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

  • $\begingroup$ 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... $\endgroup$ – mickep May 24 '15 at 19:28
  • $\begingroup$ the searched result should be $\sqrt{\pi}$ $\endgroup$ – Dr. Sonnhard Graubner May 24 '15 at 19:38
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer May 24 '15 at 19:49
  • $\begingroup$ @DanielFischer That's awesome thank you, I really should work on my complex analysis knowledge. $\endgroup$ – Anguepa May 24 '15 at 22:04
  • $\begingroup$ possible duplicate of Can I shift this integral along the complex plane? $\endgroup$ – user147263 May 25 '15 at 5:03

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.

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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*}

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  • $\begingroup$ Well that was interesting, thank you. $\endgroup$ – Anguepa Jun 29 '15 at 23:43

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