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Question Number 61976 by maxmathsup by imad last updated on 13/Jun/19

$$let{A}_n={\int }_{\frac{1}{n}}^n\frac{arctan(x^2+y^2)}{x^2+y^2}dxdy$$$$1)calculate{A}_n$$$$2)find{lim}_{n\to \mathrm{\infty }}{A}_n$$

Commented by maxmathsup by imad last updated on 13/Jun/19

$$A_n=\int {\int }_{]\frac{1}{n},n[{}^2}\frac{arctan(x^2+y^2)}{x^2+y^2}dxdy$$

Commented by maxmathsup by imad last updated on 14/Jun/19

$$1)letusethediffeomorphismx=rcos\theta andy=rsin\theta wehave$$$$\frac{1}{n}⩽x⩽nand\frac{1}{n}⩽y⩽n\Rightarrow \frac{2}{n^2}⩽x^2+y^2⩽2n^2\Rightarrow \frac{2}{n^2}⩽r^2⩽2n^2\Rightarrow$$$$\frac{\sqrt{2}}{n}⩽r⩽n\sqrt{2}\Rightarrow A_n=\int {\int }_{\frac{\sqrt{2}}{n}⩽r⩽n\sqrt{2}and0⩽\theta ⩽\frac{\pi }{2}}\frac{arctan\left(r^2\right)}{r^2}rdrd\theta$$$$={\int }_{\frac{\sqrt{2}}{n}}^{n\sqrt{2}}\frac{arctan\left(r^2\right)}{r}dr{\int }_0^{\frac{\pi }{2}}d\theta =\frac{\pi }{2}{\int }_{\frac{\sqrt{2}}{n}}^{n\sqrt{2}}\frac{arctan\left(r^2\right)}{r}drlet$$$$w_n\left(x\right)={\int }_{\frac{\sqrt{2}}{n}}^{n\sqrt{2}}\frac{arctan\left({xr}^2\right)}{r}dr(x>0)\Rightarrow w_n^{{}^{\prime }}\left(x\right)={\int }_{\frac{\sqrt{2}}{n}}^{n\sqrt{2}}\frac{r^2}{r(1+x^2{r}^4)}dr$$$$={\int }_{\frac{\sqrt{2}}{n}}^{n\sqrt{2}}\frac{rdr}{1+x^2{r}^4}letdecomposeF\left(r\right)=\frac{r}{x^2{r}^4+1}=\frac{r}{{\left(\sqrt{x}r\right)}^4+1}$$$$=\frac{r}{{(1+{xr}^2)}^2-2{xr}^2}=\frac{r}{(1+{xr}^2-\sqrt{2x}r)(1+{xr}^2+\sqrt{2x}r)}$$$$=\frac{ar+b}{{xr}^2-\sqrt{2x}r+1}+\frac{cr+d}{{xr}^2+\sqrt{2x}r+1}$$$$F(-r)=F\left(r\right)\Rightarrow c=-aandb=d\Rightarrow$$$$F\left(r\right)=\frac{ar+b}{{xr}^2-\sqrt{2x}r+1}+\frac{-ar+b}{{xr}^2+\sqrt{2x}r+1}....becontinued....$$

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