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Why does the following equation hold?

$$\sum_{i=1}^{\infty}\frac{2^i}{\sqrt{\pi}\prod_{j=1}^{i}(2j-1)}x^{\frac{2i-1}{2}}=e^x Erf(\sqrt{x})$$

in which $Erf$ is the Gauss error function.

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\begin{align} e^{x^2}\operatorname{erf}(x) &=\frac{1}{\sqrt{\pi}}\int_{-x}^{x}e^{x^2-t^2}\,dt &&\text{[definition of $\operatorname{erf}(x)$]} \\&=\frac{2x}{\sqrt{\pi}}\int_{0}^{1}e^{4x^2y(1-y)}\,dy &&\text{[substitution $t=x(1-2y)$]} \\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(4x^2)^n}{n!}\int_{0}^{1}y^n(1-y)^n\,dy &&\text{[power series expansion of $e^z$]} \\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(4x^2)^n}{n!}\frac{n!^2}{(2n+1)!} &&\text{[known "beta" integral]} \\&=\frac{2x}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{(2x^2)^n}{(2n+1)!!}=\frac{1}{\sqrt{\pi}}\sum_{n=0}^{\infty}\frac{2^{n+1}x^{2n+1}}{(2n+1)!!}. &&\text{[cancellation]} \end{align}

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  • $\begingroup$ Thank you so much for your answer. $\endgroup$ – M.Ramana Jun 19 at 16:23

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