What is $\langle v^n\rangle$, found from evaluating this integral?

Question

$$\langle v^n \rangle=\int_0^\infty v^{n+2}e^{-mv^2/2kT}dv$$

I got $(2kT/m)^{{n+3}/2}Γ({n+2}/2)$

While Wolfram alpha got what can be seen in the link.

I would like to ask how I can get what Wolfram got, and provided my result in case it shows what I did wrong. Thank you very much.

The problem is also with the units here: the k is Boltzmann's constant with units $m^2kgs^{-2}K^{-1}$, $v$, as I understand, must be velocity ($m/s$), T is in Kelvin, and $m$ in kilograms. My result ends up with the units $(m^2/s^2)^n$, but, as I understand, should be $(m/s)^n$, so it seems a factor of $1/v$ is missing inside the power of $n$ bracket of my result.

Answer 1

Redoing the calculation independently of Wolfram Alpha, for the antiderivative $$I=\int v^{n+2} e^{-\frac{m v^2}{2 k T}}\,dv=-2^{\frac{n+1}{2}} v^{n+3} \left(\frac{m v^2}{k T}\right)^{-\frac{n+3}{2} } \Gamma \left(\frac{n+3}{2},\frac{m v^2}{2 k T}\right)$$ and, for the integral $$\int_0^\infty v^{n+2} e^{-\frac{m v^2}{2 k T}}\,dv=2^{\frac{n+1}{2}} \left(\frac{m}{k T}\right)^{-\frac{n+3}{2} }\Gamma \left(\frac{n+3}{2}\right)$$

Edit

Change variable $$v=\frac{\sqrt{2kT} }{\sqrt{m}}\sqrt{x}\implies dv=\frac{\sqrt{kT} }{\sqrt{2m} \sqrt{x}}\,dx$$ to make for the antiderivative (I skipped some simplifications) $$I=\frac{k T \left(\frac{\sqrt{2k T}}{\sqrt{m}}\right)^{n+1}}{m}\int e^{-x} x^{\frac{n+1}{2}}\,dx$$ and $$\int_0^\infty e^{-x} x^{\frac{n+1}{2}}\,dx=\Gamma \left(\frac{n+3}{2}\right)$$ What is suspect is that, if you used this change of variable, may be, you missed the $\sqrt x$ in $dv$.

Answer 2

$\begin{array}\\ \int_0^\infty v^{n+2}e^{-mv^2/2kT}dv &=\int_0^\infty v^{n+2}e^{-av^2}dv \qquad a = m/2kT\\ &=\int_0^\infty (u/\sqrt{a})^{n+2}e^{-u^2}du/\sqrt{a} \qquad u=\sqrt{a}v, dv = du/\sqrt{a}\\ &=\dfrac1{a^{(n+3)/2}}\int_0^\infty u^{n+2}e^{-u^2}du\\ &=\dfrac{(2kT)^{(n+3)/2}}{m^{(n+3)/2}}\dfrac{\Gamma((n+3)/2)}{2}\\ &=\dfrac{2^{(n+1)/2}\Gamma((n+3)/2)(kT)^{(n+3)/2}}{m^{(n+3)/2}}\\ \end{array} $