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Question Number 140639 by Mathspace last updated on 10/May/21

$$let{U}_n={\int }_0^{\mathrm{\infty }}\frac{x^{n}logx}{{(x^2+1)}^2}dx$$$$1)explicite{U}_n$$$$2)fondnatureof\mathrm{\Sigma }{U}_n$$$$\left(nintegrnatural\right)$$

Answered by Dwaipayan Shikari last updated on 10/May/21

$$\mu \left(n\right)={\int }_0^{\mathrm{\infty }}\frac{x^n}{{(x^2+1)}^2}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{n+1}{2}-1}}{{(1+u)}^2}du=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }(2-\frac{n+1}{2})}{\mathrm{\Gamma }\left(2\right)}$$$$=\frac{1}{2}(1-\frac{n+1}{2})\frac{\pi }{sin\left(\frac{n+1}{2}\pi \right)}$$$${\mu }^{\prime }\left(n\right)=-\frac{\pi (n+1)}{4}(1-\frac{n+1}{2})cosec\left(\frac{n+1}{2}\pi \right)cot\left(\frac{n+1}{2}\pi \right)-\frac{\pi }{4}cosec\left(\frac{n+1}{2}\pi \right)$$$$U_n=\frac{\pi (n^2-1)}{8}.\frac{cos\left(\frac{n+1}{2}\pi \right)}{sin\left(\frac{n+1}{2}\pi \right)}-\frac{\pi }{4}cosec\left(\frac{n+1}{2}\pi \right)$$

Commented by 06 last updated on 10/May/21

$$\mathrm{See}\mathrm{my}\mathrm{question},\mathrm{please}$$

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