Derivation of Generalization of the Gamma Function

The following definite integral problem was posted on 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.

Dave Rosenman
Dave Rosenman
Senior Business Analyst