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Question Number 95215 by mathmax by abdo last updated on 24/May/20
$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}} \\ $$
Answered by mathmax by abdo last updated on 25/May/20
![let s(x) =Σ_(n=0) ^∞ (((−1)^n )/(4n+1))x^(4n+1) with ∣x∣ ≤1 and 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/(t^4 +1)) +c s(0) =0 =c ⇒s(x) =∫_0 ^x (dt/(t^4 +1)) and Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =∫_0 ^1 (dt/(t^4 +1)) =∫_0 ^1 ((1/t^2 )/(t^2 +(1/t^2 )))dt =(1/2)∫_0 ^1 ((1+(1/t^2 )−1+(1/t^2 ))/(t^2 +(1/t^2 )))dt =(1/2)∫_0 ^1 ((1+(1/t^2 ))/(t^2 +(1/t^2 )))dt−(1/2)∫_0 ^1 ((1−(1/t^2 ))/(t^2 +(1/t^2 )))dt =(1/2) ∫_0 ^1 ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt(→t−(1/t)=u)−(1/2) ∫_0 ^1 ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt(→t+(1/t)=v) =(1/2) ∫_(−∞) ^0 (du/(u^2 +2)) −(1/2) ∫_(+∞) ^2 (dv/(v^2 −2)) we have ∫_(−∞) ^0 (du/(u^2 +2)) =_(u =(√2)α) ∫_(−∞) ^0 (((√2)dα)/(2(1+α^2 ))) =(1/(√2)) [arctanα]_(−∞) ^0 =(1/(√2))((π/2)) =(π/(2(√2))) and ∫_∞ ^2 (dv/(v^2 −2)) =−∫_2 ^∞ (dv/((v−(√2))(v+(√2)))) =−(1/(2(√2)))∫_2 ^(+∞) ((1/(v−(√2)))−(1/(v+(√2))))dv =−(1/(2(√2)))[ln∣((v−(√2))/(v+(√2)))∣]_2 ^(+∞) =−(1/(2(√2)))(−ln(((2−(√2))/(2(√2))))) =(1/(2(√2)))ln((1/(√2))−(1/2)) ⇒ Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =(π/(4(√2))) −(1/(4(√2)))ln((1/(√2))−(1/2))](Q95463.png)
$$\mathrm{let}\:\mathrm{s}\left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\mathrm{x}^{\mathrm{4n}+\mathrm{1}} \:\:\mathrm{with}\:\mid\mathrm{x}\mid\:\leqslant\mathrm{1}\:\mathrm{and}\:\mathrm{x}\neq−\mathrm{1} \\ $$$$\mathrm{s}^{'} \left(\mathrm{x}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{4n}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{s}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:+\mathrm{c}\:\:\:\:\mathrm{s}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\mathrm{c}\:\Rightarrow\mathrm{s}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{x}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{and} \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}}\mathrm{dt}\left(\rightarrow\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{u}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} −\mathrm{2}}\mathrm{dt}\left(\rightarrow\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{v}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{\mathrm{0}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{+\infty} ^{\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{2}}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{−\infty} ^{\mathrm{0}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} \:+\mathrm{2}}\:=_{\mathrm{u}\:=\sqrt{\mathrm{2}}\alpha} \:\:\:\int_{−\infty} ^{\mathrm{0}} \:\frac{\sqrt{\mathrm{2}}\mathrm{d}\alpha}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left[\mathrm{arctan}\alpha\right]_{−\infty} ^{\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{and}\:\:\int_{\infty} ^{\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{2}}\:=−\int_{\mathrm{2}} ^{\infty} \:\frac{\mathrm{dv}}{\left(\mathrm{v}−\sqrt{\mathrm{2}}\right)\left(\mathrm{v}+\sqrt{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{2}} ^{+\infty} \left(\frac{\mathrm{1}}{\mathrm{v}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{v}+\sqrt{\mathrm{2}}}\right)\mathrm{dv}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{ln}\mid\frac{\mathrm{v}−\sqrt{\mathrm{2}}}{\mathrm{v}+\sqrt{\mathrm{2}}}\mid\right]_{\mathrm{2}} ^{+\infty} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(−\mathrm{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{4n}+\mathrm{1}}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$ \\ $$