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Question Number 74395 by mathmax by abdo last updated on 23/Nov/19

$$calculate{\int }_0^{\mathrm{\infty }}{e}^{-2x}\left[e^x\right]dx$$

Commented by ~blr237~ last updated on 23/Nov/19

$$letitbeI$$$$I={\int }_0^{\mathrm{\infty }}{e}^{-2x}\left[e^x\right]e^{-x}d\left(e^x\right)={\int }_1^{\mathrm{\infty }}\frac{\left[u\right]}{u^3}du$$$$I=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\int }_k^{k+1}\frac{\left[u\right]}{u^3}du=\underset{k=1}{\sum ^{\mathrm{\infty }}}{\left[\frac{-k}{2u^2}\right]}_k^{k+1}$$$$I=\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{2k}-\frac{k}{2{(k+1)}^2}]=\underset{k=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{2k}-\frac{1}{2(k+1)}+\frac{1}{{(k+1)}^2}]$$$$I=\frac{1}{2}J+\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k+1)}^2}withJ=\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k}-\frac{1}{k+1})=1$$$$I=\frac{1}{2}+\frac{{\pi }^2}{6}-1=\frac{{\pi }^2-3}{6}$$$$$$

Commented by abdomathmax last updated on 24/Nov/19

$$letI={\int }_0^{\mathrm{\infty }}{e}^{-2x}\left[e^x\right]dxvhangement{e}^x=tgive$$$$I={\int }_1^{+\mathrm{\infty }}{e}^{-2lnt}\left[t\right]\frac{dt}{t}={\int }_1^{+\mathrm{\infty }}\frac{\left[t\right]}{t^3}dt$$$$=\sum _{n=1}^{\mathrm{\infty }}{\int }_n^{n+1}\frac{n}{t^3}dt=\sum _{n=1}^{\mathrm{\infty }}n{\int }_n^{n+1}{t}^{-3}dt$$$$=\sum _{n=1}^{\mathrm{\infty }}n{\left[\frac{1}{-2}{t}^{-2}\right]}_n^{n+1}=-\sum _{n=1}^{\mathrm{\infty }}\frac{n}{2}\{{(n+1)}^{-2}-n^{-2}\}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{2}\{\frac{1}{n^2}-\frac{1}{{(n+1)}^2}\}$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{n}{{(n+1)}^2})$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\{\frac{1}{n}-\frac{n+1-1}{{(n+1)}^2}\}$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}-\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+1)}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(n+1)}^2}$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}-\frac{1}{2}\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n}+\frac{1}{2}\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n^2}$$$$=\frac{1}{2}+\frac{1}{2}(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}-1)=\frac{1}{2}\times \frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}$$

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