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Let $a$ and $b$ be positive parameters.

By integrating the complex function $$ f(z) = \exp \left( iaz^{2} - \frac{ib}{z^{2}} \right)$$ around a wedge in the first quadrant that makes an angle of $\frac{\pi}{4}$ with the positive real axis and is indented at the origin, I get

$$ \begin{align} \int_{0}^{\infty} \sin \left(ax^{2}-\frac{b}{x^{2}} \right) \, dx &= \text{Im} \int_{0}^{\infty} f(x) \, dx \\ &= \text{Im} \int_{0}^{\infty} f(te^{i \pi /4}) \ e^{i \pi /4} \, dt \\ &= \frac{1}{\sqrt{2}} \int_{0}^{\infty} \exp \left( -at^{2} - \frac{b}{t^{2}} \right) \, dt \\ &= \frac{1}{2} \sqrt{\frac{\pi}{2a}} e^{-2 \sqrt{ab}}.\end{align}$$

But when I ask Maple to evaluate the integral numerically for specific values of the parameters, the result disagrees with my result.

Is my answer incorrect, or is it that the highly oscillatory nature of the integrand makes the integral hard to evaluate numerically?

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  • $\begingroup$ Could you provide an example each for the two Wolfram|Alpha cases you describe? $\endgroup$ – joriki Mar 7 '13 at 20:20
  • $\begingroup$ Now it says it needs extra time and I don't have a subscription. But is there anything fundamentally flawed with my evaluation? $\endgroup$ – Random Variable Mar 7 '13 at 20:29
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    $\begingroup$ 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. $\endgroup$ – joriki Mar 7 '13 at 20:32
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Maple agrees with your symbolic value. Can you provide us with a numerical case of disagreement?

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  • $\begingroup$ 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 $4.630404235$. $\endgroup$ – Random Variable Mar 7 '13 at 21:17
  • $\begingroup$ Version 16.02 ... for $a=b=1$ Maple says $$-\biggl(\frac{1}{16} + \frac{i}{16}\biggr)\sqrt{\pi} \operatorname{e} ^{\bigl(-(1 - i)\sqrt[4]{-1} \sqrt{2}\bigr)} \Biggl(2 i - \sqrt{2} \sqrt[4]{-1} + (-1)^{\frac{3}{4}} \sqrt{2}\Biggr) \sqrt{2}\approx0.0848088118$$ $\endgroup$ – GEdgar Mar 7 '13 at 22:23

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