*Michael Fowler*

We’ve shown that differentiating the exponential function just multiplies it by the constant in the exponent, that is to say,

_{}

Integrating the exponential function, of course, has the opposite
effect: it *divides* by the constant in
the exponent:

_{}

as you can easily check by differentiating both sides of the equation.

An important *definite*
integral (one with limits) is

_{}

Notice the minus sign in the exponent: we need an integrand
that decreases as *x* goes towards
infinity, otherwise the integral will itself be infinite.

To visualize this result, we plot below *e***^{-x} **and

Now for something a bit more challenging: how do we evaluate the integral

_{}

(*a* has to be
positive, of course.) The integral will
definitely not be infinite: it falls off equally fast in both positive and
negative directions, and in the positive direction for *x* greater than 1, it’s smaller than *e** ^{-ax}*,
which we know converges.

To see better what this function looks like, we plot it
below for *a *= 1 (red) and *a* = 4 (blue).

Notice first how much faster than the ordinary exponential *e** ^{-x}*
this function falls away. Then note that the blue curve,

But—it’s not so easy to evaluate! There is a trick: square it. That is to say, write

_{}

Now, this product of two integrals along *lines*, the *x*-integral and the *y*-integral,
is exactly the same as an integral over a *plane*,
the (*x*, *y*) plane, stretching to infinity in all directions. We can rewrite it

_{}

where _{}, *r* is just the distance from the origin (0, 0) in the (*x*, *y*)
plane. The plane is divided up into tiny squares of area *dxdy*, and doing the integral amounts to finding the value of _{}for each tiny square, multiplying by the area of that square,
and adding the contributions from every square in the plane.

In fact, though, this approach is no easier than the
original problem—the trick is to notice that the integrand _{}has a *circular symmetry*:
for any circle centered at the origin (0, 0), it has the *same value anywhere on the circle*.
To exploit this, we shouldn’t be dividing the (*x*, *y*) plane up into
little squares at all, we should be dividing it into regions having all points
the same distance from the origin.

These are called “annular” regions: the area between two
circles, both centered at the origin, the inner one of radius *r*, the slightly bigger outer one having
radius *r* + *dr*. We take *dr* to be very small, so this is a *thin* circular strip, of length 2*πr* (the circumference of the
circle) and breadth *dr*, and therefore
its total area is 2*πrdr*
(neglecting terms like *dr*^{2},
which become negligible for *dr* small
enough).

So, the contribution from one of these annular regions is _{}, and the complete integral over the whole plane is:

_{}

*This* integral is
easy to evaluate: make the change of variable to *u* = *r*^{2}, *du* = 2*rdr* giving

_{}

so taking square roots

_{}

We can easily generate more results by differentiating *I*(*a*)
above with respect to the constant *a*!

Differentiating once:

_{}

so we have

_{}

and differentiating *this*
result with respect to *a* gives

_{}

The ratio of these two integrals comes up in the kinetic theory of gases in finding the average kinetic energy of a molecule with Maxwell’s velocity distribution.

_{}

**Finding this ratio
without doing the integrals:**

It is interesting to note that this ratio could have been
found with *much less work*, in fact
without evaluating the integrals fully, as follows:

Make the change of variable _{}, so _{} and

_{}

where *C* is a
constant *independent* of *a*, because *a* has completely disappeared in the integral over *y*.
(Of course, we know _{}, but that took a lot of work.) Now, the integral with *x*^{4} in place of *x*^{2}
is given by differentiating the *x*^{2}
integral with respect to *a*, and
multiplying by -1,
as discussed above, so, differentiating the right hand side of the above
equation, the *x*^{4} integral
is just _{}, and the *C* cancels
out in the ratio of the integrals.

However, we *do*
need to do the integrals at one point in the kinetic theory: the overall
normalization of the velocity distribution function is given by requiring that

_{}

and this in fact determines *A*, using the results we found above, giving

_{}

We didn’t need this in the kinetic theory lecture, but is seems a pity to review exponential integrals without mentioning it.

It’s easy to do the integral

_{}

It can be written

_{}

where to do the last step just change variables from *x* to *y*
= *x* -
*b*/2*a*.

This can even be used to evaluate for example

_{}

by writing the cosine as a sum of exponentials.