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Why $\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty}e^{tx}e^{-x^{2}/2}dx$

equals to $e^{t^{2}/2}$ ?

I know it is error function. but I just do not have any basic knowledge about error function and do not know how to derive it.

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  • $\begingroup$ "I know it is error function": what ? $\endgroup$ – Yves Daoust Oct 17 '18 at 11:59
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$e^{tx-x^{2}/2}=e^{-(x-t)^{2}/2} e^{t^{2}}$, Make the substitution $y=x-t$ to evaluate the integral.

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