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Question Number 141320 by mnjuly1970 last updated on 17/May/21

$$......nice......calculuus.....$$$$provethat::$$$$\mathit{\varphi }:={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{\mathcal{A}rctan\left(x^2{y}^2\right)}{x^4+y^4}dxdy=\frac{{\pi }^2\sqrt{2}}{16}$$$$.....$$

Answered by mindispower last updated on 19/May/21

$$x=rcos\left(s\right),y=rsin\left(s\right)$$$$\iff {\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{arctan\left(r^4\frac{{sin}^2\left(2a\right)}{4}\right)}{r^3(1-\frac{{sin}^2\left(2a\right)}{2})}drda$$$$y=\frac{{sin}^2\left(2a\right)}{4},likeparameterindependentwither$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(r^{4}y\right)}{r^3(1-2y)}dr=\frac{1}{2(1-2y)}{\int }_0^{\mathrm{\infty }}\frac{arctn\left(r^{4}y\right)}{r^4}.{dr}^2$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(r^{2}y\right)}{r^2}dr=\sqrt{y}{\int }_0^{\mathrm{\infty }}\frac{arcatn\left({\left(r\sqrt{y}\right)}^2\right)}{{\left(r\sqrt{y}\right)}^2}.d\left(r\sqrt{y}\right)$$$$=\sqrt{y}{\int }_0^{\mathrm{\infty }}\frac{arcatn\left(x^2\right)}{x^2}dx=\pi \sqrt{\frac{y}{2}}$$$$proof{\int }_0^{\mathrm{\infty }}arcatan\left(x^2\right)\frac{dx}{x^2}={\int }_0^{\mathrm{\infty }}\frac{2dx}{1+x^4}$$$$={\int }_0^{\mathrm{\infty }}\frac{2}{1+x}.\frac{1}{4}{x}^{-\frac{3}{4}}dx=\frac{1}{2}\beta (\frac{1}{4},\frac{3}{4})=\frac{1}{2}.\frac{\pi }{sin\left(\frac{\pi }{4}\right)}$$$$=\frac{\pi }{\sqrt{2}}..proved$$$$=\frac{\pi }{2\sqrt{\pi }}{\underset{0}{\int }}^{\frac{\pi }{2}}.\frac{sin\left(2a\right)}{2}.\frac{1}{1-\frac{{sin}^2\left(2a\right)}{2}}$$$$=\frac{\pi }{4}.\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}.\frac{sin\left(2a\right)}{2-{sin}^2\left(2a\right)}=\frac{\pi }{8}\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{-d\left(cos\left(2a\right)\right)}{1+{cos}^2\left(2a\right)}$$$$=\frac{\pi }{8}\sqrt{2}.{(-arcatan\left(cos\left(2a\right)\right)]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}.\frac{\pi }{2}.\sqrt{2}=\frac{{\pi }^2}{16}.\sqrt{2}$$$$\iff$$$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{arcatn\left(x^2{y}^2\right)}{x^4+y^4}dxdy=\frac{{\pi }^2}{16}.\sqrt{2}$$

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