I learned that a Gaussian Integral is \begin{equation} \int_{-\infty}^{\infty}xe^{-2ax^2}dx=0 \end{equation} Because of the odd function symmetry. But if I shift the $x$ to \begin{equation} \int_{-\infty}^{\infty}xe^{-2a(x-x_0)^2}dx \end{equation} then how to solve this integral?
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Add and subtract $x_0$ in the $x = x-x_0 + x_0$. Then you have two integrals you know the answer to. – Ninad Munshi Dec 25 '21 at 06:33
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@NinadMunshi I see, thank you! – ALLin Dec 25 '21 at 06:40
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Btw the Gaussian Integral is $\int_{-\infty}^{\infty}e^{-x^{2}},dx$ – Mr.Gandalf Sauron Dec 25 '21 at 06:57
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$$\int_{-\infty}^{\infty} xe^{-2a(x-x_0)^2}dx$$
$$=\int_{-\infty}^{\infty} (x-x_0)e^{-2a(x-x_0)^2}dx + x_0 \int_{-\infty}^{\infty} e^{-2a(x-x_0)^2}dx$$
$$=\int_{-\infty}^{\infty} xe^{-2ax^2}dx + x_0 \int_{-\infty}^{\infty} e^{-2ax^2}dx$$
$$=x_0\sqrt{\frac{\pi}{2a}}$$
cineel
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