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I am wondering how to integrate

$$ I(a, b) = \int_{-\infty}^{\infty}dx \frac{\sin(x^2 + a)}{x^2 + a} \exp(i[x-b]^2) $$

Thus far I tried integration by parts and contour integration but could not find a solution. This integral arises in connection with the phase-matching condition in nonlinear crystals. I suspect there is no solution in closed form but an expression for $I(a,b)$ that is suitable for efficient numerical computation would be most appreciated.

user140462
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  • This may have two residues at $z = \pm i\sqrt{a}$. You should be able to integrate around a semi-circle (which would only hold one of those two residues), part of which, is this integral (along the reals). The line integral for the semicircle's arc should go towards $0$ as $r \rightarrow \infty$ (the magnitude of the complex number) because you have $\sin$ and $e^{ix}$ over $x^2 + a \tilde{} r^2 + a$ so you're dividing two numbers bounded by $1$ by a very large number. – Jared Apr 04 '14 at 19:13
  • Actually, I don't think it's true that the integral at the arc goes away. I forgot that you have to put in a complex number and $\sin(z)$ and $e^{iz}$ aren't bounded by $1$ for complex $z$ (anything with a non-zero imaginary part). – Jared Apr 04 '14 at 19:46
  • You are right, that the integral does not vanish on a (upper/lower) semicircle due to the square in the argument of the exponential. The integral is zero from 0 to $\pi/2$ but is then unbounded on the second half of the semicircle (analogous for the lower semicircle - the details depend on a and b). – user140462 Apr 07 '14 at 08:25

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