Why $\int_{-\infty-\frac{i\omega}{2k^2}}^{{\infty-\frac{i\omega}{2k^2}}} e^{-k^2x^2}dx=\int_{-\infty}^{{\infty}} e^{-k^2x^2}dx$?

Question

In complex analysis lecture, my professor said the following

$$\int_{-\infty-\frac{i\omega}{2k^2}}^{{\infty-\frac{i\omega}{2k^2}}} e^{-k^2x^2}dx=\int_{-\infty}^{{\infty}} e^{-k^2x^2}dx$$

the equality holds because of Cauchy Theorem and this is because the function is analytic.

My question is that how do I use Cauchy Theorem to see this? Both end points are still not the same.

Answer 1

Take the rectangle with corners at $R,-r,-r-\frac{i\omega}{2k^2},R-\frac{i\omega}{2k^2}$. The integrand is analytic, so by Cauchy-Goursat theorem, the integral around the rectangle is zero. Let $R \to \infty$ and $-r \to -\infty$. The integral along the two vertical line segments will vanish, which can be shown using the ML-inequality. The remaining terms will give you the expression above.