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Question Number 65581 by aliesam last updated on 31/Jul/19

Commented by mathmax by abdo last updated on 31/Jul/19

$$letA=\int e^{(\frac{1}{x}-x)}dxwehave{e}^u=\sum _{n=0}^{\mathrm{\infty }}\frac{u^n}{n!}withradiusinfinite\Rightarrow$$$$e^{(\frac{1}{x}-x)}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{(\frac{1}{x}-x)}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\left(\sum _{k=0}^n{C}_n^k{x}^k{\left(\frac{1}{x}\right)}^{n-k}\right)$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\sum _{k=0}^n{C}_n^k{x}^k.x^{k-n}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\sum _{k=0}^n{C}_n^k{x}^{2k-n}\Rightarrow$$$$A=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}(\sum _{k=0}^n{C}_n^k\int x^{2k-n}dx)$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}(\sum _{k=0}^n\frac{C_n^k}{2k-n+1}{x}^{2k-n+1}+\lambda )$$$$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^n\frac{C_n^k}{n!(2k-n+1)}{x}^{2k-n+1}+\lambda e$$$$$$

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