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Question Number 182432 by SANOGO last updated on 09/Dec/22

$${\int }_R\frac{t^2}{{(t^2+a^2)}^{n+1}}dt=?/a\ne o,nϵN^{\ast }$$

Answered by dre23 last updated on 10/Dec/22

$$t=\mid a\mid s$$$$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{s^2\mid a\mid }{a^{2n}{(1+s^2)}^{n+1}}ds,s=tg\left(z\right)$$$$=\frac{2\mid a\mid }{a^{2n}}{\int }_0^{\frac{\pi }{2}}{sin}^2\left(z\right){cos}^{2n}\left(z\right)dz=\mathit{A}$$$$\beta (x,y)=2{\int }_0^{\frac{\pi }{2}}{cos}^{2x-1}\left(t\right){sin}^{2y-1}\left(\mathit{t}\right)\mathit{d}\mathit{t}$$$$A=\frac{\mid a\mid }{a^{2n}}\beta (2n+\frac{1}{2},1+\frac{1}{2})=\frac{\mid a\mid }{a^{2n}}.\frac{\mathrm{\Gamma }(2n+\frac{1}{2})\mathrm{\Gamma }(1+\frac{1}{2})}{\mathrm{\Gamma }(2n+2)}$$$$=\frac{\mid a\mid \left[\underset{k=0}{\prod ^{2n-1}}(k+\frac{1}{2})\right]\mathrm{\Gamma }\left(\frac{1}{2}\right).\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)}{a^{2n}(2n+1)!}$$$$=\frac{\pi \mid a\mid \underset{k=0}{\prod ^n}(2k+1)}{2^{2n+1}{a}^{2n}(2n+1)!}=\frac{\pi \mid a\mid }{2^{2n+1}{a}^{2n}{2}^{n}n!}$$$$=\frac{\pi \mid a\mid }{2^{3n+1}{a}^{2n}.n!}$$$$$$

Commented by SANOGO last updated on 11/Dec/22

$$merci$$

Commented by dre23 last updated on 14/Dec/22

$$jevousenPrie$$

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