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Question Number 122120 by sina1377 last updated on 14/Nov/20

$$\underset{0}{\stackrel{3}{\int }}\frac{1}{\sqrt{y}}.e^{y}dy$$

Answered by Bird last updated on 14/Nov/20

$$I={\int }_0^3\frac{e^y}{\sqrt{y}}dy\Rightarrow I{=}_{\sqrt{y}=x}{\int }_0^{\sqrt{3}}\frac{e^{x^2}}{x}\left(2xdx\right)$$$$=2{\int }_0^{\sqrt{3}}{e}^{x^2}dx=2{\int }_0^{\sqrt{3}}\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{n!}dx$$$$=2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{\int }_0^{\sqrt{3}}{x}^{2n}dx$$$$=2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{\left[\frac{1}{2n+1}{x}^{2n+1}\right]}_0^{\sqrt{3}}$$$$=2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)n!}{\left(\sqrt{3}\right)}^{2n+1}$$$$=2\sqrt{3}\sum _{n=0}^{\mathrm{\infty }}\frac{3^n}{(2n+1)n!}$$

Answered by Dwaipayan Shikari last updated on 14/Nov/20

$${\int }_0^3\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{y^{n-\frac{1}{2}}}{n!}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^3\frac{y^{n-\frac{1}{2}}}{n!}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{3^{n+\frac{1}{2}}}{n!(2n+1)}$$$$2\sqrt{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{3^n}{n!(2n+1)}$$

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