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I'm self studying complex analysis, and I've found the following problem which I can't solve.

Prove that $$\int_{-\infty}^{\infty} e^{-x^2}Im(e^{-2ix}p(x+i))dx = 0$$

Where p(z) is a polynomial of real coefficients. As a hint the books says to consider the function $f(z)=e^{-z^2}p(z)$

I tried doing the integral $$\int e^{-z^2}p(z) dz $$ with the half circle countour of radius $R$ and doing $R \to \infty$ so that the integral fro $-\infty$ to $\infty$ appears, but I haven't had any luck. Any help solving this as well as the intuition behind the solution will be greatly appreciated.

Apo
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  • You need to consider a contour over the rectangle that starts at -R on the real axis, from there to R on the real axis, from there to R + i, parallel to the imaginary axis, from there to -R +i parallel to the real axis, and from there back to -R parallel to the imaginary axis. By Cauchy's theorem the contour integral over this rectangle is zero, you can show that the integrals parallel to the imaginary axis tend to zero in the limit of R to infinity, so you then see that the two integrals parallel to the real axis from minus to plus infinity cancel. – Count Iblis Feb 17 '18 at 18:59
  • This means that the integral from -R + i to R + i in the limit of R to infinity has the same value as the integral along the real axis, and that means that this integral has a real value, the imaginary part is thus zero. – Count Iblis Feb 17 '18 at 19:01

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