Derivation of Generalization of the Gamma Function

The following definite integral problem was posted on stackexchange.com today. $\begin{equation} 1.35 \times10^-7\int e^{-0.03x}x^4dx \tag{1} \end{equation}$

The version of the gamma function, a useful function for solving improper integrals, that I learned as an undergraduate physics major is $\begin{equation} \int_{0}^{\infty} e^{-x}x^ndx = n! \tag{2} \end{equation}$

At first glance, it appears that the gamma function (equation 2) can be used to solve equation 1, since a constant times a definite integral is equal to the constant times the definite integral. However, since $-0.03x$ is not equal to $x$, equation 2 cannot be used to solve equation 1; the integrands are not of the same form. But it is possible to derive a formula for solving improper integrals of the same form as equation 1 using u-subtitution and the gamma function.

Equation 1 has the form $\begin{equation} c\int_{0}^{\infty} e^{-ax}x^ndx \tag{3} \end{equation}$

(with $c$ and $a$ being constants)

Let $\begin{equation} u = ax \tag{4} \end{equation}$

If follows from equation 4 that $\begin{equation} x = \frac{u}{a} \tag{5} \end{equation}$

and $\begin{equation} dx = \frac{du}{a} \tag{6} \end{equation}$

Applying equations 4, 5, and 6 to equation 3: $\begin{equation} c \int_{0}^{\infty} e^{-ax}x^ndx = c\int_{0}^{\infty}e^{-u}(\frac{u}{a})^n\frac{du}{a} \end{equation}$

$\begin{equation} c \int_{0}^{\infty} e^{-ax}x^ndx = \frac{c}{a^{n+1}}\int_{0}^{\infty}e^{-u}u^ndu \tag{7} \end{equation}$

In equation 7, $\int_{0}^{\infty}e^{-u}u^ndu$ is just the gamma function and is thus equal to $n!$.

One important thing to be aware of is that the bounds of integration have not changed, since when $x$ goes to zero, from equation 4, u goes to zero as well. And when $x$ goes to infinity, $u$ goes to infinity for any non-zero constant $a$.

Finally, from equations 7 and the gamma function (equation 2): $\begin{equation} c \int_{0}^{\infty}e^{-ax}x^ndx = \frac{c}{a^{n+1}}n! \tag{8} \end{equation}$

So to solve our problem equation 1, with $c = 1.35 \times 10^{-7}$ and $a = 0.03$, we get: $\begin{equation} 1.35 \times 10^{-7}\int_{0}^{\infty}e^{-0.03x}x^ndx = \frac{1.35 \times 10^{-7}}{0.03^5}4! = \frac{400}{3} \end{equation}$

Without using equation 8, the integral could be solved, but it would require using integration by parts multiples times (unless there is another method that I am unaware of), which is an unnecessarily long and arduous way to evaluate the integral. 