Question Number 32025 by abdo imad last updated on 18/Mar/18
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:. \\ $$
Commented by abdo imad last updated on 20/Mar/18
![let put S=Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) and S(x)= Σ_(n=0) ^∞ (−1)^n (x^(4n+1) /(4n+1)) the radius of S(x)is r=1 and for ∣x∣ ≤1 S^′ (x)=Σ_(n=0) ^∞ (−1)^n x^(4n) =Σ_(n=0) ^∞ (−x^4 )^n = (1/(1+x^4 )) ⇒ S(x)= ∫_0 ^x (dt/(1+t^4 )) +λ but λ=S(0)=0 ⇒S(x)=∫_0 ^x (dt/(1+t^4 )) let decompose f(t) = (1/(1+t^4 )) f(t)= (1/((1+t^2 )^2 −2t^2 )) = (1/((t^2 +1 −(√2) t)(t^2 +1 +(√2) t))) = ((at +b)/((t^2 −(√2) t +1))) + ((ct+d)/(t^2 +(√2) t +1)) ′ f(−t)=f(t) ⇒ ((−at +b)/(t^2 +(√2) t +1)) + ((−ct+d)/(t^2 −(√2)t +1)) =f(t) ⇒c=−a and b=d ⇒ f(t)= ((at+b)/(t^2 −(√2)t +1)) +((−at +b)/(t^2 +(√2)t +1)) f(0)=1= 2b ⇒ b=(1/2) f(1) =(1/2) = ((a+b)/(2−(√2))) + ((b−a)/(2+(√2))) ⇔(2+(√2))(a+b)+(2−(√2))(b−a)=1 ⇒2a +2b +(√2) a +(√2) b +2b −2a −(√2)b +(√2) a=1 ⇒ 2(√2) a +4b =1 ⇒2(√2) a +2=1 ⇒2(√2) a =−1 ⇒a= ((−1)/(2(√(2 )))) . f(t)= ((((−1)/(2(√2)))t +(1/2))/(t^2 −(√2)t +1)) + (((1/(2(√2)))t +(1/2))/(t^2 +(√2) t +1)) f(t) = (1/(2(√2))) ((−t+(√2))/(t^2 −(√2) t +1)) +(1/(2(√2))) ((t+(√2))/(t^2 +(√2) t +1)) ∫_0 ^x (dt/(1+t^4 )) =(1/(2(√2))) ∫_0 ^x ((−t +(√2))/(t^2 −(√2) t +1)) dt +(1/(2(√2))) ∫_0 ^x ((t+(√2))/(t^2 +(√2) t +1))dt ∫_0 ^(x ) ((−t +(√2))/(t^2 −(√2) t+1)) dt=− (1/2) ∫_0 ^x ((2t −2(√2))/(t^2 −(√2) t +1)) dt =−(1/2) ∫_0 ^x ((2t−(√2))/(t^2 −(√2) t +1)) +((√2)/2) ∫_0 ^x (dt/(t^2 −(√2) t +1)) ∫_0 ^x ((2t−(√2))/(t^2 −(√2) t+1))dt =[ln(t^2 −(√2) t+1)]_0 ^x =ln(x^2 −(√2) x +1) ∫_0 ^x (dt/(t^2 −(√2) t+1)) = ∫_0 ^x (dt/(t^2 −2((√2)/2) t +(1/2) +(3/2))) = ∫_0 ^x (dt/((t −((√2)/2))^2 +(3/4))) =_(t−((√2)/2)= ((√3)/2) u) ∫_(−((√2)/( (√3)))) ^((2/( (√3)))(x−((√2)/2))) (1/((3/4)(1+u^2 ))) ((√3)/2)du =(4/3) ((√3)/2) ∫_((√2)/( (√3))) ^( ((2x−(√2))/( (√3)))) (du/(1+u^2 )) =((2(√3))/3) (arctan(((2x−(√2))/( (√3))))−arctan(((√2)/( (√3))))) ∫_0 ^x ((t+(√2))/(t^2 +(√2) t +1)) dt= (1/2) ∫_0 ^(x ) ((2t +2(√2))/(t^2 +(√2) t +1)) dt = (1/2) ∫_0 ^x ((2t +(√2))/(t^2 +(√2) t+1)) dt +((√2)/2) ∫_0 ^x (dt/(t^2 +(√2) t +1)) but ∫_0 ^x ((2t +(√2))/(t^2 +(√2) t+1)) dt =[ln(t^2 +(√2) t+1)]_0 ^x =ln(x^2 +(√2) x +1) ∫_0 ^x (dt/(t^2 +(√2) t+1)) =((2(√3))/3)(arctan(((2x+(√2))/( (√3))))−arctan(((√2)/( (√3))))) 2(√2) ∫_0 ^x (dt/(1+t^4 )) =−(1/2)ln(x^2 −(√2) x+1) + ((√6)/3)(arctan(((2x−(√2))/( (√3))))−arctan(((√2)/( (√3))))) +(1/2)ln(x^2 +(√2) x+1) +((√6)/3)(arctan(((2x+(√2))/( (√3))))−arctan(((√2)/( (√3)))))](https://www.tinkutara.com/question/Q32116.png)
$${let}\:{put}\:{S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:{and}\:{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\:\frac{{x}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{radius}}\:\boldsymbol{{of}}\:\boldsymbol{{S}}\left({x}\right){is}\:{r}=\mathrm{1}\:\:{and}\:{for}\:\mid{x}\mid\:\leqslant\mathrm{1} \\ $$$${S}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{4}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{x}^{\mathrm{4}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$${S}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:+\lambda\:\:\:{but}\:\lambda={S}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{S}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$${let}\:{decompose}\:{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$${f}\left({t}\right)=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}{t}^{\mathrm{2}} }\:=\:\:\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\:−\sqrt{\mathrm{2}}\:{t}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\:+\sqrt{\mathrm{2}}\:{t}\right)} \\ $$$$=\:\frac{{at}\:+{b}}{\left({t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}\right)}\:+\:\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:'\:{f}\left(−{t}\right)={f}\left({t}\right)\:\Rightarrow \\ $$$$\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:+\:\frac{−{ct}+{d}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:={f}\left({t}\right)\:\Rightarrow{c}=−{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${f}\left({t}\right)=\:\frac{{at}+{b}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}=\:\mathrm{2}{b}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:+\:\frac{{b}−{a}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:\Leftrightarrow\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left({a}+{b}\right)+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left({b}−{a}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{a}\:+\mathrm{2}{b}\:+\sqrt{\mathrm{2}}\:{a}\:+\sqrt{\mathrm{2}}\:{b}\:+\mathrm{2}{b}\:−\mathrm{2}{a}\:−\sqrt{\mathrm{2}}{b}\:+\sqrt{\mathrm{2}}\:{a}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\:{a}\:+\mathrm{4}{b}\:=\mathrm{1}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}\:{a}\:+\mathrm{2}=\mathrm{1}\:\Rightarrow\mathrm{2}\sqrt{\mathrm{2}}\:{a}\:=−\mathrm{1}\:\Rightarrow{a}=\:\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}}\:. \\ $$$${f}\left({t}\right)=\:\:\frac{\frac{−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:\:+\:\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}} \\ $$$${f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\frac{−{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{−{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}{dt} \\ $$$$\int_{\mathrm{0}} ^{{x}\:} \:\:\:\frac{−{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}\:{dt}=−\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{2}{t}\:−\mathrm{2}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{\mathrm{2}{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{\mathrm{2}{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}{dt}\:=\left[{ln}\left({t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{{x}} \\ $$$$={ln}\left({x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{x}\:+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{{x}} \:\:\:\:\:\:\:\frac{{dt}}{\left({t}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{t}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{u}} \:\:\:\int_{−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} ^{\:\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\left({arctan}\left(\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{dt}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}\:} \:\:\frac{\mathrm{2}{t}\:+\mathrm{2}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{\mathrm{2}{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}\:{dt}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}\:+\mathrm{1}}\:{but} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{\mathrm{2}{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}\:{dt}\:=\left[{ln}\left({t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{{x}} \:={ln}\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{x}\:+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{t}+\mathrm{1}}\:=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left({arctan}\left(\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{x}+\mathrm{1}\right)\:+\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\left({arctan}\left(\frac{\mathrm{2}{x}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}\:{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\left({arctan}\left(\frac{\mathrm{2}{x}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$
Commented by abdo imad last updated on 20/Mar/18
$${be}\:{continued}…. \\ $$