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I have a heat equation for which in the solution I need to evaluate the following integral: $$\int_{-\infty}^{\infty}\sin(k_0 \xi)e^{-\frac{(x-\xi)^2}{4a^2 t}}d\xi$$ Except of the common gaussian integral ($\int_{-\infty}^\infty exp(-\alpha s^2 +\beta s)ds=\sqrt{\pi/\alpha}e^{\beta^2/4\alpha}$) by break the sine function to 2 exponents. I got a very long expression. Any help would be appreciated.

E Be
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  • do you just want the result or the proof ? and if you understand analytic continuation you can deduce $\int_{-\infty}^\infty e^{- (x-a)^2}dx$ for $a \in \mathbb{C}$ from the result for $a \in \mathbb{R}$, after what you'll modify it slightly to get your integral. – reuns Feb 10 '16 at 19:40

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Observe first that $$ \int_{-\infty}^{+\infty}\sin(k_0\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}\,d\xi =\Im\left[\int_{-\infty}^{+\infty}e^{ik_0\xi}e^{-\frac{(x-\xi)^2}{4a^2t}}\,d\xi\right] $$ Then notice that \begin{align*} ik_0\xi-\frac{(x-\xi)^2}{4a^2t} &=-\frac1{4a^2t}\left[\xi^2-2\left(ik_02a^2t+x\right)\xi+x^2\right]\\ &=-\frac1{4a^2t}\left[\xi-(ik_02a^2t+x)\right]^2+\frac{(ik_02a^2t+x)^2+x^2}{4a^2t}\\ &=-\left(\frac{\xi-\alpha}{2a\sqrt t}\right)^2+\beta \end{align*} (provided $t$ real and positive) where we have set $$ \alpha:=ik_02a^2t+x\\ \beta:=\frac{\alpha^2+x^2}{4a^2t} $$ thus we got \begin{align*} \int_{-\infty}^{+\infty}e^{ik_0\xi}e^{-\frac{(x-\xi)^2}{4a^2t}}\,d\xi &=\int_{-\infty}^{+\infty}e^{-\left(\frac{\xi-\alpha}{2a\sqrt t}\right)^2+\beta}\,d\xi\\ &=e^{\beta}\int_{-\infty}^{+\infty}e^{-\left(\frac{\xi-\alpha}{2a\sqrt t}\right)^2}\,d\xi\\ \end{align*} Then you can conclude by the substitution $$ \frac{\xi-\alpha}{2a\sqrt t}=\eta $$ using the Residue Theorem.

Joe
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