Question Number 141320 by mnjuly1970 last updated on 17/May/21
$$\:\:\:\:\:\:\:......{nice}\:......{calculuus}..... \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \frac{\mathscr{A}\:{rctan}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{dxdy}=\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$$$..... \\ $$
Answered by mindispower last updated on 19/May/21
![x=rcos(s),y=rsin(s) ββ«_0 ^(Ο/2) β«_0 ^β ((arctan(r^4 ((sin^2 (2a))/4)))/(r^3 (1β((sin^2 (2a))/2))))drda y=((sin^2 (2a))/4),like parameter independent withe r β«_0 ^β ((arctan(r^4 y))/(r^3 (1β2y)))dr=(1/(2(1β2y)))β«_0 ^β ((arctn(r^4 y))/r^4 ).dr^2 β«_0 ^β ((arctan(r^2 y))/r^2 )dr=(βy)β«_0 ^β ((arcatn((r(βy))^2 ))/((r(βy))^2 )).d(r(βy)) =(βy)β«_0 ^β ((arcatn(x^2 ))/x^2 )dx=Ο(β(y/2)) proof β«_0 ^β arcatan(x^2 )(dx/x^2 )=β«_0 ^β ((2dx)/(1+x^4 )) =β«_0 ^β (2/(1+x)).(1/4)x^(β(3/4)) dx=(1/2)Ξ²((1/4),(3/4))=(1/2).(Ο/(sin((Ο/4)))) =(Ο/( (β2)))..proved =(Ο/(2(βΟ)))β«^(Ο/2) _0 .((sin(2a))/( 2)).(1/(1β((sin^2 (2a))/2))) =(Ο/4).(1/( (β2)))β«_0 ^(Ο/2) .((sin(2a))/(2βsin^2 (2a)))=(Ο/8)(β2)β«_0 ^(Ο/2) ((βd(cos(2a)))/(1+cos^2 (2a))) =(Ο/8)(β2).(βarcatan(cos(2a))]_0 ^(Ο/2) =(Ο/8).(Ο/2).(β2)=(Ο^2 /(16)).(β2) β β«_0 ^β β«_0 ^β ((arcatn(x^2 y^2 ))/(x^4 +y^4 ))dxdy=(Ο^2 /(16)).(β2)](Q141397.png)
$${x}={rcos}\left({s}\right),{y}={rsin}\left({s}\right) \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({r}^{\mathrm{4}} \frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{a}\right)}{\mathrm{4}}\right)}{{r}^{\mathrm{3}} \left(\mathrm{1}β\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{a}\right)}{\mathrm{2}}\right)}{drda} \\ $$$${y}=\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{a}\right)}{\mathrm{4}},{like}\:{parameter}\:{independent}\:{withe}\:{r} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({r}^{\mathrm{4}} {y}\right)}{{r}^{\mathrm{3}} \left(\mathrm{1}β\mathrm{2}{y}\right)}{dr}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}β\mathrm{2}{y}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{arctn}\left({r}^{\mathrm{4}} {y}\right)}{{r}^{\mathrm{4}} }.{dr}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({r}^{\mathrm{2}} {y}\right)}{{r}^{\mathrm{2}} }{dr}=\sqrt{{y}}\int_{\mathrm{0}} ^{\infty} \frac{{arcatn}\left(\left({r}\sqrt{{y}}\right)^{\mathrm{2}} \right)}{\left({r}\sqrt{{y}}\right)^{\mathrm{2}} }.{d}\left({r}\sqrt{{y}}\right) \\ $$$$=\sqrt{{y}}\int_{\mathrm{0}} ^{\infty} \frac{{arcatn}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}=\pi\sqrt{\frac{{y}}{\mathrm{2}}} \\ $$$${proof}\:\int_{\mathrm{0}} ^{\infty} {arcatan}\left({x}^{\mathrm{2}} \right)\frac{{dx}}{{x}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}}{\mathrm{1}+{x}}.\frac{\mathrm{1}}{\mathrm{4}}{x}^{β\frac{\mathrm{3}}{\mathrm{4}}} {dx}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}..{proved} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\pi}}\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} .\frac{{sin}\left(\mathrm{2}{a}\right)}{\:\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}β\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{a}\right)}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{4}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} .\frac{{sin}\left(\mathrm{2}{a}\right)}{\mathrm{2}β{sin}^{\mathrm{2}} \left(\mathrm{2}{a}\right)}=\frac{\pi}{\mathrm{8}}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{β{d}\left({cos}\left(\mathrm{2}{a}\right)\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left(\mathrm{2}{a}\right)} \\ $$$$=\frac{\pi}{\mathrm{8}}\sqrt{\mathrm{2}}.\left(β{arcatan}\left({cos}\left(\mathrm{2}{a}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{8}}.\frac{\pi}{\mathrm{2}}.\sqrt{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}.\sqrt{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{arcatn}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)}{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{dxdy}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}.\sqrt{\mathrm{2}} \\ $$