# another Fresnel-like integral

$\displaystyle \int_{0}^{\infty} \sin \left(ax^{2}-\frac{b}{x^{2}} \right) \ dx$

Maple returns numerical results for different values of the parameters that don't agree at all with my answer, and Wolfram Alpha returns numerical results that sometimes appear to be the same as my answer and sometimes appear to be half of my answer.

Let $\displaystyle f(z) = \exp \left( iaz^{2} - \frac{ib}{z^{2}} \right)$ and integrate around a closed sector in the first quadrant that makes an angle of $\frac{\pi}{4}$ with the positive real axis.

As you let the radius of the sector go to $\infty$, the integral will evaluate to zero along the arc of the sector.

So we have $\displaystyle \int_{0}^{\infty} f(x) \ dx = \int_{0}^{\infty} f(te^{\frac{i \pi}{4}}) \ e^{i\frac{\pi}{4}} dt$

$\displaystyle = \int_{0}^{\infty} \exp \left(iat^{2}e^{i\frac{\pi}{2}} - \frac{ib}{t^{2}e^{i\frac{\pi}{2}}} \right) \ dt$

$\displaystyle = e^{\frac{i \pi}{4}} \int_{0}^{\infty} \exp \left( -at^{2} - \frac{b}{t^{2}} \right) \ dt = e^{i\frac{\pi}{4}} \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{-2 \sqrt{ab}}$

I'm sure most people have seen that integral. You first need to complete the "square."

$\displaystyle \therefore \int_{0}^{\infty} \sin \left(ax^{2}-\frac{b}{x^{2}} \right) \ dx = \text{Im} \int_{0}^{\infty} f(x) \ dx = \frac{1}{2} \sqrt{\frac{\pi}{2a}} e^{-2 \sqrt{ab}}$

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Could you provide an example each for the two Wolfram|Alpha cases you describe? –  joriki Mar 7 '13 at 20:20
Now it says it needs extra time and I don't have a subscription. But is there anything fundamentally flawed with my evaluation? –  Random Variable Mar 7 '13 at 20:29
I don't see any errors. I wouldn't worry too much if the numerical results are completely different, since this is a tough integral to evaluate numerically; but if Wolfram|Alpha has the results right but sometimes differs by exactly a factor of $2$ then that seems to require an explanation. –  joriki Mar 7 '13 at 20:32
I let $a=2$ and $b=3$. And so that it would return something, I made the upper limit 150. Curiously it returns $0.00166135$ while my answer is $0.00330305$. –  Random Variable Mar 7 '13 at 21:14
Well, this is interesting. It returned $0.00321557$ for the upper integration limit of 175. I think the oscillations are making this integral very hard to approximate numerically. –  Random Variable Mar 7 '13 at 21:23

Maple agrees with your symbolic value. Can you provide us with a numerical case of disagreement?

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What version are you using? I'm using 14. For $a=b=1$, it returns $\left(\frac{-1}{16}-i \frac{1}{16} \right)e^{2}\sqrt{\pi}\Big(2i-\sqrt{2}(-1)^{1/4}+(-1)^{3/4} \sqrt{2}\Big) \sqrt{2}$ which is approximately $4.630404235$. –  Random Variable Mar 7 '13 at 21:17
Version 16.02 ... for $a=b=1$ Maple says $$-\biggl(\frac{1}{16} + \frac{i}{16}\biggr)\sqrt{\pi} \operatorname{e} ^{\bigl(-(1 - i)\sqrt[4]{-1} \sqrt{2}\bigr)} \Biggl(2 i - \sqrt{2} \sqrt[4]{-1} + (-1)^{\frac{3}{4}} \sqrt{2}\Biggr) \sqrt{2}\approx0.0848088118$$ –  GEdgar Mar 7 '13 at 22:23