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Question Number 130137 by mathmax by abdo last updated on 22/Jan/21

$$\mathrm{calculate}\mathrm{for}\mathrm{n}\mathrm{integr}\mathrm{natural}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}(\mathrm{n}⩾2)$$

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

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{x^a}{1+x^n}dx$$$$I\left(a\right)=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{a-n+1}{n}}}{1+u}du{x}^n=u$$$$I\left(a\right)=\frac{1}{n}{\int }_0^1{t}^{\left(\frac{a+1}{n}\right)-1}{(1-t)}^{-\left(\frac{a+1}{n}\right)}dt=\frac{1}{n}.\mathrm{\Gamma }\left(\frac{a+1}{n}\right)\mathrm{\Gamma }(1-\frac{a+1}{n})$$$$=\frac{\pi }{nsin\left(\frac{a+1}{n}\pi \right)}$$$$I^{\prime }\left(a\right)=-\frac{{\pi }^2}{n^2}cosec\left(\frac{a+1}{n}\pi \right)cot\left(\frac{a+1}{n}\pi \right)$$$$I^{\prime }\left(0\right)=-\frac{{\pi }^2}{n^2}.\frac{cos\left(\frac{\pi }{n}\right)}{{sin}^2\left(\frac{\pi }{n}\right)}={\int }_0^{\mathrm{\infty }}\frac{logx}{1+x^n}dx$$$$$$

Commented by Lordose last updated on 22/Jan/21

$$\mathrm{Nice}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 23/Jan/21

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{1+{\mathrm{x}}^{\mathrm{n}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{chamgement}\mathrm{x}={\mathrm{t}}^{\frac{1}{\mathrm{n}}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}\ln \left(\mathrm{t}\right)}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{n}}^2{\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\frac{1}{\mathrm{n}}-1}}{1+\mathrm{t}}\mathrm{lnt}\mathrm{dt}\mathrm{let}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{\pi }{\sin \left(\pi \mathrm{a}\right)}(\mathrm{o}<\mathrm{a}<1)\mathrm{we}\mathrm{have}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial \mathrm{a}}\left(\frac{{\mathrm{e}}^{(\mathrm{a}-1)\mathrm{lnt}}}{1+\mathrm{t}}\right)\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnt}{\mathrm{t}}^{\mathrm{a}-1}}{1+\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{n}}^2{\mathrm{A}}_{\mathrm{n}}={\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{\mathrm{n}}\right)\mathrm{wehave}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{{\pi }^2\cos \left(\pi \mathrm{a}\right)}{{\sin }^2\left(\pi \mathrm{a}\right)}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{\mathrm{n}}\right)=-{\pi }^2\frac{\cos \left(\frac{\pi }{\mathrm{n}}\right)}{{\sin }^2\left(\frac{\pi }{\mathrm{n}}\right)}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=-\frac{{\pi }^2}{{\mathrm{n}}^2}\times \frac{\cos \left(\frac{\pi }{\mathrm{n}}\right)}{{\sin }^2\left(\frac{\pi }{\mathrm{n}}\right)}$$

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