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An old complex analysis exam question:

Evaluate

$$\large I(a) = \int_{-\infty}^{\infty} e^{-\frac{1}{2}x^2+iax}dx$$

So far, I have completed the square in exponent, and now I have the integral

$$\large I(a) = \int_{-\infty}^{\infty} e^{(x+\frac{ia}{2})^2-\frac{3}{2}x^2+\frac{a^2}{4}}dx$$

Perhaps completing the square above gives me an integral where I now don't have to worry about whether a is positive or negative -- and so I don't need to break the integration out into two cases.

But I'm not sure what to do from here. Most integration that I come across in old exam questions is an application of the Residue Theorem, but currently the integrand looks pole-free / entire, so there'd be no residues to compute.

One last thing I thought of so far is that the integral looks a bit like the Gaussian integrals. Would this be the better / correct path to follow? If so, what should I start with? The $iax$ term makes it tricky to know what sort of substitution I could go with to perhaps get something like $e^{-ax^2}$

Any hints or suggestions are welcome.

Thanks,

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    $\begingroup$ This is not completing the square: $\frac{1}{2}(x^2 + 2iax) = \frac{1}{2}(x+ia)^2 + \frac{a^2}{2}$, this is... Now do you see how to proceed? $\endgroup$ – Hamed Nov 10 '15 at 0:53
  • $\begingroup$ Hi @Hamed, thanks for your comment - I have completed the square correctly now. But I am not sure how to proceed. I am trying integration along a semi-circular contour but am not getting anywhere with it unfortunately. What do you think could be a good choice of contour? Thanks, $\endgroup$ – User001 Nov 10 '15 at 1:37
  • $\begingroup$ ah, I think a box in the upper half plane, that sits on the real axis could do the job, @Hamed ... I will try now ...thanks $\endgroup$ – User001 Nov 10 '15 at 1:42
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    $\begingroup$ Simply change the variable to $u=x+ia$ then you have simple Gaussian. If you are really worried that this integral is actually a real integral, note that this change of variable is equivalent to a countour with imaginary part constant $ia$. Since Gaussian has no poles in the rectangle (x axis and $z=ia$ with perpendicular lines at infinity) the Gaussian integral along the x axis (simple real Gaussian) has the same value as the new integral along $z=ia$. $\endgroup$ – Hamed Nov 10 '15 at 1:46
  • $\begingroup$ Yes, agreed @Hamed. I just finished the problem, using the Cauchy-Goursat theorem. Thanks so much for your help :-). Have a great night. $\endgroup$ – User001 Nov 10 '15 at 2:44