Question Number 53795 by maxmathsup by imad last updated on 25/Jan/19
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{9}}\:−\frac{\mathrm{1}}{\mathrm{13}}\:+…. \\ $$
Commented by maxmathsup by imad last updated on 26/Jan/19
![let S(x)=Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) x^(4n+1) with∣x∣<1 due to uniform convergence we have (d/dx)S(x) =Σ_(n=0) ^∞ (−1)^n x^(4n) =Σ_(n=0) ^∞ (−x^4 )^n =(1/(1+x^4 )) ⇒S(x)=∫_0 ^x (dt/(t^4 +1)) +c c=S(0)=0 ⇒S(x)=∫_0 ^x (dt/(t^4 +1)) and Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =S(1)=∫_0 ^1 (dt/(t^4 +1)) let decompose F(t) =(1/(t^4 +1)) ⇒ F(t)=(1/((t^2 +1)^2 −2t^2 )) =(1/((t^2 +(√2)t +1)(t^2 −(√2)t +1))) =((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)) +((−bt +c)/(t^2 +(√2)t +1)) =F(t) ⇒c=−a and d=b ⇒ 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))) +((−a+b)/(2−(√2))) ⇒ (1/2) =(((2−(√2))(a+(1/2))+(2+(√2))(−a+(1/2)))/2) ⇒ −2(√2)a +(1/(2 ))(4) =1 ⇒ −2(√2)a =−1 ⇒a =(1/(2(√2))) ⇒ F(t) = (((t/(2(√2)))+(1/2))/(t^2 +(√2)t +1)) +((−(t/(2(√2)))+(1/2))/(t^2 −(√2)t +1)) =(1/(2(√2))){ ((t+(√2))/(t^2 +(√2)t +1)) +((−t +(√2))/(t^2 −(√2)t +1))} ⇒ ∫_0 ^1 (dt/(t^4 +1)) =((√2)/4){ ∫_0 ^1 ((t+(√2))/(t^2 +(√2)t +1))dt − ∫_0 ^1 ((t−(√2))/(t^2 −(√2)t +1))} =((√2)/4){ (1/2) ∫_0 ^1 ((2t +(√2) +(√2))/(t^2 +(√2)t +1)) dt −(1/2)∫_0 ^1 ((2t−(√2)−(√2))/(t^2 −(√2)t +1)) dt} =((√2)/8){ [ln(t^2 +(√2)t +1)]_0 ^1 −[ln(t^2 −(√2)t +1)]_0 ^1 +(√2)∫_0 ^1 (dt/(t^2 +(√2)t +1)) +(√2)∫_0 ^1 (dt/(t^2 −(√2)t +1))} =((√2)/8){ ln(2+(√2))−ln(2−(√2))} +(1/4){ ∫_0 ^1 (dt/(t^2 +(√2)t +1)) +∫_0 ^1 (dt/(t^2 −(√2)t +1))} but ∫_0 ^1 (dt/(t^2 +(√2)t +1)) = ∫_0 ^1 (dt/(t^2 +2 ((√2)/2)t +(1/2)+(1/2))) =∫_0 ^1 (dt/((t+(1/( (√2))))^2 +(1/2))) =_(t +(1/( (√2)))=(u/( (√2)))) ∫_1 ^(1+(√2)) (du/( (√2)(1/2)(1+u^2 ))) =(√2) ∫_1 ^(1+(√2)) (du/u) =(√2){arctan((√2)+1)−(π/4)} ∫_0 ^1 (dt/(t^2 −(√2)t +1)) = ∫_0 ^1 (dt/((t−(1/( (√2))))^2 +(1/2))) =_(t−(1/( (√2)))=(u/( (√2)))) ∫_(−1) ^((√2)−1) (du/( (√2)(1/2)(1+u^2 ))) =(√2){arctan((√2)−1) +(π/4)} ⇒ ∫_0 ^1 (dt/(t^4 +1)) =((√2)/8)ln(((2+(√2))/(2−(√2)))) +((√2)/4){ arctan((√2)+1) +arctan((√2)−1)} but arctan((√2)−1) =arctan( (1/( (√2)+1)))=(π/2) −arctan((√2)+1) ⇒ ∫_0 ^1 (dt/(t^4 +1)) =((√2)/8)ln(((2+(√2))/(2−(√2)))) +((π(√2))/8) ⇒ Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =((√2)/8)ln(((2+(√2))/(2−(√2))))+((π(√2))/8) .](https://www.tinkutara.com/question/Q53887.png)
$${let}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:{x}^{\mathrm{4}{n}+\mathrm{1}} \:\:{with}\mid{x}\mid<\mathrm{1}\:\:{due}\:{to}\:{uniform}\:{convergence}\:{we}\:{have} \\ $$$$\frac{{d}}{{dx}}{S}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{4}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{4}} \right)^{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow{S}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:+{c} \\ $$$${c}={S}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{S}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:={S}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)\:=\frac{\mathrm{1}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow\:{F}\left({t}\right)=\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)} \\ $$$$=\frac{{at}+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${F}\left(−{t}\right)\:={F}\left({t}\right)\:\Rightarrow\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{bt}\:+{c}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:={F}\left({t}\right)\:\Rightarrow{c}=−{a}\:{and}\:{d}={b}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{at}\:+{b}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\mathrm{2}{b}\:\Rightarrow{b}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{a}\:+{b}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{−{a}+{b}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left(−{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$−\mathrm{2}\sqrt{\mathrm{2}}{a}\:\:+\frac{\mathrm{1}}{\mathrm{2}\:}\left(\mathrm{4}\right)\:=\mathrm{1}\:\Rightarrow\:−\mathrm{2}\sqrt{\mathrm{2}}{a}\:=−\mathrm{1}\:\Rightarrow{a}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\:\frac{\frac{{t}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−\frac{{t}}{\mathrm{2}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\:\:\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\frac{−{t}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}{dt}\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\right\} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left\{\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{t}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:{dt}\right\} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left\{\:\:\left[{ln}\left({t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\left[{ln}\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:+\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\right. \\ $$$$\left.+\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\right\} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\left\{\:{ln}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)−{ln}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\right\}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{t}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{{u}}{\:\sqrt{\mathrm{2}}}} \:\:\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{du}}{\:\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\sqrt{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{\mathrm{2}}} \:\frac{{du}}{{u}}\:=\sqrt{\mathrm{2}}\left\{{arctan}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right\} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}\:+\mathrm{1}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:=_{{t}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{{u}}{\:\sqrt{\mathrm{2}}}} \:\:\:\:\int_{−\mathrm{1}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{{du}}{\:\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\sqrt{\mathrm{2}}\left\{{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:+\frac{\pi}{\mathrm{4}}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left\{\:{arctan}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:+{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right\}\:{but} \\ $$$${arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:={arctan}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\:+\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:\Rightarrow\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)+\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:. \\ $$
Answered by Smail last updated on 26/Jan/19
![(1/(1−x))=Σ_(n=0) ^∞ x^n So (1/(1+x^4 ))=Σ_(n=0) ^∞ (−1)^n x^(4n) let p(x)=∫_0 ^x (dt/(1+t^4 ))=Σ_(n=0) ^∞ (−1)^n ∫_0 ^x t^(4n) dt=Σ_(n=0) ^∞ (((−1)^n )/(4n+1))x^(4n+1) p(1)=Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) p(x)=∫_0 ^x (dt/(1+t^4 ))=∫_0 ^x (dt/((t^2 +(√2)t+1)(t^2 −(√2)t+1))) (1/(1+t^4 ))=((at+b)/(t^2 +(√2)t+1))+((ct+d)/(t^2 −(√2)t+1)) a=−c=((√2)/4) ; d=b=(1/2) p(x)=((√2)/4)∫_0 ^x (((t+(√2))/(t^2 +(√2)t+1))−((t−(√2))/(t^2 −(√2)t+1)))dt p(1)=((√2)/2)∫_0 ^1 (((t+(√2))/(((√2)t+1)^2 +1))−((t−(√2))/(((√2)t−1)^2 +1)))dt u=(√2)t+1 ⇒du=(√2)dt y=(√2)t−1⇒dy=(√2)dt p(1)=((√2)/4)(∫_1 ^((√2)+1) ((u+1)/(u^2 +1))dy−∫_(−1) ^((√2)−1) ((y−1)/(y^2 +1))dy) =((√2)/4)([(1/2)ln(u^2 +1)+tan^(−1) (u)]_1 ^((√2)+1) −[(1/2)ln(y^2 +1)−tan^(−1) (y)]_(−1) ^((√2)−1) ) =((√2)/4)((1/4)ln2+(1/2)ln((√2)+1)+tan^(−1) ((√2)+1)−(π/4)−((1/4)ln(2)+(1/2)ln((√2)−1)−tan^(−1) ((√2)−1)−(π/4))) =((√2)/4)(ln((√2)+1)+tan^(−1) ((√2)+1)+tan^(−1) ((√2)−1)) =((√2)/4)(ln((√2)+1)+((3π)/8)+(π/8)) Σ_(n=0) ^∞ (((−1)^n )/(4n+1))=p(1)=((√2)/4)(ln((√2)+1)+(π/2))](https://www.tinkutara.com/question/Q53810.png)
$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \\ $$$${So}\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{4}{n}} \\ $$$${let}\:{p}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{4}{n}} {dt}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}{x}^{\mathrm{4}{n}+\mathrm{1}} \\ $$$${p}\left(\mathrm{1}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}} \\ $$$${p}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }=\int_{\mathrm{0}} ^{{x}} \frac{{dt}}{\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }=\frac{{at}+{b}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}+\frac{{ct}+{d}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}} \\ $$$${a}=−{c}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:;\:\:{d}={b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}\left({x}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \left(\frac{{t}+\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\mathrm{1}}−\frac{{t}−\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\mathrm{1}}\right){dt} \\ $$$${p}\left(\mathrm{1}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{t}+\sqrt{\mathrm{2}}}{\left(\sqrt{\mathrm{2}}{t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}−\frac{{t}−\sqrt{\mathrm{2}}}{\left(\sqrt{\mathrm{2}}{t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$${u}=\sqrt{\mathrm{2}}{t}+\mathrm{1}\:\Rightarrow{du}=\sqrt{\mathrm{2}}{dt} \\ $$$${y}=\sqrt{\mathrm{2}}{t}−\mathrm{1}\Rightarrow{dy}=\sqrt{\mathrm{2}}{dt} \\ $$$${p}\left(\mathrm{1}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}+\mathrm{1}} \frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}{dy}−\int_{−\mathrm{1}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{{y}−\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}{dy}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} +\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left({u}\right)\right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}+\mathrm{1}} −\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left({y}\right)\right]_{−\mathrm{1}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}−\left(\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left({ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left({ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\frac{\mathrm{3}\pi}{\mathrm{8}}+\frac{\pi}{\mathrm{8}}\right) \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{1}}={p}\left(\mathrm{1}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left({ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\frac{\pi}{\mathrm{2}}\right) \\ $$
Commented by maxmathsup by imad last updated on 26/Jan/19
$${thank}\:{you}\:{sir}\:. \\ $$