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Question Number 119157 by mathdave last updated on 22/Oct/20
Answered by mindispower last updated on 23/Oct/20
x=(1/t)⇒ =∫_0 ^∞ ((tan^− ((1/t)))/(1+(1/t))).(dt/(t^2 .(1/( (√t)))))=∫_0 ^∞ ((tan^− ((1/t)))/(t+1)).(dt/( (√t))) 2∫_0 ^∞ ((tan^− (x))/(1+x)).(dx/( (√x)))=∫_0 ^∞ ((tan^− (x)+tan^− ((1/x)))/(x+1)).(dx/( (√x))) =(π/2)∫_0 ^∞ 2.(d(√x)/((1+((√x))^2 ))) =π[arctan((√x))]_0 ^∞ =(π^2 /2) ∫_0 ^∞ ((tan^− (x))/(1+x)).(dx/( (√x)))=(π^2 /4) i complete answer 2∫_0 ^∞ ((tan^− (x))/(1+x)).(dx/( (√x)))=(π/4)in previous
$${x}=\frac{\mathrm{1}}{{t}}\Rightarrow \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left(\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} .\frac{\mathrm{1}}{\:\sqrt{{t}}}}=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left(\frac{\mathrm{1}}{{t}}\right)}{{t}+\mathrm{1}}.\frac{{dt}}{\:\sqrt{{t}}} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}\right)}{\mathrm{1}+{x}}.\frac{{dx}}{\:\sqrt{{x}}}=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}\right)+{tan}^{−} \left(\frac{\mathrm{1}}{{x}}\right)}{{x}+\mathrm{1}}.\frac{{dx}}{\:\sqrt{{x}}} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{2}.\frac{{d}\sqrt{{x}}}{\left(\mathrm{1}+\left(\sqrt{{x}}\right)^{\mathrm{2}} \right)}\:=\pi\left[{arctan}\left(\sqrt{{x}}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}\right)}{\mathrm{1}+{x}}.\frac{{dx}}{\:\sqrt{{x}}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$${i}\:{complete}\:{answer} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}\right)}{\mathrm{1}+{x}}.\frac{{dx}}{\:\sqrt{{x}}}=\frac{\pi}{\mathrm{4}}{in}\:{previous} \\ $$
Commented by mathdave last updated on 22/Oct/20
Commented by mathmax by abdo last updated on 22/Oct/20
let try parametric method we have I =_((√x)=t) ∫_0 ^∞ ((arctan(t^2 ))/(t(1+t^2 )))(2t)dt =∫_(−∞) ^∞ ((arctan(t^2 ))/(t^2 +1))dt let f(a) =∫_(−∞) ^∞ ((arctan(at^2 ))/(t^2 +1))dt (a>0) ⇒ f^′ (a) =∫_(−∞) ^∞ (t^2 /((1+a^2 t^4 )(t^2 +1)))dt let ϕ(z)=(z^2 /((a^2 z^4 +1)(z^2 +1))) ⇒ϕ(z) =(z^2 /(a^2 (z^4 +(1/a^2 ))(z^2 +1))) =(z^2 /(a^2 (z^2 −(i/a))(z^2 +(i/a))(z^2 +1))) =(z^2 /(a^2 ( z−(√(i/a)))(z+(√(i/a)))(z−(√((−i)/a)))(z+(√((−i)/a)))(z−i)(z+i))) =(z^2 /(a^2 (z−(1/( (√a)))e^((iπ)/4) )(z+(1/( (√a)))e^((iπ)/4) )(z−(1/( (√a)))e^(−((iπ)/4)) )(z+(1/( (√a)))e^(−((iπ)/4)) )(z−i)(z+i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(1/( (√a)))e^((iπ)/4) ) +Res(ϕ,−(1/( (√a)))e^(−((iπ)/4)) ) +Res(ϕ,i)} Res(ϕ,i) =((−1)/(2i(a^2 +1))) Res(ϕ,(1/( (√a)))e^((iπ)/4) ) =(1/a)(i/(a^2 ((2/( (√a)))e^((iπ)/4) )(((2i)/a))((i/a)+1))) =((a^2 (√a)e^(−((iπ)/4)) )/(4a^3 (a+i))) ...be continued...
$$\mathrm{let}\:\mathrm{try}\:\mathrm{parametric}\:\mathrm{method}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{I}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\left(\mathrm{2t}\right)\mathrm{dt}\:=\int_{−\infty} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\mathrm{dt} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{−\infty} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{at}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\mathrm{dt}\:\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{−\infty} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{t}^{\mathrm{4}} \right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dt}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{a}^{\mathrm{2}} \mathrm{z}^{\mathrm{4}} +\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\frac{\mathrm{i}}{\mathrm{a}}\right)\left(\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{i}}{\mathrm{a}}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \left(\:\mathrm{z}−\sqrt{\frac{\mathrm{i}}{\mathrm{a}}}\right)\left(\mathrm{z}+\sqrt{\frac{\mathrm{i}}{\mathrm{a}}}\right)\left(\mathrm{z}−\sqrt{\frac{−\mathrm{i}}{\mathrm{a}}}\right)\left(\mathrm{z}+\sqrt{\frac{−\mathrm{i}}{\mathrm{a}}}\right)\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\varphi,\mathrm{i}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\frac{−\mathrm{1}}{\mathrm{2i}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{1}}{\mathrm{a}}\frac{\mathrm{i}}{\mathrm{a}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2i}}{\mathrm{a}}\right)\left(\frac{\mathrm{i}}{\mathrm{a}}+\mathrm{1}\right)}\:=\frac{\mathrm{a}^{\mathrm{2}} \sqrt{\mathrm{a}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{4a}^{\mathrm{3}} \left(\mathrm{a}+\mathrm{i}\right)} \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$
Commented by mindispower last updated on 23/Oct/20
yes thank you sir
$${yes}\:{thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 23/Oct/20
you are welcome sir
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 22/Oct/20
I =∫_0 ^∞ ((arctanx)/( (√x)(1+x)))dx changement (√x)=t give I =∫_0 ^∞ ((arctan(t^2 ))/(t(1+t^2 )))(2t)dt =2∫_0 ^∞ ((arctan(t^2 ))/(1+t^2 ))dt =∫_(−∞) ^(+∞) ((arctan(t^2 ))/(1+t^2 ))dt let ϕ(z) =((arctan(z^2 ))/(z^2 +1)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) =2iπ×((∣arctan(−1)∣)/(2i)) =π×(π/4)=(π^2 /4)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctanx}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\mathrm{x}\right)}\mathrm{dx}\:\:\mathrm{changement}\:\sqrt{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\left(\mathrm{2t}\right)\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{arctan}\left(\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{2i}\pi×\frac{\mid\mathrm{arctan}\left(−\mathrm{1}\right)\mid}{\mathrm{2i}} \\ $$$$=\pi×\frac{\pi}{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$

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