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Series with Gamma

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Post Fri Aug 12, 2016 9:57 am

Posts: 46
Location: http://www.tolaso.com.gr/
Prove that

$$\sum_{n=0}^{\infty} \frac{\Gamma \left( \frac{n+1}{2} \right)}{n!}=\sqrt{\pi} \sqrt[4]{e} \left ({\rm efri} \left(\frac{1}{2} \right) +1 \right)$$

where $\Gamma$ stands for the Euler's Gamma function and ${\rm erfi}$ is the error function.

Post Fri Aug 12, 2016 1:41 pm

Posts: 38
$$\begin{align*}
S&=\sum_{n=0}^{\infty}\frac{\Gamma\left(\frac{n+1}{2}\right)}{n!} \\
&=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\int_{0}^{\infty}x^{\frac{n-1}{2}}e^{-x}\text{d}x\right) \\
&=\int_{0}^{\infty}e^{-x}\left(\sum_{n=0}^{\infty}\frac{x^{\frac{n-1}{2}}}{n!}\right)\text{d}x \\
&=\int_{0}^{\infty}e^{-x}\left(\frac{e^{\sqrt{x}}}{\sqrt{x}}\right)\text{d}x \\
&=\int_{0}^{\infty}\frac{e^{\sqrt{x}-x}}{\sqrt{x}}\text{d}x \\
&=2\int_{0}^{\infty}e^{x-x^{2}}\text{d}x \\
&=2e^{\frac{1}{4}}\int_{-\frac{1}{2}}^{\infty}e^{-x^{2}}\text{d}x \\
&=\sqrt{\pi}e^{\frac{1}{4}}\left(1+\operatorname{erf}\left(\frac{1}{2}\right)\right)
\end{align*}$$


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