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Question Number 68470 by mathmax by abdo last updated on 11/Sep/19

find the value of ∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$

Commented by mathmax by abdo last updated on 11/Sep/19

let I =∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx changement x =2t give I =∫_0 ^∞ ((artan(8t^2 ))/(4(1+t^2 ))) (2dt) =(1/2) ∫_0 ^∞ ((arctan(8t^2 ))/(1+t^2 ))dt =(1/4)∫_(−∞) ^(+∞) ((arctan(8t^2 ))/(t^2 +1))dt let f(x) =∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(t^2 +1))dt with x≥0 we have f^′ (x) =∫_(−∞) ^(+∞) (t^2 /((1+x^2 t^4 )(t^2 +1)))dt let W(z)=(z^2 /((z^2 +1)(x^2 z^4 +1))) W(z) =(z^2 /((z^2 +1)x^2 (z^4 +(1/x^2 )))) =(z^2 /(x^2 (z−i)(z+i)(z^2 −(i/x))(z^2 +(i/x)))) =(z^2 /(x^2 (z−i)(z+i)(z−(1/((√x) ))e^((iπ)/4) )(z+(1/(√x))e^((iπ)/4) )(z−(1/(√x))e^(−((iπ)/4)) )(z+(1/(√x))e^(−((iπ)/4)) ))) so the poles of W are +^− i and +^− (1/(√x))e^((iπ)/4) and +^− (1/(√x))e^(−((iπ)/4)) ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,i)+Res(W,(1/(√x))e^((iπ)/4) )+Res(W,−(1/(√x))e^(−((iπ)/4)) )} Res(W,i) =((−1)/(x^2 (2i)(1+(1/x^2 )))) =((−1)/(2i(x^2 +1))) Res(W,(1/(√x))e^((iπ)/4) ) =(i/(x^3 (i+1)((2/(√x))e^((iπ)/4) )(i+(i/x)))) =(1/((1+i)x^3 (1+(1/x))(2/(√x))))e^(−((iπ)/4)) =(((√x)e^(−((iπ)/4)) )/(2(1+i)(x^3 +x^2 ))) Res(W,−(1/(√x))e^(−((iπ)/4)) ) =((−i)/(x^3 (−i+1)(−(2/(√x))e^(−((iπ)/4)) )(−i+(i/x)))) =((√x)/(x^3 (i−1)(1−(1/x))2 e^(−((iπ)/4)) )) =(((√x)e^((iπ)/4) )/(2(i−1)(x^3 −x^2 ))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{ −(1/(2i(x^2 +1))) +(((√x)e^(−((iπ)/4)) )/(2(1+i)(x^3 +x^2 )))−(((√x)e^((iπ)/4) )/(2(1−i)(x^3 −x^2 )))} =2iπ{ −(1/(2i(x^2 +1))) +(((√x)(1−i))/(4(x^3 +x^2 )))e^(−((iπ)/4)) −(((√x)(1+i)e^((iπ)/4) )/(4(x^3 −x^2 )))} =−(π/(x^(2 ) +1)) +((iπ(1−i)(√x))/(2(x^3 +x^2 )))e^(−((iπ)/4)) −((iπ(1+i)(√x)e^((iπ)/4) )/(2(x^3 −x^2 ))) ....be continued...

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\:{changement}\:{x}\:=\mathrm{2}{t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{artan}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{\mathrm{4}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:\left(\mathrm{2}{dt}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{8}{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:\:{let}\:{f}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:{with}\:{x}\geqslant\mathrm{0} \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} \right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:\:{let}\:{W}\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)} \\ $$$${W}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right){x}^{\mathrm{2}} \left({z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}^{\mathrm{2}} −\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} +\frac{{i}}{{x}}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}−\frac{\mathrm{1}}{\sqrt{{x}}\:}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:\overset{−} {+}{i}\:\:{and}\:\overset{−} {+}\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:{and}\:\overset{−} {+}\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left({W},{i}\right)+{Res}\left({W},\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left({W},−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left({W},{i}\right)\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{2}{i}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\frac{−\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${Res}\left({W},\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{{i}}{{x}^{\mathrm{3}} \left({i}+\mathrm{1}\right)\left(\frac{\mathrm{2}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({i}+\frac{{i}}{{x}}\right)}\:=\frac{\mathrm{1}}{\left(\mathrm{1}+{i}\right){x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\frac{\mathrm{2}}{\sqrt{{x}}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=\frac{\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\left(\mathrm{1}+{i}\right)\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \right)} \\ $$$${Res}\left({W},−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{−{i}}{{x}^{\mathrm{3}} \left(−{i}+\mathrm{1}\right)\left(−\frac{\mathrm{2}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(−{i}+\frac{{i}}{{x}}\right)} \\ $$$$=\frac{\sqrt{{x}}}{{x}^{\mathrm{3}} \left({i}−\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\mathrm{2}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=\frac{\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\left({i}−\mathrm{1}\right)\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:−\frac{\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\left(\mathrm{1}+{i}\right)\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \right)}−\frac{\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\left(\mathrm{1}−{i}\right)\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)}\right\} \\ $$$$=\mathrm{2}{i}\pi\left\{\:−\frac{\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\sqrt{{x}}\left(\mathrm{1}−{i}\right)}{\mathrm{4}\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \right)}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:−\frac{\sqrt{{x}}\left(\mathrm{1}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}^{\mathrm{3}} \:−{x}^{\mathrm{2}} \right)}\right\} \\ $$$$=−\frac{\pi}{{x}^{\mathrm{2}\:} +\mathrm{1}}\:+\frac{{i}\pi\left(\mathrm{1}−{i}\right)\sqrt{{x}}}{\mathrm{2}\left({x}^{\mathrm{3}} \:+{x}^{\mathrm{2}} \right)}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:−\frac{{i}\pi\left(\mathrm{1}+{i}\right)\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{2}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)} \\ $$$$....{be}\:{continued}... \\ $$

Commented by mathmax by abdo last updated on 11/Sep/19

another easy method but need a proof let I =∫_0 ^∞ ((arctan(2x^2 ))/(x^2 +4))dx ⇒2I =∫_(−∞) ^(+∞) ((arctan(2x^2 ))/(x^2 +4))dx let ϕ(z) =((arctan(2z^2 ))/(z^2 +4)) ⇒ϕ(z) =((arctan(z^2 ))/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) and Res(ϕ,2i) =((∣arctan(−4)∣)/(4i)) =((arctan(4))/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×((arctan(4))/(4i)) =(π/2) arctan(4).

$${another}\:{easy}\:{method}\:{but}\:{need}\:{a}\:{proof}\: \\ $$$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left(\mathrm{2}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\:{let} \\ $$$$\varphi\left({z}\right)\:=\frac{{arctan}\left(\mathrm{2}{z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{arctan}\left({z}^{\mathrm{2}} \right)}{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\right)\:\:{and}\:{Res}\left(\varphi,\mathrm{2}{i}\right)\:=\frac{\mid{arctan}\left(−\mathrm{4}\right)\mid}{\mathrm{4}{i}} \\ $$$$=\frac{{arctan}\left(\mathrm{4}\right)}{\mathrm{4}{i}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{{arctan}\left(\mathrm{4}\right)}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:{arctan}\left(\mathrm{4}\right). \\ $$

Commented by mathmax by abdo last updated on 12/Sep/19

⇒I =(π/4)arctan(4).

$$\Rightarrow{I}\:=\frac{\pi}{\mathrm{4}}{arctan}\left(\mathrm{4}\right). \\ $$

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