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Question Number 52482 by maxmathsup by imad last updated on 08/Jan/19

find the value or ∫_0 ^∞ ((arctan(x^2 ))/(1+x^4 ))dx .

$${find}\:{the}\:{value}\:{or}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:. \\ $$

Commented by Abdo msup. last updated on 09/Jan/19

let f(t)=∫_0 ^∞ ((arctan(tx^2 ))/(1+x^4 ))dx with t>0 we have f^′ (t)= ∫_0 ^∞ (x^2 /((1+t^2 x^4 )(1+x^4 )))dx ⇒ 2f^′ (x)=∫_(−∞) ^(+∞) (x^2 /((1+t^2 x^4 )(1+x^4 )))dx let consider the complex function ϕ(z)=(z^2 /((t^2 z^4 +1)(z^4 +1))) poles of ϕ? we have ϕ(z)=(z^2 /((tz^2 −i)(tz^2 +i)(z^2 −i)(z^2 +i))) =(z^2 /(((√t)z−e^((iπ)/4) )((√t)z+e^((iπ)/4) )((√t)z−e^(−((iπ)/4)) )((√t)z +e^(−((iπ)/4)) )(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) =(z^2 /(t^2 (z−(1/(√t))e^((iπ)/4) )(z+(1/(√t))e^((iπ)/4) )(z−(1/(√t))e^(−((iπ)/4)) )(z+(1/(√t))e^(−((iπ)/4)) )(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− (1/(√t)) e^((iπ)/4) ,+^− (1/(√t)) e^(−((iπ)/4)) ,+^− e^((iπ)/4) ,+^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Σ_(im(z_k )>0) Res(ϕ,z_k ) =2iπ{ Res(ϕ,(1/(√t))e^((iπ)/4) ) +Res(ϕ,−(1/(√t)) e^(−((iπ)/4)) )+Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )}

$${let}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({tx}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:{with}\:{t}>\mathrm{0}\:{we}\:{have} \\ $$$${f}^{'} \left({t}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{dx}\:{let}\:{consider}\:{the} \\ $$$${complex}\:{function}\:\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)\left({z}^{\mathrm{4}} +\mathrm{1}\right)} \\ $$$${poles}\:{of}\:\varphi?\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({tz}^{\mathrm{2}} −{i}\right)\left({tz}^{\mathrm{2}} +{i}\right)\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} +{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{\left(\sqrt{{t}}{z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\sqrt{{t}}{z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\sqrt{{t}}{z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(\sqrt{{t}}{z}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({z}−\frac{\mathrm{1}}{\sqrt{{t}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{t}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\sqrt{{t}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{t}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:\frac{\mathrm{1}}{\sqrt{{t}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:,\overset{−} {+}\frac{\mathrm{1}}{\sqrt{{t}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} ,\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} ,\overset{−} {+}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\sum_{{im}\left({z}_{{k}} \right)>\mathrm{0}} \:{Res}\left(\varphi,{z}_{{k}} \right) \\ $$$$=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\frac{\mathrm{1}}{\sqrt{{t}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−\frac{\mathrm{1}}{\sqrt{{t}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$

Commented by Abdo msup. last updated on 09/Jan/19

Res(ϕ,(1/(√t)) e^((iπ)/4) ) = (1/t^3 ) (i/(((2/(√t))e^((iπ)/4) )(((2i)/(√t))sin((π/4)))((2/(√t))cos((π/4)))((i/t)−i)((i/t)+i))) =(1/(8t^3 (1/(t(√t))) (1/2)(−(1/t^2 )+1)))e^(−((iπ)/4)) =(((√t)e^(−((iπ)/4)) )/(4 t^2 (((t^2 −1)/t^2 )))) =(((√t)e^(−((iπ)/4)) )/(4(t^2 −1))) ....be continued...

$${Res}\left(\varphi,\frac{\mathrm{1}}{\sqrt{{t}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\:\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:\frac{{i}}{\left(\frac{\mathrm{2}}{\sqrt{{t}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2}{i}}{\sqrt{{t}}}{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\frac{\mathrm{2}}{\sqrt{{t}}}{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right)\left(\frac{{i}}{{t}}−{i}\right)\left(\frac{{i}}{{t}}+{i}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{8}{t}^{\mathrm{3}} \:\frac{\mathrm{1}}{{t}\sqrt{{t}}}\:\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}\right)}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=\frac{\sqrt{{t}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\:{t}^{\mathrm{2}} \left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:=\frac{\sqrt{{t}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}\:\:....{be}\:{continued}... \\ $$

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