Gaussian integral with offset, and other cases 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. 
$$\int_{ki-\infty}^{ki+\infty} e^{-f(x)} \ dx = \int_{\infty}^{\infty} e^{-f(x)} \ dx$$
 A: 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.
A: Under the change of variable $x+c = y$ where we compute $\mathrm{d}x = \mathrm{d}y$, all that changes is the range/path of integration.  (In your example, this depends on the fact that the integrand has no poles in the complex plane -- it is an entire function.  For an example where the integral is not fixed under translation of the path, see the inverse Laplace transform.)
$$\int_{k\mathrm{i}-\infty}^{k\mathrm{i}+\infty} \mathrm{e}^{-x^2} \ \mathrm{d}x 
= \int_{-\infty}^{\infty} \mathrm{e}^{-(x+k\mathrm{i})^2} \ \mathrm{d}x 
= \int_{-\infty}^{\infty} \mathrm{e}^{-y^2} \ \mathrm{d}y
= \sqrt{\pi}$$
You comment on $\exp(-(k \mathrm{i})^2)$, but you neglect that $(x+k \mathrm{i})^2 = x^2 - k^2 + 2 x k \mathrm{i} $ and the $2 x k \mathrm{i} $ matters. \begin{align*}
\exp(-(k \mathrm{i})^2) &= \exp(-(x^2 - k^2 + 2 x k \mathrm{i})) \\
 &= \exp(k^2-x^2) \exp(2xk\mathrm{i}) \\
 &= \exp(k^2-x^2) (\cos(2xk) + \mathrm{i} \sin(2xk))
\end{align*} where the last line uses Euler's formula.
Since polynomials also have no singularities, exactly the same change of variable in the integrand will show that the value of the integral is independent of imaginary shift of the path.
