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Question Number 59182 by maxmathsup by imad last updated on 05/May/19
let f(x) =arctan(1+ix) determine Re(f(x)) and Im(f(x))dx
$${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{1}+{ix}\right) \\ $$$${determine}\:{Re}\left({f}\left({x}\right)\right)\:{and}\:{Im}\left({f}\left({x}\right)\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 10/Jun/19
we have f^′ (x) =(i/(1+(1+ix)^2 )) =(i/(1+1+2ix −x^2 )) =(i/(−x^2 +2ix +2)) =((−i)/(x^2 −2ix −2)) =((−i)/(x^2 −2−2ix)) =((−i(x^2 −2+2ix))/((x^2 −2)^2 +4x^2 ))=((2x)/((x^2 −2)^2 +4x^2 )) −i(x^2 /((x^2 −2)^2 +4x^2 )) ⇒f(x) =∫ ((2xdx)/((x^2 −2)^2 +4x^2 )) +c_0 −i (∫ ((x^2 dx)/((x^2 −2)^2 +4x^2 )) +c_1 ) ⇒ Re(f(x)) =∫ ((2xdx)/((x^2 −2)^2 +4x^2 )) +c_0 and Im(f(x)) =∫ ((x^2 dx)/((x^2 −2)^2 +4x^2 )) +c_1 c_0 and c_1 reals ∫ ((2xdx)/((x^2 −2)^2 +4x^2 )) =∫ ((2xdx)/(x^4 −4x^2 +4 +4x^2 )) = ∫ ((2x dx)/(x^4 +4)) x^4 +4 =(x^2 )^2 +2^2 =(x^2 +2)^2 −4x^2 =(x^2 +2−2x)(x^2 +2+2x) ⇒ F(x) =((2x)/(x^4 +4)) =((2x)/((x^2 −2x+2)(x^2 +2x +2))) =((ax +b)/(x^2 −2x +2)) +((cx+d)/(x^2 +2x+2)) ⇒ ∫ F(x)dx[=∫ ((ax+b)/(x^2 −2x+2)) dx +∫ ((cx +d)/(x^2 +2x +2))dx =....be continued ....
$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\frac{{i}}{\mathrm{1}+\left(\mathrm{1}+{ix}\right)^{\mathrm{2}} }\:=\frac{{i}}{\mathrm{1}+\mathrm{1}+\mathrm{2}{ix}\:−{x}^{\mathrm{2}} }\:=\frac{{i}}{−{x}^{\mathrm{2}} \:+\mathrm{2}{ix}\:+\mathrm{2}} \\ $$$$=\frac{−{i}}{{x}^{\mathrm{2}} −\mathrm{2}{ix}\:−\mathrm{2}}\:=\frac{−{i}}{{x}^{\mathrm{2}} −\mathrm{2}−\mathrm{2}{ix}}\:=\frac{−{i}\left({x}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}{ix}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }=\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:−{i}\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({x}\right)\:=\int\:\:\:\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:+{c}_{\mathrm{0}} −{i}\:\left(\int\:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:+{c}_{\mathrm{1}} \right)\:\Rightarrow \\ $$$${Re}\left({f}\left({x}\right)\right)\:=\int\:\:\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:+{c}_{\mathrm{0}} \:\:{and}\:{Im}\left({f}\left({x}\right)\right)\:=\int\:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:+{c}_{\mathrm{1}} \\ $$$${c}_{\mathrm{0}} {and}\:{c}_{\mathrm{1}} \:{reals} \\ $$$$\int\:\:\:\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{xdx}}{{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}\:\:+\mathrm{4}{x}^{\mathrm{2}} }\:=\:\int\:\:\:\frac{\mathrm{2}{x}\:{dx}}{{x}^{\mathrm{4}} \:+\mathrm{4}} \\ $$$${x}^{\mathrm{4}} \:\:+\mathrm{4}\:=\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:=\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} \:=\left({x}^{\mathrm{2}} \:+\mathrm{2}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{2}{x}}{{x}^{\mathrm{4}} \:+\mathrm{4}}\:=\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}\right)}\:=\frac{{ax}\:+{b}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{2}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\left[=\int\:\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}\:{dx}\:+\int\:\:\:\frac{{cx}\:+{d}}{{x}^{\mathrm{2}} +\mathrm{2}{x}\:+\mathrm{2}}{dx}\:=….{be}\:{continued}\:….\right. \\ $$
Commented by mathmax by abdo last updated on 22/Jun/19
∫ ((2xdx)/((x^2 −2)^2 +4x^2 )) =_(x^2 =t) ∫ (dt/((t−2)^2 +4t)) =∫ (dt/(t^2 −4t +4 +4t)) =∫ (dt/(4+t^2 )) =_(t =2u) ∫ ((2du)/(4+4u^2 )) =(1/2) ∫ (du/(1+u^2 )) =(1/2) arctan(u)+c =(1/2) arctan((x^2 /2)) +c ⇒ Re(f(x)) =(1/2) arctan((x^2 /(2 )))+c_0
$$\int\:\:\:\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{x}^{\mathrm{2}} }\:=_{{x}^{\mathrm{2}} \:={t}} \:\:\:\:\int\:\:\frac{{dt}}{\left({t}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{4}{t}}\:=\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{4}{t}\:+\mathrm{4}\:+\mathrm{4}{t}}\:=\int\:\:\:\frac{{dt}}{\mathrm{4}+{t}^{\mathrm{2}} } \\ $$$$=_{{t}\:=\mathrm{2}{u}} \:\:\:\:\int\:\:\:\frac{\mathrm{2}{du}}{\mathrm{4}+\mathrm{4}{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({u}\right)+{c}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\:+{c}\:\Rightarrow \\ $$$${Re}\left({f}\left({x}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}\:}\right)+{c}_{\mathrm{0}} \\ $$$$ \\ $$

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