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

let U_n =∫_0 ^∞ ((x^n logx)/((x^2 +1)^2 ))dx 1) explicite U_n 2) fond nature of Σ U_n (n integr natural)

$${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} {logx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{explicite}\:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{fond}\:{nature}\:{of}\:\Sigma\:{U}_{{n}} \\ $$$$\left({n}\:{integr}\:{natural}\right) \\ $$

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

μ(n)=∫_0 ^∞ (x^n /((x^2 +1)^2 ))dx=(1/2)∫_0 ^∞ (u^(((n+1)/2)−1) /((1+u)^2 ))du=(1/2) ((Γ(((n+1)/2))Γ(2−((n+1)/2)))/(Γ(2))) =(1/2)(1−((n+1)/2))(π/(sin(((n+1)/2)π))) μ′(n)=−((π(n+1))/4)(1−((n+1)/2))cosec(((n+1)/2)π)cot(((n+1)/2)π)−(π/4)cosec(((n+1)/2)π) U_n =((π(n^2 −1))/8).((cos(((n+1)/2)π))/(sin(((n+1)/2)π)))−(π/4)cosec(((n+1)/2)π)

$$\mu\left({n}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{{n}+\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{du}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{2}−\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\frac{\pi}{{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right)} \\ $$$$\mu'\left({n}\right)=−\frac{\pi\left({n}+\mathrm{1}\right)}{\mathrm{4}}\left(\mathrm{1}−\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){cosec}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right){cot}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right)−\frac{\pi}{\mathrm{4}}{cosec}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right) \\ $$$${U}_{{n}} =\frac{\pi\left({n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{8}}.\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right)}{{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right)}−\frac{\pi}{\mathrm{4}}{cosec}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\pi\right) \\ $$

Commented by 06 last updated on 10/May/21

See my question, please

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

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