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Some Useful Integrals of Exponential Functions

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Some Useful Integrals of Exponential Functions

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.

-3 x . Note that the green line forms the hypotenuse of a

right-angled triangle of area 1, and it is very plausible from the graph that the total area under the e -x

curve is the same, that is, 1, as it must be. The e

-3 x curve has area 1/3 under it, ( a = 3).

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

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To visualize this result, we plot below e -x and e

Some Useful Integrals of Exponential Functions

file:///D:/Kwanty/e_do_minus_x_kwadrat/ExpIntegrals.htm

( 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, a = 4, has about half the total area of the a = 1 curve. In fact, the area goes as . The

green lines help see that the area under the red curve (positive plus negative) is somewhat less than 2, in

fact it’s

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

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approximately.

Some Useful Integrals of Exponential Functions

file:///D:/Kwanty/e_do_minus_x_kwadrat/ExpIntegrals.htm

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 2p r (the circumference of the circle) and breadth dr , and therefore its total area is

2p rdr (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

Some Integrals Useful in the Kinetic Theory of Gases

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

Differentiating once:

so we have

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Some Useful Integrals of Exponential Functions

file:///D:/Kwanty/e_do_minus_x_kwadrat/ExpIntegrals.htm

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

One last trick…

We didn’t need this in the kinetic theory lecture, but is seems a pity to review exponential integrals

without mentioning it.

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Some Useful Integrals of Exponential Functions

file:///D:/Kwanty/e_do_minus_x_kwadrat/ExpIntegrals.htm

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.

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