## Approximation to divergent integral

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Hi everyone,

I'm a physicist working on stochastic processes and I've come up against an integral that I'm not able to approximate using steepest descent (I don't have a large or small parameter), integration by parts, or any of the other common techniques. So I'd very much appreciate the input of any applied mathematicians!

The integral is $$f(x) = \int_{x}^{\infty} \frac{\Phi(t)}{t^{5}}dt$$ with $\Phi(t) = e^{i \pi t^{2} / 2}[C(t) + i S(t)]$. Here, $C(t)$ and $S(t)$ are the Fresnel integrals defined by $$C(t) + i S(t) = \int_{0}^{t} e^{i \pi u^{2} / 2} du\ .$$ What I really want is the behaviour of $f(x)$ for small $x$. But, the integral is formally divergent if $x = 0$.

Made a little progress with integration by parts, but I wasn't able to entirely separate my integral into convergent pieces.

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2
You say that you don't have a small parameter, but then also say that you want the behavior for small $x$. What happens if you define $x=\epsilon y$ and then look for the leading terms for fixed $y$ and small $\epsilon$? – Yossi Farjoun Feb 4 2010 at 11:15

## 2 Answers

Hi David,

That answer was perfect -- thanks so much. So it can be approximated directly (with a messy constant), but it doesn't have a Laurent series or anything similar.

That is fine for my needs though.

Cheers,

Irwin

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First, cut off the tail towards infinity:

$$f(x) = \int_{x}^1 \frac{\Phi(t)}{t^5} dt + \int_1^{\infty} \frac{\Phi(t)}{t^5} dt.$$

The second term is a constant, so you can compute it numerically once and for all.

Write $$e^{i \pi u^2/2} = 1 + \frac{i \pi}{2} u^2 +R(u)$$ and $$\int_{0}^t e^{i \pi u^2/2} du = t + \frac{i \pi}{6} t^3 + \int_{0}^t R(u) du.$$

So $$\frac{\Phi(t)}{t^5} = \left( t^{-4} + \frac{i \pi}{6} t^{-2} + t^{-5} \int_{0}^t R(u) du \right) \left( 1 + \frac{i \pi}{2} t^2 + R(t) \right)=$$ $$t^{-4} + \frac{2 \pi i}{3} t^{-2} + \left( t^{-4} R(t) - \frac{\pi^2}{12} + \int_{0}^t R(u) du \right).$$

So $$\int_{x}^1 \frac{\Phi(t)}{t^5} dt = \frac{1}{3}\left( x^{-3} - 1 \right) + \frac{2 \pi i}{3} \left( x^{-1} -1 \right) + \int_{x}^1 \left( t^{-4} R(t) - \frac{\pi^2}{12} + \int_{0}^t R(u) du \right) du.$$

The integrands in the last term are bounded functions, and they are being integrated over bounded domains, so there is no problem approximating them numerically.

If you want an asymptotic formula, instead of a numerical approximation, you should be able to keep taking more terms out to get a formula like $$f(x) = \frac{1}{3} x^{-3} + \frac{2 \pi i}{3} x^{-1} + C + a_1 x + a_2 x^2 + \cdots + a_n x^n + O(x^{n+1}) \quad \mathrm{as} \ x \to 0.$$ You probably won't be able to get the constant $C$ in closed form, because it involves all those convergent integrals. The other $a_i$ will be gettable in closed form, although they will get worse and worse as you compute more of them.

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