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Question Number 29077 by abdo imad last updated on 04/Feb/18
let give g(x)=∫_0 ^∞ ((arctan(x(1+t^2 )))/(1+t^2 ))dt find a simple form of g^′ (x) without integral.
$${let}\:{give}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{find}\:{a}\:{simple} \\ $$$${form}\:{of}\:\:{g}^{'} \left({x}\right)\:{without}\:{integral}. \\ $$
Commented by abdo imad last updated on 10/Feb/18
g^′ (x)= ∫_0 ^∞ (dt/(1+x^2 (1+t^2 )^2 ))=(1/2) ∫_(−∞) ^(+∞) (dt/(1+x^2 (1+t^2 )^2 )) let introduce the complex function ϕ(z) =(1/(1+x^2 (1+z^2 )^2 )) poles ofϕ? we have ϕ(z)= (1/((x(1+z^2 )+i)(x(1+z^2 )−i))) = (1/((xz^2 +x+i)( xz^2 +x−i)))= (1/(x^2 (z^2 +1 +(i/x))(z^2 +1−(i/x)))) roots of z^2 +1 +(i/x)=0?⇔ z^2 =−1−(i/x)=(i(√(1+(i/x))))^2 ∣1+(i/x)∣=(√(1+(1/x^2 ) )) =((√(1+x^2 ))/(∣x∣))and 1+(i/x) =((√(1+x^2 ))/(∣x∣))((1/((√(1+x^2 ))/(∣x∣))) + (1/(x((√(1+x^2 ))/(∣x∣))))i)=r e^(iθ) ⇒cosθ=((∣x∣)/( (√(1+x^2 )))) and sinθ =((∣x∣)/(x(√(1+x^2 )))) .let suppose x>0⇒cosθ= (x/( (√(1+x^2 )))) and sinθ= (1/( (√(1+x^2 ))))⇒tanθ= (1/x) ⇒θ=arctan((1/x))and 1+(i/x)=((√(1+x^2 ))/x) e^(iarctan((1/x))) ⇒(√(1+(i/x))) =(((1+x^2 )^(1/4) )/( (√x))) e^((i/2)arctan((1/x))) ⇒ z= +−i (((1+x^2 )^(1/4) )/( (√x))) e^((i/2)arctan((1/x))) the poles are z_0 =i(((1+x^2 )^(1/4) )/( (√x)))(cos((1/2)arctan((1/x)) +isin((1/2)arctan((1/x))) z_1 =−i(((1+x^2 )^(1/4) )/( (√x)))(cos((1/2)arctan((1/x))+isin((1/2)arctan((1/x))) roots of z^2 +1 −(i/x)=0 be continued.....
$${g}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{poles}\:{of}\varphi?\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)=\:\:\:\:\frac{\mathrm{1}}{\left({x}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)+{i}\right)\left({x}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)−{i}\right)} \\ $$$$=\:\:\:\:\:\:\:\frac{\mathrm{1}}{\left({xz}^{\mathrm{2}} +{x}+{i}\right)\left(\:{xz}^{\mathrm{2}} \:+{x}−{i}\right)}=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+\mathrm{1}\:+\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\mathrm{1}−\frac{{i}}{{x}}\right)} \\ $$$${roots}\:{of}\:{z}^{\mathrm{2}} \:+\mathrm{1}\:+\frac{{i}}{{x}}=\mathrm{0}?\Leftrightarrow\:{z}^{\mathrm{2}} \:=−\mathrm{1}−\frac{{i}}{{x}}=\left({i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\right)^{\mathrm{2}} \\ $$$$\mid\mathrm{1}+\frac{{i}}{{x}}\mid=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:}\:\:=\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mid{x}\mid}{and}\:\mathrm{1}+\frac{{i}}{{x}} \\ $$$$=\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mid{x}\mid}\left(\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mid{x}\mid}}\:+\:\frac{\mathrm{1}}{{x}\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mid{x}\mid}}{i}\right)={r}\:{e}^{{i}\theta} \:\:\Rightarrow{cos}\theta=\frac{\mid{x}\mid}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${and}\:{sin}\theta\:=\frac{\mid{x}\mid}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:.{let}\:{suppose}\:{x}>\mathrm{0}\Rightarrow{cos}\theta=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${and}\:{sin}\theta=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\Rightarrow{tan}\theta=\:\frac{\mathrm{1}}{{x}}\:\Rightarrow\theta={arctan}\left(\frac{\mathrm{1}}{{x}}\right){and} \\ $$$$\mathrm{1}+\frac{{i}}{{x}}=\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\:\:=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\:\sqrt{{x}}}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\Rightarrow\:\:{z}=\:+−{i}\:\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\:\sqrt{{x}}}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:{the}\:{poles}\:{are} \\ $$$${z}_{\mathrm{0}} ={i}\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\:\sqrt{{x}}}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\:+{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\right.\right. \\ $$$${z}_{\mathrm{1}} =−{i}\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{\:\sqrt{{x}}}\left({cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)+{isin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\right.\right. \\ $$$${roots}\:{of}\:{z}^{\mathrm{2}} \:+\mathrm{1}\:−\frac{{i}}{{x}}=\mathrm{0}\:\:\:{be}\:{continued}….. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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