Derivation of Generalization of the Gamma Function

The following definite integral problem was posted on stackexchange today.

(1)   \begin{equation*} 1.35 \times {10^{ - 7}}\int {{e^{ - 0.03x}}{x^4}dx} \end{equation*}


The version of the gamma function, a useful function for solving improper integrals, that I learned as an undergraduate physics major is

(2)   \begin{equation*}\int_0^\infty {{e^{ - x}}{x^n}dx = n!} \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 \neq x, equation (2) can’t 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 (2) using u-substitution and the gamma function.

Equation (1) has the form

(3)   \begin{equation*} c \int_0^\infty {{e^{ - ax}}{x^n}dx} \end{equation*}

(with c and a being constants being any constant).


(4)   \begin{equation*} u = ax \end{equation*}

It follows from equation (4) that

(5)   \begin{equation*} x = \frac{u}{a} \end{equation*}


(6)   \begin{equation*} dx = \frac{du}{a} \end{equation*}

Applying equations (4), (5), and (6) to equation (3):

(7)   \begin{equation*}c\int_0^\infty {{e^{ - ax}}{x^n}dx} = c\int_0^\infty {{e^{ - u}}{{\left( {\frac{u}{a}} \right)}^n}\frac{{du}}{a}} \end{equation*}

(8)   \begin{equation*} c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}\int_0^\infty {{e^{ - u}}{u^n}du}\end{equation*}

In equation (7), \int_0^\infty {{e^{ - u}}{u^n}du} is just the gamma function, and is thus equal to n!.

Note: 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 (2) (the gamma function):

(9)   \begin{equation*} \boxed{c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}n!}\end{equation*}

From (1), c = 1.35 \times 10^{-7} and a = 0.03. So the solution to (1) is:

    \[\boxed{1.35 \times {10^{ - 7}}\int_0^\infty {{e^{ - 0.03x}}{x^n}dx} = \frac{{1.35 \times {{10}^{ - 7}}}}{{{{0.03}^5}}}4! = \frac{{400}}{3}}\]

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


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