Calculate using Euler integrals $\int \limits_0^{+\infty} \frac{1}{x^{n+1}} e^{-\frac{\alpha}{2x^2}}dx$ Find existence region and express the function in terms of Euler's  integrals 
$$\int \limits_0^{+\infty} \frac{1}{x^{n+1}} e^{-\frac{\alpha}{2x^2}}dx, \alpha > 0, n \in N$$
I have tried $t = \frac{1}{x}, dx = - \frac{1}{t^2}dt$
So I got
$$\int \limits_0^{+\infty}t^{n-1}e^{-\frac{\alpha t^2}{2}}dt$$
Is it correct substitution, how to make the next step?
 A: Now substitute again: $u=\frac{\alpha t^2}{2}$ so that $du=\alpha tdt$, you get:
\begin{align} \frac{\left(\frac{2}{\alpha}\right)^\frac{n}{2}}{2}\int^{\infty}_0 u^{\frac{n}{2}-1}e^{-u}\mathrm d u=\frac{\left(\frac{2}{\alpha}\right)^\frac{n}{2}}{2}\Gamma\left(\frac{n}{2}\right) \end{align}
Where $\Gamma(z):=\int_0^{\infty}t^{z-1}e^{-t}\mathrm d t$ is the gamma function.
Furthermore for $n$ multiple of 2 you get: $\Gamma(\frac{n}{2})=(\frac{n}{2}-1)!$
A: Well, we have that:
$$\mathscr{I}\left(\text{a},\text{n},x\right):=\int\frac{\exp\left(-\frac{\text{a}}{2x^2}\right)}{x^{1+\text{n}}}\space\text{d}x\tag1$$
Substitute:
$$\text{u}=\frac{\text{a}^\frac{\text{n}}{2}}{2^\frac{\text{n}}{2}\cdot x^\text{n}}\tag2$$
So, we get that:
$$\mathscr{I}\left(\text{a},\text{n},x\right)=-\frac{2^\frac{\text{n}}{2}}{\text{n}\cdot\text{a}^\frac{\text{n}}{2}}\int\exp\left(-\text{u}^\frac{2}{\text{n}}\right)\space\text{d}\text{u}\tag3$$
And, for the integral we use the incomplete gamma function:
$$\int\exp\left(-\text{u}^\frac{2}{\text{n}}\right)\space\text{d}\text{u}=\text{C}-\frac{\text{n}\Gamma\left(\frac{\text{n}}{2},\text{u}^\frac{2}{\text{n}}\right)}{2}\tag4$$
So, for $(1)$ we get:
$$\mathscr{I}\left(\text{a},\text{n},x\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)+\text{C}\tag5$$
Now, for the boundaries:


*

*When $x=0$:
$$\mathscr{I}\left(\text{a},\text{n},0\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\lim_{x\to0}\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)\tag6$$

*When $x\to\infty$:
$$\lim_{x\to\infty}\mathscr{I}\left(\text{a},\text{n},x\right)=\frac{2^{\frac{\text{n}}{2}-1}}{\text{a}^\frac{\text{n}}{2}}\cdot\lim_{x\to\infty}\Gamma\left(\frac{\text{n}}{2},\frac{\text{a}}{2x^2}\right)\tag7$$

