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

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  • $\begingroup$ 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
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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\left(z\right)=e^{-z^{2}} $. Then by Cauchy theorem we have $$\oint_{\Gamma}f\left(z\right)dz=0\,\,\,\,\,\,\,\,\,(1) $$ and $$\oint_{\Gamma}f\left(z\right)dz=\int_{-T}^{T}f\left(x\right)dx+\int_{0}^{a}f\left(T+iy\right)dy+\int_{T}^{-T}f\left(x+ia\right)dx+\int_{a}^{0}f\left(-T+ia\right)dy= $$ $$=I_{1}+I_{2}+I_{3}+I_{4}. $$ As $T\rightarrow\infty $ we have $$I_{1}=\sqrt{\pi} $$ and $$I_{2}=I_{4}=0 $$ because $e^{-\left(x+ia\right)^{2}}\rightarrow0 $ if $\left|x\right|\rightarrow\infty $. And so by $(1) $ $$-I_3 = \int_{-\infty}^{\infty}f\left(x+ia\right)dx=\sqrt{\pi}. $$ 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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