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The Fresnel integral

July 29th, 2009 — Mark Przepiora
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Here’s a deceptively simple-looking integral:

\int_0^{\infty} sin(x^2)dx.

At first glance, it’s not even obvious whether the above converges.

On the other hand, it is easy to see that \int_0^{\infty} sin(x)dx does not exist (at least not in in the traditional sense) and is analogous to the alternating sequence { 1, -1, 1, -1, … }. But what of squaring the x? The function sin(x^2) alternates between positive and negative as well, so one might be tempted to say that its integral also diverges.

But before jumping to conclusions, let’s take a look at its graph:

sin(x^2)

As x increases, the areas under the positive and negative parts of the curve become smaller and smaller. The question is, do the areas become small fast enough for the integral to converge? The answer turns out to be yes, but proving so (and evaluating its value) turns out to be less straight-forward than the problem seems.

To prove convergence, we can show that \int_1^{\infty} sin(x^2)dx converges. By substituting u=x^2 the integral becomes \frac{1}{2}\int_1^{\infty} \frac{sin(u)}{\sqrt u} du, and integrating by parts (ignoring the constant factor of 1/2) yields

-[cos(u)u^{-1/2}]^{\infty}_1 + \frac{1}{2}\int_1^{\infty} cos(u)u^{-3/2}

It’s easy to see that the left term vanishes at infinity, so we need only worry about the integral in the right term. But this converges as well—in fact, absolutely, because

\int_1^{\infty} |cos(u)|u^{-3/2}\le\int_1^{\infty} u^{-3/2}

where the right-hand side converges by the “p-test.”

So we’ve proved that the bugger converges, but what about evaluating it?

This also turns out to be more difficult than it has any right to be. The antiderivative of sin(x^2) doesn’t have any nice, closed form. The standard method for evaluating it involves observing that e^{iz^2} = \cos(z^2) + i\sin(z^2) and considering the contour integral of this function around a pizza slice-shaped region in the complex plane which includes a line on the positive x-axis.

Pizza slice contour(Image stolen from the Wikipedia entry.)

As the radius R of the pizza slice goes to infinity, evaluating the line integral of e^{iz^2} along γ1 and taking its imaginary part would give us the value of the integral we want. Unfortunately, this integral is no easier to evaluate directly. However, e^{iz^2} has no singularities, so by the residue theorem its path integral all the way around the above path must be zero. Furthermore, we can prove that its integral along the crust goes to zero as R tends to infinity.

Therefore, the path integral along the diagonal γ3 (from the origin outward) must be equal to the integral along γ1 (also outward from the origin.)

We can parametrize the diagonal by g(t) = te^{i\pi / 4}, so the path integral across the diagonal is

\int_0^{\infty} e^{ig(t)^2}g\prime (t)dt = e^{i\pi / 4} \int_0^{\infty} e^{-t^2}dt

Taking only the imaginary part, we arrive at

\int_0^{\infty} sin(x^2) dx = \frac{1}{\sqrt 2} \int_0^{\infty} e^{-t^2}dt = \sqrt{2\pi}/4

where the second integral is simply half of the Gaussian integral.

The contribution of complex analysis is to prove the first identity above. My question is, is there a more elementary way of establishing it? It seems as though we shouldn’t need the theory of complex functions to evaluate something like this, but I can’t find any other way.

  1. Abel
    August 2nd, 2009 at 11:25 | #1
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    I believe there is a typo in “the path integral along the diagonal γ2 (from the origin outward) must be equal to the integral along γ1″ … maybe it should be γ3 instead of γ2

  2. Mark
    August 3rd, 2009 at 17:07 | #2
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    @Abel Right you are. Thanks for spotting this.

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