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Question Number 130135 by mathmax by abdo last updated on 22/Jan/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{(\mathrm{x}+1)}^6}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 22/Jan/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^{a+1-1}}{{(1+x)}^{6+a+1-a-1}}dx\Rightarrow I\left(a\right)=\frac{\mathrm{\Gamma }(a+1)\mathrm{\Gamma }(5-a)}{\mathrm{\Gamma }\left(6\right)}$$$$I^{\prime }\left(a\right)=\frac{1}{120}({\mathrm{\Gamma }}^{\prime }(a+1)\mathrm{\Gamma }(5-a)-{\mathrm{\Gamma }}^{\prime }(5-a)\mathrm{\Gamma }(a+1))$$$$I^{\prime }\left(0\right)=-\frac{\gamma }{5}-\frac{1}{120}(-24\gamma +24(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}))$$$$=-\frac{5}{12}$$$$MeansinGeneral{\int }_0^{\mathrm{\infty }}\frac{logx}{{(1+x)}^n}dx=-\frac{1}{n}\underset{n=1}{\sum ^{n-2}}\frac{1}{n}$$

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