Question Number 34771 by abdo mathsup 649 cc last updated on 10/May/18
$${let}\:{A}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{ix}^{\mathrm{2}} \right){dx} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right)\:\:\:\:\left({x}\in{R}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 13/May/18
![we have 1+ix^2 =(√(1+x^4 )) ((1/( (√(1+x^4 )))) +i(x^2 /( (√(1+x^4 ))))) =r e^(iθ) ⇒r= (√(1+x^4 )) and cosθ =(1/( (√(1+x^4 )))) ,sinθ=(x^2 /( (√(1+x^2 )))) ⇒tanθ =x^2 ⇒θ =arctan(x^2 ) ⇒ 1+ix^2 =(√(1+x^4 )) e^(iarctan(x^2 )) ⇒ ln(1+ix^2 ) = (1/2)ln(1+x^4 ) +i arctan(x^2 ) ⇒ A(x) = (1/2)∫_0 ^1 ln(1+x^4 )dx +i ∫_0 ^1 arctan(x^2 )dx A +iB by parts B = [x arctan(x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 )/(1+x^4 ))dx B = (π/4) − ∫_0 ^1 ((2x^2 )/(1+x^4 ))dx is calculated after decomposition of F(x)= ((2x^2 )/(1+x^4 )) .... ln^′ (1+t) =(1/(1+t)) =Σ_(n=0) ^∞ (−1)^n t^n ⇒ ln(1+t)= Σ_(n=0) ^∞ (((−1)^n t^(n+1) )/(n+1)) =Σ_(n=1) ^∞ (((−1)^(n−1) t^n )/n) ⇒ln(1+x^4 ) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) ⇒ A =(1/2) ∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^(4n) =(1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) (1/(4n+1)) (A/2) = Σ_(n=1) ^∞ (((−1)^(n−1) )/(4n(4n+1))) = Σ_(n=1) ^∞ (−1)^(n−1) {(1/(4n)) −(1/(4n+1))} = (1/4) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −Σ_(n=1) ^∞ (((−1)^(n−1) )/(4n+1)) =(1/4)ln(2) + Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) let w(x) = Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) x^(4n+1) w^′ (x) = Σ_(n=1) ^∞ (−1)^n x^(4n) = Σ_(n=1) ^∞ (−x^4 )^n = (1/(1+x^4 )) ⇒w(x)= ∫_0 ^x (dt/(1+t^4 )) + c but c=w(0)=0 ⇒ Σ_(n=1) ^∞ (((−1)^n )/(4n+1)) = ∫_0 ^1 (dt/(1+t^4 )) ...the value of this integral is known after decoposition... =](https://www.tinkutara.com/question/Q34936.png)
$${we}\:{have}\:\mathrm{1}+{ix}^{\mathrm{2}} =\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:+{i}\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:\:{and}\:\:{cos}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:\:,{sin}\theta=\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{tan}\theta\:={x}^{\mathrm{2}} \:\Rightarrow\theta\:={arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\mathrm{1}+{ix}^{\mathrm{2}} \:=\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:{e}^{{iarctan}\left({x}^{\mathrm{2}} \right)} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{ix}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:+{i}\:{arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${A}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx}\:\:+{i}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{2}} \right){dx} \\ $$$${A}\:+{iB} \\ $$$${by}\:{parts}\:\:{B}\:=\:\left[{x}\:{arctan}\left({x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${B}\:=\:\frac{\pi}{\mathrm{4}}\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:\:{is}\:{calculated}\:{after}\: \\ $$$${decomposition}\:{of}\:{F}\left({x}\right)=\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:…. \\ $$$${ln}^{'} \left(\mathrm{1}+{t}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{t}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {t}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{t}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{t}^{{n}} }{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{x}^{\mathrm{4}} \right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{4}{n}} \:\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{\mathrm{4}{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}} \\ $$$$\frac{{A}}{\mathrm{2}}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{4}{n}\left(\mathrm{4}{n}+\mathrm{1}\right)}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{4}{n}}\:−\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{1}}\right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\:\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)\:\:\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}} \\ $$$${let}\:{w}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:{x}^{\mathrm{4}{n}+\mathrm{1}} \\ $$$${w}^{'} \left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{4}{n}} \:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−{x}^{\mathrm{4}} \right)^{{n}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow{w}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:+\:{c}\:\:\:\:{but}\:{c}={w}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:…{the}\:{value}\:{of}\:{this} \\ $$$${integral}\:{is}\:{known}\:{after}\:{decoposition}… \\ $$$$= \\ $$$$ \\ $$$$ \\ $$