Consider the Gaussian Integral $$ \int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$$
Numerically, it seems that for any arbitrary imaginary offset, ki,
$$\int_{ki-\infty}^{ki+\infty} e^{-x^2} \ dx = \int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$$
Why is this so edit and how would you prove it? $\exp(-(x+ki)^2)=\exp(k^2) \times \exp(-x^2 – 2kix)$ gets large very quickly, yet the overall integral does not change, no matter what the value of k is.
I have another related question. It appears that for f, an arbitrary polynomial with an even number of terms, that this might still hold, if f(x) gets arbitrarily large positive for real x as the absolute value of x gets arbitrarily large.
$$\int_{ki-\infty}^{ki+\infty} e^{-f(x)} \ dx = \int_{\infty}^{\infty} e^{-f(x)} \ dx$$
Best Answer
Consider a rectangle in the complex plane with vertices at $z=-R, z=R, z=R+ ik,$ and $z=-R+ik$.
Let $ f(z) = \displaystyle e^{-z^{2}}$ and integrate counterclockwise around the rectangle.
Then letting $R$ go to infinity,
$$ \int_{-\infty}^{\infty} e^{-x^{2}} \ dx + \lim_{R \to \infty} \int_{0}^{k} f(R+ it) \ i \ dt + \int_{\infty+ik}^{-\infty+ik} e^{-z^{2}} \ dz + \lim_{R \to \infty} \int_{k}^{0} f(-R+it) \ i \ dt = 0 . $$
If we can show that the second and fourth integrals vanish, then we have the result.
Notice that $$ \begin{align} \Big| \int_{0}^{k} f(R+it) \ i \ dt \Big| & \le \int _{0}^{k} \Big|e^{-(R+it)^{2}} \Big| \ dt \\ &= e^{-R^{2}}\int_{0}^{k} e^{t^{2}} \ dt \end{align} $$
which vanishes as $ R \to \infty$ since $k$ is finite.
Similarly,
$$ \begin{align} \Big| \int_{0}^{k} f(-R+it) \ i \ dt \Big| &\le \int _{0}^{k} \Big| e^{-(-R+it)^{2}} \Big| dt \\ &= e^{-R^{2}}\int_{0}^{k} e^{t^{2}} \ dt \end{align}$$ which vanishes as $R \to \infty$ for the same reason.