Mathematics for the Maxwell-Boltzmann Distribution

Mathematics Section: Integrals

In our study of the MBS distribution we will encounter many integrals of the following form:

\begin{displaymath}I_n(c)=\int_0^\infty u^n e^{-cu^2}\ du\end{displaymath}

All of these integrals have an exponential term exp(-cu2) multiplied by n powers of u and integrated from 0 to infinity. The value of the integral will depend on c and on n.  We begin with I0(c):

\begin{displaymath}I_0(c)=\int_0^\infty e^{-cu^2}\ du=\frac{\sqrt{\pi}}{2}c^{-1/2}\end{displaymath}

This one integral we just take as a definition. We call this integral I0(c) because the exponential function is multiplied by no powers of u. We can now use it to derive the answers to other integrals.

The Differentiation Method

We begin with the above equation and take the derivative of both sides with respect to c:

\begin{displaymath}\int_0^\infty \frac{\partial}{\partial c} e^{-cu^2}\ du=\frac{\sqrt{\pi}}{2}\frac{\partial}{\partial c}c^{-1/2}\end{displaymath}

performing the derivatives yields

\begin{displaymath}\int_0^\infty -u^2e^{-cu^2}\ du=-\frac{\sqrt{\pi}}{4}c^{-3/2}\end{displaymath}

We have now derived an expression for I2(c), the integral with 2 powers of u:

\begin{displaymath}I_2(c)=\int_0^\infty u^2e^{-cu^2}\ du=\frac{\sqrt{\pi}}{4}c^{-3/2}\end{displaymath}

We can repeat the same procedure as many times as we want. For example, differentiating the expression for I2(c) allows us to find this expression for I4(c): (Make sure you know how to do this for yourself!)

\begin{displaymath}I_4(c)=\int_0^\infty u^4e^{-cu^2}\ du=\frac{3\sqrt{\pi}}{8}c^{-5/2}\end{displaymath}

We could continue on and find expressions for I6(c), I8(c), etc. What about In(c) for odd n? Well, we begin with  I1(c) and use the substitution x=u2:

\begin{displaymath}I_1(c)=\int_0^\infty u^1e^{-cu^2}\ du=\frac{1}{2}\int_0^\infty e^{-cx}\ dx=\frac{1}{2}c^{-1}\end{displaymath}

Now, we can perform the same differentiation method to convert I1(c) to I3(c):

\begin{displaymath}I_3(c)=\int_0^\infty u^3e^{-cu^2}\ du=\frac{1}{2}c^{-2}\end{displaymath}

Once again, we can continue on to find In(c) for any odd value of n. Thus, as long as we are given the expression for I0(c), we can derive expressions for all other n.

© Andrew M. Rappe