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It is known that $ \int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}.$ What about $\int_{-\infty}^\infty e^{-(x-ti)^2} dx, $ where $t \in \mathbb{R}$, $i= \sqrt{-1}$.

Thanks.

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    To substitute (formally, as they say) $x-it$ in the formula one gets for real $x$ and to pretend that $x-it$ is real, or that it is not but hey, who cares? is not what I call a proof. So no, this is not the *one-line proof* mentioned in my comment.2012-03-24

3 Answers 3

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Note that $ \frac{d}{dt}\int_{-\infty}^\infty e^{-(x-it)^2}\,dx = -i\int_{-\infty}^\infty -2(x-it)e^{-(x-it)^2}\,dx = -i e^{-(x-it)^2}\big|_{x=-\infty}^\infty = 0. $ So the integral is constant as a function of $t$, and you can set $t=0$ to find the constant.

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    Beautiful Solution!2012-03-25
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We start with $ \int_{-\infty}^\infty \mathrm{e}^{-(x-ti)^2} \mathrm{d}x = \mathrm{e}^{t^2} \cdot \int_{-\infty}^\infty \mathrm{e}^{-x^2} \cdot \cos(2 t x) \mathrm{d}x $ Now, let $\mathcal{I}(t) = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \cdot \cos(2 t x) \mathrm{d}x$. Then $ \mathcal{I}^{\prime}(t) = -2\int_{-\infty}^\infty x \cdot \mathrm{e}^{-x^2} \cdot \sin(2 t x) \mathrm{d}x = \int_{-\infty}^\infty \sin(2 t x) \cdot \mathrm{d} \left( \mathrm{e}^{-x^2} \right) \stackrel{\text{by parts}}{=}\\ -2t \int_{-\infty}^\infty \cos(2 t x) \cdot \mathrm{e}^{-x^2} \mathrm{d} x = -2 t \mathcal{I}(t) $ Thus $ \mathcal{I}(t) = \mathcal{I}(0) \cdot \mathrm{e}^{-t^2} = \sqrt{\pi} \cdot \mathrm{e}^{-t^2} $

Hence $ \int_{-\infty}^\infty \mathrm{e}^{-(x-ti)^2} \mathrm{d}x = \sqrt{\pi} $

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    @KCd you are right. I posted an answer and fixed it there.2012-03-24
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First show that $ F(z) := \int_{-\infty}^\infty e^{-(x-z)^2} dx $ exists for all $z \in \mathbb C$, and that it is analytic in $z$. Then evaluate it for some convenient values (say $z$ real) that have a limit point. Then you may conclude the value for all $z$.