Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is easy to evaluate $\int_{-\infty}^\infty x\cdot\exp(-x^2+ix)\,dx$ without using complex analysis, i.e.,

$\int_{-\infty}^\infty x\cdot\exp(-x^2+ix)\,dx=\exp(-\frac{1}{4})\int_{-\infty}^\infty x\cdot\exp(-(x-\frac{1}{2}i)^2)\,dx$ and then use substitution to get the answer $\exp(-\frac{1}{4})\frac{i}{2}\sqrt{\pi}$.

If we want to evaluate it using complex analysis, I know we need to construct a contour so that I can apply Cauchy integral formula (since there is no pole here so that we cannot use Residue theorem). However, I cannot find a contour and the functions to integrate on the complex plane so that the contour follows the part of the complex plane that describes the real-valued integral. Any suggestion or hint will be highly appreciated.

share|improve this question
    
A related problem. –  Mhenni Benghorbal Apr 23 '13 at 3:30
add comment

1 Answer

up vote 2 down vote accepted

Your integrand is entire and so the residue theorem isn't of use here. What you want to do is use the fact that contour integral of a holomorphic function is zero. Technically, the change of variable you talked about is not well-posed in the sense of standard calculus because it's a change of variable with a complex number. The better way to view it is that you're doing a contour integral of that integrand around a rectangular box in the complex plane. The box has a bottom side on the real line and top side on the line $\frac{1}{2}i$. The two side edges go off to infinity and it's easy to argue that their contribution to the contour integral is zero. Then you can equate the two integrals (which justifies the change of variable) and you're done.

share|improve this answer
    
Thanks for your answer, I got the correct solution! –  Langma Apr 23 '13 at 15:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.