Solve the following integral $f(x) = \int_{0}^{\infty} \frac{1}{16}\frac{m^{2+x}e^{\frac{-3}{2}m}}{x!} \, dm \; .$ I am working on a statistics problem.  I get the solution as far as the following integral:
$$
  f(x) = \int_{0}^{\infty} \frac{m^{2+x}}{16}\frac{e^{\frac{-3}{2}m}}{x!} \, dm \\
   = \frac{1}{16 \cdot x!} \int_{0}^{\infty} m^{2+x}e^{\frac{-3}{2}m} \, dm \; .
$$
    Where do I go from here?  I believe this converts to the gamma function somehow as
$$   \Gamma(\alpha) = \int_{0}^{\infty} x^{\alpha-1}e^{-x} dx
$$
    for Re $\alpha > 0$.  How would I go about doing this?  Any help would be much appreciated.
-IdleMathGuy

 A: Integrating by parts (which actually is overkill; see alternate method below) three times gives
$$\begin{align*}
\int_0^\infty\frac{m^{x+2}e^{-3m/2}}{16x!}\,\mathrm dm&=\frac1{16x!}\left(-\frac23m^{x+2}e^{-3m/2}\bigg|_0^\infty+\frac23(x+2)\int_0^\infty m^{x+1}e^{-3m/2}\,\mathrm dm\right)\\[1ex]
&=\frac{x+2}{24x!}\int_0^\infty m^{x+1}e^{-3m/2}\,\mathrm dm\\[1ex]
&=\frac{x+2}{24x!}\left(-\frac23m^{x+1}e^{-3m/2}\bigg|_0^\infty+\frac23(x+1)\int_0^\infty m^xe^{-3m/2}\,\mathrm dm\right)\\[1ex]
&=\frac{(x+2)(x+1)}{36x!}\int_0^\infty m^xe^{-3m/2}\,\mathrm dm\\[1ex]
&=\frac{(x+2)(x+1)}{36x!}\left(-\frac23m^xe^{-3m/2}\bigg|_0^\infty+\frac23x\int m^{x-1}e^{-3m/2}\,\mathrm dm\right)\\[1ex]
&=\frac{(x+2)(x+1)x}{54x!}\int_0^\infty m^{x-1}e^{-3m/2}\,\mathrm dm
\end{align*}$$
Now replace $m=\frac{2n}3$ and $\mathrm dm=\frac23\,\mathrm dn$:
$$\left(\frac23\right)^x\frac{(x+2)(x+1)x}{54x!}\int_0^\infty n^{x-1}e^{-n}\,\mathrm dn=\left(\frac23\right)^x\frac{(x+2)(x+1)}{54(x-1)!}\Gamma(x)$$
but $\Gamma(x)=(x-1)!$, so the integral reduces to $\left(\frac23\right)^x\frac{(x+2)(x+1)}{54}$. WolframAlpha probably leaves the result in terms of $\Gamma$ because it doesn't assume that $x$ is an integer.

On the other hand, with $y=x+3$ and $m=\frac{2n}3$, we have
$$\int_0^\infty\frac{m^{x+2}e^{-3m/2}}{16x!}\,\mathrm dm=\int_0^\infty\frac{\left(\frac{2n}3\right)^{y-1}e^{-n}}{16(y-3)!}\frac23\,\mathrm dn=\left(\frac23\right)^y\frac1{16(y-3)!}\underbrace{\int_0^\infty n^{y-1}e^{-n}\,\mathrm dn}_{\Gamma(y)}$$
and so the integral is
$$\left(\frac23\right)^{x+3}\frac{\Gamma(x+3)}{16x!}=\left(\frac23\right)^x\frac{(x+2)!}{54x!}$$
and the same result is obtained.
