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Question Number 28811 by abdo imad last updated on 30/Jan/18

$$find{\int }_0^{\mathrm{\infty }}ln(1+e^{-xt})dxwitht>0thengivethevalueof$$ $${\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx.$$

Commented byabdo imad last updated on 01/Feb/18

$$wehavefor\mid u\mid <1\frac{d}{du}\left(ln(1+u)\right)=\frac{1}{1+u}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{u}^n\Rightarrow$$ $$ln(1+u)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{u}^{n+1}+k(k=ln\left(1\right)=0)so$$ $$ln(1+u)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{u}^{n}and$$ $${\int }_0^{\mathrm{\infty }}ln(1+e^{-xt})dx={\int }_0^{\mathrm{\infty }}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{e}^{-nxt}\right)dx$$ $$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^{\mathrm{\infty }}{e}^{-ntx}dx=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{A}_n$$ $$A_n={\int }_0^{\mathrm{\infty }}{e}^{-ntx}dx=\frac{-1}{nt}{\left[e^{-ntx}\right]}_{x=0}^{\mathrm{\infty }}=\frac{1}{nt}so$$ $${\int }_0^{\mathrm{\infty }}ln(1+e^{-xt})dx=\frac{1}{t}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=-\frac{1}{t}\delta \left(2\right)$$ $$\delta \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^x}wehaveprovedthat$$ $$\delta \left(x\right)=(2^{1-x}-1)\xi \left(x\right)\Rightarrow \delta \left(2\right)=(\frac{1}{2}-1)\xi \left(2\right)=-\frac{1}{2}\frac{{\pi }^2}{6}$$ $$=-\frac{{\pi }^2}{12}and{\int }_0^{\mathrm{\infty }}ln(1+e^{-xt})dx=\frac{{\pi }^2}{12t}lettaket=1$$ $${\int }_0^{\mathrm{\infty }}ln(1+e^{-x})dx=\frac{{\pi }^2}{12}.$$ $$$$

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