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Question Number 159592 by mnjuly1970 last updated on 19/Nov/21

calculate: I :=∫_0 ^( ∞) (((arctan(x))/x))^3 dx=?

$$ \\ $$$$\:\:\:{calculate}: \\ $$$$ \\ $$$$\:\:\:\mathcal{I}\::=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{arctan}\left({x}\right)}{{x}}\right)^{\mathrm{3}} {dx}=? \\ $$$$ \\ $$

Answered by mindispower last updated on 19/Nov/21

byart =(3/2)∫_0 ^∞ ((arctan^2 (x))/(x^2 (1+x^2 )))dx =3∫_0 ^∞ ((arctan(x))/(x(1+x^2 )))dx−(1/2).((π/2))^3 =3∫_0 ^(π/2) ((xcos(x))/(sin(x)))−(π^3 /(16)) −3∫_0 ^(π/2) ln(sin(x))dx−(π^3 /(16)) ((3π)/2)ln(2)−(π^3 /(16))

$${byart}\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{arctan}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}−\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xcos}\left({x}\right)}{{sin}\left({x}\right)}−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$−\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$

Commented by mnjuly1970 last updated on 19/Nov/21

thanks alot sir power

$${thanks}\:{alot}\:{sir}\:{power} \\ $$

Commented by mindispower last updated on 19/Nov/21

withe pleasur sir

$${withe}\:{pleasur}\:{sir} \\ $$

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