2

First of all, substitute $u=\frac{x}{2}$ and $\text{d}u=\frac{1}{2}\space\text{d}x$: $$\mathcal{I}=\int_\text{a}^\infty e^{-\frac{x}{2}}\cdot x^{y-1}\space\text{d}x=2\lim_{\text{n}\to\infty}\int_{\frac{\text{a}}{2}}^{\frac{\text{n}}{2}}e^{-u}\cdot\left(2u\right)^{y-1}\space\text{d}u=2^y\lim_{\text{n}\to\infty}\int_{\frac{\text{a}}{2}}^{\frac{\text{n}}{2}}e^{-u}\cdot u^{y-1}\space\text{d}u$$

Now, substitute $s=u^y$ and $\text{d}s=yu^{y-1}\space\text{d}u$: $$\mathcal{I}=\frac{2^y}{y}\lim_{\text{n}\to\infty}\int_{\left(\frac{\text{a}}{2}\right)^y}^{\left(\frac{\text{n}}{2}\right)^y}e^{-s^{\frac{1}{y}}}\space\text{d}s$$

This is a special integral, the incomplete gamma function:

$$\mathcal{I}=\int e^{-s^{\frac{1}{y}}}\space\text{d}s=-y\Gamma\left(y,s^{\frac{1}{y}}\right)$$