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Question Number 124919 by mathmax by abdo last updated on 07/Dec/20

$$\mathrm{find}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\arctan \left(\mathrm{x}\right)\mathrm{dx}\mathrm{with}\mathrm{n}\mathrm{integr}\mathrm{nstural}$$

Commented by mindispower last updated on 07/Dec/20

$$bypart={\left[\frac{x^{n+1}}{n+1}{\tan }^{-1}\left(x\right)\right]}_0^1-\frac{1}{n+1}{\int }_0^1\frac{x^{n+1}}{1+x^2}dx$$$$=\frac{\pi }{4(n+1)}-\frac{1}{n+1}{\int }_0^1\sum _{k⩾0}{(-1)}^k{x}^{n+1+2k}dx$$$$=\frac{\pi }{4(n+1)}-\frac{1}{(n+1)}\sum _{k⩾0}\frac{{(-1)}^k}{n+2+2k}$$$$=\frac{\pi }{4(n+1)}-\frac{1}{n+1}\sum _{k⩾0}\frac{(n+2+2(2k+1))-(n+2+4k)}{(n+2+2.2k)(n+2+2(2k+1)}$$$$=\frac{\pi }{4(n+1)}-\frac{1}{n+1}\sum _{k⩾1}\frac{2}{(n-2+4k)(n+4k)}$$$$=\frac{\pi }{4(n+1)}-\frac{1}{8(n+1)}\sum _{k⩾1}\frac{1}{(\frac{n-1}{4}+k)(\frac{n}{4}+k)}$$$$=\frac{\pi }{4(n+1)}-\frac{1}{8(n+1)}.\frac{\mathrm{\Psi }\left(\frac{n}{4}\right)-\mathrm{\Psi }\left(\frac{n-1}{4}\right)}{\frac{n}{4}-\frac{n-1}{4}}$$$$=\frac{\pi }{4(n+1)}-\frac{1}{2(n+1)}(\mathrm{\Psi }\left(\frac{n}{4}\right)-\mathrm{\Psi }\left(\frac{n-1}{4}\right)),n⩾1$$$$n=0$$$$weget\frac{\pi }{4}-{\int }_0^1\frac{x}{1+x^2}=\frac{\pi }{4}-ln\left(\sqrt{2}\right)$$$$$$$$$$

Commented by Bird last updated on 07/Dec/20

$$thankssir$$

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