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Question Number 151819 by mnjuly1970 last updated on 23/Aug/21

$$$$$$1.provethat:$$$${\int }_0^{\mathrm{\infty }}\frac{e^t.ln\left(t\right)}{{(1+e^t)}^2}dt=\frac{1}{2}(ln\left(\frac{\pi }{2}\right)-\gamma )$$$$$$

Answered by Lordose last updated on 23/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{t}}\log \left(\mathrm{t}\right)}{{(1+{\mathrm{e}}^{\mathrm{t}})}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}\log \left(\mathrm{t}\right)}{{(1+{\mathrm{e}}^{-\mathrm{t}})}^2}\mathrm{dt}$$$$\mathrm{\Omega }\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{t}}}{{(1+{\mathrm{e}}^{-\mathrm{t}})}^2}\mathrm{dt}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}+1}\mathrm{n}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}$$$$\mathrm{\Omega }\left(\mathrm{a}\right)\stackrel{\mathrm{t}=\frac{\mathrm{x}}{\mathrm{n}}}{=}\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}+1}}{{\mathrm{n}}^{\mathrm{a}}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{a}+1-1}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}=\mathbf{\Gamma }(\mathrm{a}+1)\mathit{\eta }\left(\mathrm{a}\right)$$$${\mathrm{\Omega }}^{\prime }\left(\mathrm{a}\right)=\mathbf{\Gamma }(\mathrm{a}+1)({\mathit{\eta }}^{\prime }\left(\mathrm{a}\right)+\mathit{\eta }\left(\mathrm{a}\right){\mathit{\psi }}^{\left(0\right)}(\mathrm{a}+1))$$$$\mathrm{\Omega }={\mathrm{\Omega }}^{\prime }\left(0\right)$$$${\mathit{\psi }}^{\left(0\right)}\left(1\right)=-\mathit{\gamma },\mathit{\eta }\left(0\right)=\frac{1}{2},{\mathit{\eta }}^{\prime }\left(0\right)=\frac{1}{2}\mathbf{log}\left(\frac{\mathit{\pi }}{2}\right)$$$$\mathbf{\Omega }=\frac{1}{2}(\mathbf{log}\left(\frac{\mathit{\pi }}{2}\right)-\mathit{\gamma })$$

Commented by mnjuly1970 last updated on 23/Aug/21

$$bravomrlordoseexcellent...$$$$grateful..$$

Commented by Tawa11 last updated on 23/Aug/21

$$\mathrm{Weldone}\mathrm{sir}$$

Commented by Tawa11 last updated on 23/Aug/21

$$\mathrm{Sir},\mathrm{please}\mathrm{prescribe}\mathrm{a}\mathrm{book}\mathrm{I}\mathrm{can}\mathrm{learn}\mathrm{some}\mathrm{stuffs}\mathrm{like}\mathrm{this}.$$$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{just}\mathrm{want}\mathrm{to}\mathrm{be}\mathrm{learning}\mathrm{this}\mathrm{sir}.$$

Commented by Lordose last updated on 23/Aug/21

$$\mathrm{Advanced}\mathrm{calculus}\mathrm{explored}$$$$\mathrm{Advanced}\mathrm{Engineering}\mathrm{Mathematics}$$

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