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Question Number 130512 by MJS_new last updated on 26/Jan/21

$$\int x{\mathrm{e}}^{\frac{1}{2x}}dx=?$$

Commented by Dwaipayan Shikari last updated on 26/Jan/21

$$\frac{1}{2x}=-t\Rightarrow \frac{1}{2x^2}=\frac{dt}{dx}$$$$=-2\int x^3{e}^{-t}dt=-\frac{1}{4}\int \frac{e^{-t}}{t^3}dt=-\int \frac{1}{4}(\frac{1}{t^3}-\frac{1}{t^2}+\frac{1}{2t}-\frac{1}{6}+\frac{t}{24}-\frac{t^2}{120}...)dt$$$$=-(-\frac{1}{2t^2}+\frac{1}{4t}+\frac{1}{8}log\left(t\right)-\frac{t}{6}+\frac{1}{4}\int \frac{t}{24}-\frac{t^2}{120}+\frac{t^3}{720}-\frac{t^4}{5040}+..)$$$$=(\frac{1}{2t^2}-\frac{1}{4t}+\frac{t}{6}-\frac{t^2}{192}+\frac{t^3}{1440}+...)-\frac{1}{8}log\left(t\right)+C$$$$=+(\frac{1}{t^3}.\frac{t}{2}+\frac{1}{t^3}\left(\frac{t}{2}\right)(-\frac{t}{2})+\frac{1}{t^3}\left(\frac{t}{2}\right)(-\frac{t}{2})\left(\frac{-2t}{3}\right)+\frac{1}{t^3}\left(\frac{t}{2}\right)(-\frac{t}{2})\left(\frac{-2t}{3}\right)(-\frac{t}{32})+...)-\frac{1}{8}log\left(t\right)$$$$=\frac{\frac{1}{t^3}}{1-\frac{\frac{t}{2}}{1+\frac{t}{2}+\frac{\frac{t}{2}}{1-\frac{t}{2}+\frac{\frac{2t}{3}}{1-\frac{2t}{3}+\frac{\frac{t}{32}}{1-\frac{t}{32}+...}}}}}-\frac{1}{8}log\left(t\right)+C$$$$t=\frac{1}{2x}$$

Commented by Dwaipayan Shikari last updated on 26/Jan/21

https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula I have learnt it from here

Answered by Ar Brandon last updated on 26/Jan/21

$$\frac{1}{2\mathrm{x}}=\mathrm{u}\Rightarrow -\frac{\mathrm{dx}}{{2\mathrm{x}}^2}=\mathrm{du}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{{2\mathrm{u}}^2}$$$$\mathcal{I}=-\int \frac{{\mathrm{e}}^{\mathrm{u}}}{{4\mathrm{u}}^3}\mathrm{du}=-\frac{1}{4}\int \underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{u}}^{\mathrm{k}-3}}{\mathrm{k}!}\mathrm{du}$$$$=-\frac{1}{4}\{-\frac{2}{{\mathrm{u}}^2}-\frac{1}{\mathrm{u}}+\frac{\mathrm{lnu}}{2}+\underset{\mathrm{k}=3}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{u}}^{\mathrm{k}-2}}{(\mathrm{k}!)(\mathrm{k}-2)}\}+\mathcal{C},\mathrm{u}=\frac{1}{2\mathrm{x}}$$

Commented by MJS_new last updated on 26/Jan/21

$$\mathrm{nice}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{defined}\mathrm{for}k=2$$

Commented by Ar Brandon last updated on 26/Jan/21

You're right Sir

Answered by MJS_new last updated on 26/Jan/21

$$\int x{\mathrm{e}}^{\frac{1}{2x}}dx=$$$$[t=\frac{1}{2x}\to dx=-2x^{2}dt]$$$$=-\frac{1}{4}\int \frac{{\mathrm{e}}^t}{t^3}dt$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{{\mathrm{e}}^t}{8t^2}-\frac{1}{8}\int \frac{{\mathrm{e}}^t}{t^2}dt=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$=\frac{{\mathrm{e}}^t}{8t^2}+\frac{{\mathrm{e}}^t}{8t}-\frac{1}{8}\int \frac{{\mathrm{e}}^t}{t}dt=$$$$=\frac{{\mathrm{e}}^t}{8t^2}+\frac{{\mathrm{e}}^t}{8t}-\frac{1}{8}\mathrm{Ei}\left(t\right)=$$$$=\frac{1}{4}x(2x+1){\mathrm{e}}^{\frac{1}{2x}}-\frac{1}{8}\mathrm{Ei}\left(\frac{1}{2x}\right)+C$$

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