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Question Number 221099 by Nicholas666 last updated on 24/May/25
Prove: ∫_( −π) ^( π) Σ_(n=0) ^∞ ((cos^(n + 1) x)/((n + 1)(1 + e^x^(2n +1) ))) dx = π ln2
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}: \\ $$$$\:\:\underset{\:−\pi} {\overset{\:\pi} {\int}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{cos}^{{n}\:+\:\mathrm{1}} \:{x}}{\left({n}\:+\:\mathrm{1}\right)\left(\mathrm{1}\:+\:{e}^{{x}^{\mathrm{2}{n}\:+\mathrm{1}} } \right)}\:\:\mathrm{d}{x}\:=\:\pi\:\mathrm{ln2}\:\:\:\:\: \\ $$$$ \\ $$
Answered by MrGaster last updated on 25/May/25
=Σ_(n=0) ^∞ (1/(n+1))∫_(−π) ^π ((cos^(n+1) x)/(1+e^x^(2n+1) ))dx =Σ_(n=1) ^∞ (1/(n+1))(∫_0 ^π ((cos^(n+1) x)/(1+e^x^(2n+1) ))dx+∫_(−π) ^0 ((cos^(n+1) x)/(1+e^x^(2n+1) )))dx =Σ_(n=1) ^∞ (1/(n+1))∫_0 ^π cos^(n+1) x((1/(1+e^x^(2n+1) ))+(1/(1+e^(−x^(2n+1) ) )))dx =Σ_(n=1) ^∞ (1/(n+1))∫_(0 ) ^π cos^(n+1) x dx =Σ_(n=0) ^∞ (1/k)∫_0 ^π cos^k x dx (k=n+1) =Σ_(m=1) ^∞ (1/(2m))∫_0 ^π cos^(2m) x dx (odd-order terms are zero) =πΣ_(m=1) ^∞ (((2m−1)!!)/(2m∙2^(2m) m!!)) =πΣ_(m=1) ^∞ (1/(2m))∙(((2m)!)/(2^(2m) (m!)^2 )) =(π/2)Σ_(m=1) ^∞ (((2m)!)/(m∙4m(m!)^2 )) =(π/2)∙ 2 ln 2 =π ln 2
$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{−\pi} ^{\pi} \frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}{n}+\mathrm{1}} } }{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}{n}+\mathrm{1}} } }{dx}+\int_{−\pi} ^{\mathrm{0}} \frac{\mathrm{cos}^{{n}+\mathrm{1}} {x}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}{n}+\mathrm{1}} } }\right){dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{{n}+\mathrm{1}} {x}\left(\frac{\mathrm{1}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}{n}+\mathrm{1}} } }+\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}^{\mathrm{2}{n}+\mathrm{1}} } }\right){dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}\:} ^{\pi} \mathrm{cos}^{{n}+\mathrm{1}} {x}\:{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{{k}} {x}\:{dx}\:\left({k}={n}+\mathrm{1}\right) \\ $$$$=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{m}}\int_{\mathrm{0}} ^{\pi} \mathrm{cos}^{\mathrm{2}{m}} {x}\:{dx}\:\left(\mathrm{odd}-\mathrm{order}\:\mathrm{terms}\:\mathrm{are}\:\mathrm{zero}\right) \\ $$$$=\pi\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{m}−\mathrm{1}\right)!!}{\mathrm{2}{m}\centerdot\mathrm{2}^{\mathrm{2}{m}} {m}!!} \\ $$$$=\pi\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{m}}\centerdot\frac{\left(\mathrm{2}{m}\right)!}{\mathrm{2}^{\mathrm{2}{m}} \left({m}!\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{m}\right)!}{{m}\centerdot\mathrm{4}{m}\left({m}!\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2}}\centerdot\:\mathrm{2}\:\mathrm{ln}\:\mathrm{2} \\ $$$$=\pi\:\mathrm{ln}\:\mathrm{2} \\ $$
Commented by Nicholas666 last updated on 25/May/25
nice ones your solution Sir
$$\mathrm{nice}\:\mathrm{ones}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{Sir} \\ $$

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