$$ \int_{-\infty}^\infty e^{-(x+ia)^2} \text{d}x $$ where $a\in \mathbb{R}$.
I don't know where to start but have reasons to believe the answer is $\sqrt{\pi}$. Namely $\int_{-\infty}^\infty e^{-x^2} \text{d}x$. Problem is I feel insecure performing changes of variables when suddenly the variables range in $\mathbb{C}$. I don't even know if this can be done properly, least of all how.
I saw this used once. Basically, we show that the value of the integral doesn't change with respect to $a$. It requires differentiation under the integral.
$$ I(a) = \int_{-\infty}^{\infty} \exp(-(x+ia)^2) dx $$
\begin{align*} \frac{dI}{da} &= \int_{-\infty}^{\infty} \frac{d}{da} \exp(-(x+ia)^2) dx \\\\ &= \int_{-\infty}^{\infty} -2i(x+ia)\exp(-(x+ia)^2) dx \\ &= i \int_{-\infty}^{\infty} \frac{d}{dx} \exp(-(x+ia)^2) dx \\ &= i \exp(-(x+ia)^2) |_{x=\pm \infty} = 0 \end{align*}
If $a=0$ the result is clear. Take $a>0$ and take the closed rectangular contour $\Gamma$ counterclockwise oriented in the complex plane from $-T$ to $T$, a vertical segment from $T$ to $T+ia,$ a segment from $T+ia$ to $-T+ia $ and the last segment from $-T+ia$ to $-T$. Take $f(z)=e^{-z^2}.$ Then by Cauchy's theorem we have \begin{align} \oint_\Gamma f(z) \, dz = {} & 0\tag1 \\[8pt] \text{ and } \oint_\Gamma f(z) \, dz = {} & \int_T^Tf(x) \, dx + \int_0^a f(T+iy) \, dy \\ & {} + \int_T^{-T} f(x+ia) \, dx + \int_a^0 f(-T+ia) \, dy \\[8pt] = {} &I_1+I_2+I_3+I_4. \end{align} As $T\rightarrow\infty$ we have $$I_1=\sqrt{\pi}$$ and $$I_2=I_4=0$$ because $e^{-(x+ia)^2}\rightarrow0$ if $\left|x\right|\rightarrow\infty.$ And so by $(1)$ $$-I_3 = \int_{-\infty}^\infty f(x+ia) \, dx=\sqrt{\pi}.$$ If $a<0$ the proof is similar.
