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Question Number 42605 by maxmathsup by imad last updated on 28/Aug/18
let f(x) = ∫_(−∞) ^(+∞) ((arctan (xt^2 ))/(1+2t^2 ))dt 1) find a explicite form of f(x) 2) calculate ∫_0 ^∞ ((arctan(t^2 ))/(1+2t^2 ))dt and ∫_0 ^∞ ((arctan(2t^2 ))/(1+2t^2 ))dt
$${let}\:{f}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{arctan}\:\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicite}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 30/Aug/18
1) we have f^′ (x) = ∫_(−∞) ^(+∞) (t^2 /((1+x^2 t^4 )(1+2t^2 )))dt let ϕ(z) = (z^2 /((1 +x^2 z^4 )(1+2z^2 ))) we have ϕ(z) = (z^2 /((xz^2 −i)(xz^2 +i)((√2)z−i)((√2)z+i))) =(z^2 /(2x^2 (z^2 −(i/x))(z^2 +(i/x))(z−(i/( (√2))))(z+(i/( (√2)))))) (x>0) = (z^2 /(2x^2 (z−(1/( (√x)))e^((iπ)/4) )(z+(1/( (√x)))e^((iπ)/4) )(z−(1/( (√x)))e^(−((iπ)/4)) )(z+(1/( (√x)))e^(−((iπ)/4)) )(z−(i/( (√2))))(z+(i/( (√2)))))) so the poles of ϕ are +^− (1/( (√x))) e^((iπ)/4) , +^− (1/( (√x)))e^(−((iπ)/4)) and +^− (i/( (√2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,(1/( (√x)))e^((iπ)/4) ) +Res(ϕ,−(1/( (√x))) e^(−((iπ)/4)) ) +Res(ϕ,(i/( (√2))))} Res(ϕ,(e^((iπ)/4) /( (√x)))) = (i/(2x^3 ((2/( (√x)))e^((iπ)/4) )(((2i)/x))(1+((2i)/x)))) = ((x^2 (√x))/(8 x^3 (x+2i))) e^(−((iπ)/4)) = ((√x)/(8x(x+2i))) e^(−((iπ)/4)) Res(ϕ,−(e^(−((iπ)/4)) /( (√x)))) = ((−i)/(2x^3 (((−2)/( (√x))) e^(−((iπ)/4)) )(((−2i)/( (√x))))(1−((2i)/x)))) =−((√x)/(8x(x−2i))) e^((iπ)/4) Res(ϕ,(i/( (√2)))) = ((−1)/(4x^2 (((2i)/( (√2))))(−(1/2)−(i/x))(−(1/2) +(i/x)))) = ((−1)/(4i(√2)x^2 ((1/4)+(1/x^2 )))) =((−1)/(i(√2)(4x^2 )(((x^2 +4)/(4x^2 ))))) = (i/( (√2)(x^2 +4))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { ((x^2 (√x))/(8x(x+2i))) e^(−((iπ)/4)) −((√x)/(8x(x−2i))) e^((iπ)/4) +(i/( (√2)(x^2 +4)))} ....be continued...
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)}{dt}\:\:{let}\: \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}\:+{x}^{\mathrm{2}} {z}^{\mathrm{4}} \right)\left(\mathrm{1}+\mathrm{2}{z}^{\mathrm{2}} \right)}\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left({xz}^{\mathrm{2}} −{i}\right)\left({xz}^{\mathrm{2}} +{i}\right)\left(\sqrt{\mathrm{2}}{z}−{i}\right)\left(\sqrt{\mathrm{2}}{z}+{i}\right)}\:=\frac{{z}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} −\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)}\:\:\left({x}>\mathrm{0}\right) \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} \left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)}\:{so}\:{the} \\ $$$${poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:,\:\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\:{and}\:\overset{−} {+}\:\frac{{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\right\} \\ $$$${Res}\left(\varphi,\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\:\:=\:\frac{{i}}{\mathrm{2}{x}^{\mathrm{3}} \left(\frac{\mathrm{2}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2}{i}}{{x}}\right)\left(\mathrm{1}+\frac{\mathrm{2}{i}}{{x}}\right)}\:=\:\:\:\frac{{x}^{\mathrm{2}} \sqrt{{x}}}{\mathrm{8}\:{x}^{\mathrm{3}} \left({x}+\mathrm{2}{i}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$=\:\frac{\sqrt{{x}}}{\mathrm{8}{x}\left({x}+\mathrm{2}{i}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi,−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}\right)\:=\:\frac{−{i}}{\mathrm{2}{x}^{\mathrm{3}} \left(\frac{−\mathrm{2}}{\:\sqrt{{x}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{−\mathrm{2}{i}}{\:\sqrt{{x}}}\right)\left(\mathrm{1}−\frac{\mathrm{2}{i}}{{x}}\right)}\:=−\frac{\sqrt{{x}}}{\mathrm{8}{x}\left({x}−\mathrm{2}{i}\right)}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi,\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\:\frac{−\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} \left(\frac{\mathrm{2}{i}}{\:\sqrt{\mathrm{2}}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{i}}{{x}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{{x}}\right)}\:=\:\frac{−\mathrm{1}}{\mathrm{4}{i}\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{−\mathrm{1}}{{i}\sqrt{\mathrm{2}}\left(\mathrm{4}{x}^{\mathrm{2}} \right)\left(\frac{{x}^{\mathrm{2}} \:+\mathrm{4}}{\mathrm{4}{x}^{\mathrm{2}} }\right)}\:=\:\frac{{i}}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:\:\frac{{x}^{\mathrm{2}} \sqrt{{x}}}{\mathrm{8}{x}\left({x}+\mathrm{2}{i}\right)}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:−\frac{\sqrt{{x}}}{\mathrm{8}{x}\left({x}−\mathrm{2}{i}\right)}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:+\frac{{i}}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\right\} \\ $$$$….{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 30/Aug/18
∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (−2i Im (((√x)/(8x(x−2i))) e^((iπ)/4) )) −((2π)/( (√2)(x^2 +4))) =4π Im(((√x)/(8x(x−2i))) e^((iπ)/4) ) −((2π)/( (√2)(x^2 +4))) =((π(√x))/(2x)) Im((e^((iπ)/4) /(x−2i)))−((2π)/( (√2)(x^2 +4))) but (e^((iπ)/4) /(x−2i)) =(((x+2i) e^((iπ)/4) )/(x^2 +4)) =(((x+2i)((1/( (√2))) +(i/( (√2)))))/(x^2 +4)) =(1/( (√2))) (((x+2i)(1+i))/((x^2 +4))) =(1/( (√2))) ((x+ix +2i −2)/(x^2 +4)) =(1/( (√2))) ((x−2 +i(x+2))/(x^2 +4)) ⇒Im(...) =((x+2)/( (√2)(x^2 +4))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((π(√x))/(2x)) ((x+2)/( (√2)(x^2 +4))) −((2π)/( (√2)(x^2 +4))) =f^′ (x) ⇒ f^′ (x) = (π/(2(√2)(√x))) ((x+2)/((x^2 +4))) −((2π)/( (√2)(x^2 +4))) ⇒f(x) =(π/(2(√2))) ∫ ((x+2)/( (√x)(x^2 +4))) −π(√2)∫ (dx/(x^2 +4)) +c
$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left(−\mathrm{2}{i}\:{Im}\:\left(\frac{\sqrt{{x}}}{\mathrm{8}{x}\left({x}−\mathrm{2}{i}\right)}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right)\:−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)} \\ $$$$=\mathrm{4}\pi\:{Im}\left(\frac{\sqrt{{x}}}{\mathrm{8}{x}\left({x}−\mathrm{2}{i}\right)}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:=\frac{\pi\sqrt{{x}}}{\mathrm{2}{x}}\:{Im}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}−\mathrm{2}{i}}\right)−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:{but} \\ $$$$\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}−\mathrm{2}{i}}\:=\frac{\left({x}+\mathrm{2}{i}\right)\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=\frac{\left({x}+\mathrm{2}{i}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\frac{\left({x}+\mathrm{2}{i}\right)\left(\mathrm{1}+{i}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\frac{{x}+{ix}\:+\mathrm{2}{i}\:−\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\frac{{x}−\mathrm{2}\:+{i}\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow{Im}\left(…\right)\:=\frac{{x}+\mathrm{2}}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:\:=\frac{\pi\sqrt{{x}}}{\mathrm{2}{x}}\:\frac{{x}+\mathrm{2}}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:={f}^{'} \left({x}\right)\:\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\sqrt{{x}}}\:\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:−\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{x}+\mathrm{2}}{\:\sqrt{{x}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}\:−\pi\sqrt{\mathrm{2}}\int\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:+{c} \\ $$
Commented by maxmathsup by imad last updated on 30/Aug/18
changement (√x) =tgive ∫ ((x+2)/( (√x)(x^2 +4)))dx =∫ ((t^2 +2)/(t(t^4 +4)))(2t)dt = ∫ ((2t^2 +4)/(t^4 +4)) dt ⇒f(x) = ∫ ((2t^2 +4)/(t^4 +4))dt−π arctanx +c ...be continued...
$${changement}\:\sqrt{{x}}\:={tgive}\:\int\:\:\:\frac{{x}+\mathrm{2}}{\:\sqrt{{x}}\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}{dx}\:=\int\:\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{2}}{{t}\left({t}^{\mathrm{4}} \:+\mathrm{4}\right)}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{4}}{{t}^{\mathrm{4}} \:+\mathrm{4}}\:{dt}\:\Rightarrow{f}\left({x}\right)\:=\:\int\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{4}}{{t}^{\mathrm{4}} \:+\mathrm{4}}{dt}−\pi\:{arctanx}\:+{c}\:…{be}\:{continued}… \\ $$

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