# On the integral $\int_{-\infty}^\infty e^{-(x-ti)^2} dx$

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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Please avoid titles that are entirely in $\LaTeX$. Thanks. – J. M. Mar 24 '12 at 15:26
Consider a counter-clockwised rectangle with four corners $\pm R$ and $\pm R - it$. With a simple application of Cauchy integral formula, you will reach the conclusion: $$\int_{-\infty}^{\infty} e^{-(x - it)^2} \; dx = \sqrt{\pi}.$$ – Sangchul Lee Mar 24 '12 at 15:27
There is a one-line proof that the answer is $\sqrt\pi$ for every real $t$ if complex analysis is allowed, and a somewhat longer one if only real analysis tools are allowed. Here is a typical example where answers to the questions what do you know? what did you try? where are you stuck? should be made clear in the post. – Did Mar 24 '12 at 15:27
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. – Did Mar 24 '12 at 15:54

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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I'm just astounded... – Sangchul Lee Mar 24 '12 at 16:40
Beautiful Solution! – Daniel Montealegre Mar 25 '12 at 7:29

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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Very thanks for answer. – Richard Mar 24 '12 at 15:55
I usually preferred real analysis methods to complex analysis ones when calculating integrals on real lines. Now I'm a bit alarmed by your answer, since it gives me a feeling that I began to forget real analysis methods. I threw you an upvote for that. – Sangchul Lee Mar 24 '12 at 16:01
There is no need to decompose the exponential into a sine and cosine part. Write the integral on the right side at the top as $\int_{-\infty}^{\infty} e^{-x^2+2xti}dx$ and call it $I(t)$. Then $I'(t) = -2tI(t)$ using integration by parts (Sasha, you lost a factor of 2 when you get just $-tI(t)$) with $u = -ie^{2xti}$ and $dv = -2xe^{-x^2}dx$. – KCd Mar 24 '12 at 16:04
@sos440: why would you prefer real analysis methods to complex analysis methods to evaluate real integrals, after seeing how incredible the residue theorem is for such computations? – KCd Mar 24 '12 at 16:06
Why even expand out the $(x-ti)^2$? If $J(t) = \int_{-\infty}^\infty e^{-(x-ti)^2}\,dt$, then $$J'(t) = -i\int_{-\infty}^\infty -2(x-ti)e^{-(x-ti)^2}\,dt = -i e^{-(x-ti)^2}\big|_{x=-\infty}^{\infty} = 0.$$ So $J(t)$ is constant. – Nick Strehlke Mar 24 '12 at 16:20

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 $z$.

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