# Gaussian integral with a shift in the complex plane

$$\int_{-\infty}^\infty e^{-(x+ia)^2} \text{d}x$$ 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.

• You could always expand the square and look at it as a Fourier transform, do the calculations, and see if you get what you think that you should get... May 24, 2015 at 19:28
• the searched result should be $\sqrt{\pi}$ May 24, 2015 at 19:38
• You can use Cauchy's integral theorem to get rid of the $ia$, by shifting the contour from the line $\operatorname{Im} z = a$ to the real axis. May 24, 2015 at 19:49
• @DanielFischer That's awesome thank you, I really should work on my complex analysis knowledge. May 24, 2015 at 22:04
• possible duplicate of Can I shift this integral along the complex plane?
– user147263
May 25, 2015 at 5:03

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.
$$I(a) = \int_{-\infty}^{\infty} \exp(-(x+ia)^2) dx$$
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.