Avoid having to integrate by parts twice

Posted on by John

Suppose f(x) and g(x) are functions that are each proportional to their second derivative. These include exponential, circular, and hyperbolic functions. Then the integral of f(x) g(x) can be computed in closed form with a moderate amount of work.

The first time you see how such integrals are computed, it’s an interesting trick. I wrote up an example here. It may seem like you’re going in circles—and if you’re not careful you will go in circles—but the integral pops out.

After you’ve done this kind of integration a few times, the novelty wears off. You know how the calculation is going to end, and it’s a bit tedious and error-prone to get there.

There’s a formula that can compute all these related integrals in one fell swoop [1].

Suppose

f''(x) = hf(x)

and

g''(x) = kg(x)

for constants h and k. All the functions mentioned at the top of the post are of this form. Then

\int f(x)g(x)\,dx = \frac{1}{h-k} \bigg(f'(x)g(x) - f(x)g'(x)\bigg) + C.

So, for example, let

f(x) = e^{20x}

and

g(x) = \sin(23x).

Then h = 400, k = −529, and

\int e^{20x} \sin(23x) \, dx = \frac{1}{929} \bigg(20 e^{20x} \sin (23x) - 23e^{20x}\cos(23x) \bigg) + C.

Here’s another example.

Let

f(x) = \cos(x)

and

g(x) = \sinh(30x).

Then h = -1, k = 900, and

\int \cos(x) \sinh(30x) \,dx = \frac{1}{901}\bigg(\sin(x) \sinh(30x) + 30 \cos(x)\cosh(30x)\bigg) + C

Implicit in the formula above is the requirement h ≠ k. If h does equal k then the integral can be done by more basic techniques.

Related posts

[1] Donald K. Pease. A useful integral formula. American Mathematical Monthly. December 1959. p. 908.

One thought on “Avoid having to integrate by parts twice”

  1. John

    Note that the right hand side of the integral formula is the negative of the Wronskian of f and g. This measures how close f and g are to being linearly dependent.

    Roughly speaking, as h and k get close, f and g become more similar. The left side is approaching f^2, but the Wronskian is approaching zero. But dividing by h-k keeps the right side from going to zero.

    This is dancing around the fact that the differential equation y” = ky is well-posed, i.e. solutions depend continuously on the parameter k (and on initial conditions which we haven’t made explicit.)

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