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Question Number 197099 by Erico last updated on 07/Sep/23
∫^( (π/2)) _( 0) ((ln(cost))/(sint)) dt=???
$$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{cos}{t}\right)}{\mathrm{sin}{t}}\:\mathrm{d}{t}=??? \\ $$
Answered by witcher3 last updated on 07/Sep/23
=∫_0 ^(π/2) ((sin(t)ln(cos(t)))/(1−cos^2 (t)))dt,cos(t)=y =∫_0 ^1 ((ln(y))/(1−y^2 ))dt=Σ_(n≥0) ∫_0 ^1 y^(2n) ln(t)dt =−Σ_(n≥0) (1/((2n+1)^2 ))−(ζ(2)−(1/4)ζ(2))=−(π^2 /8)
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\left(\mathrm{t}\right)\mathrm{ln}\left(\mathrm{cos}\left(\mathrm{t}\right)\right)}{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \left(\mathrm{t}\right)}\mathrm{dt},\mathrm{cos}\left(\mathrm{t}\right)=\mathrm{y} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{y}\right)}{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\mathrm{dt}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}^{\mathrm{2n}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$=−\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }−\left(\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Answered by Mathspace last updated on 09/Sep/23
tan((t/2))=x ⇒ I=∫_0 ^1 ((ln(((1−x^2 )/(1+x^2 ))))/((2x)/(1+x^2 )))((2dx)/(1+x^2 )) =∫_0 ^1 ((ln(1−x^2 )−ln(1+x^2 ))/x)dx ln^′ (1−u)=((−1)/(1−u))=−Σ_(n=0) ^∞ u^n ⇒ln(1−u)=−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) ⇒ln(1−x^2 )=−Σ_(n=1) ^∞ (x^(2n) /n) ln^′ (1+u)=(1/(1+u))=Σ_(n=0) ^∞ (−1)^n u^n ⇒ln(1+u)=Σ_(n=0) ^∞ (−1)^n (u^(n+1) /(n+1)) =Σ_(n=1) ^∞ (−1)^(n−1) (u^n /n) and ln(1+x^2 )=Σ_(n=1) ^∞ (−1)^(n−1) (x^(2n) /n) ⇒((ln(1−x^2 )−ln(1+x^2 ))/x) =−(1/x)Σ_(n=1) ^∞ (x^(2n) /n)+(1/x)Σ_(n=1) ^∞ (−1)^n (x^(2n) /n) =Σ_(n=1) ^∞ ((−1)^n −1)(x^(2n−1) /n) =−2Σ_(n=1) ^∞ (x^(2(2n+1)−1) /(2n+1)) =−2Σ_(n=1) ^∞ (x^(4n+1) /(2n+1)) and I=−2Σ_(n=1) ^∞ (1/((2n+1)))∫_0 ^1 x^(4n+1) dx =−2Σ_(n=1) ^∞ (1/((2n+1)(4n+2))) =−Σ_(n=1) ^∞ (1/((2n+1)^2 )) Σ(1/n^2 )=(1/4)Σ(1/n^2 )+Σ(1/((2n+1)^2 )) ⇒(3/4)Σ(1/n^2 )=Σ(1/((2n+1)^2 )) ⇒(3/4).(π^2 /6)=Σ(1/((2n+1)^2 )) ⇒(π^2 /8)=Σ(1/((2n+1)^2 )) ⇒ ∫_0 ^(π/2) ((ln(cosx))/(sinx))=−(π^2 /8)
$${tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:\Rightarrow \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$${ln}^{'} \left(\mathrm{1}−{u}\right)=\frac{−\mathrm{1}}{\mathrm{1}−{u}}=−\sum_{{n}=\mathrm{0}} ^{\infty} {u}^{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}−{u}\right)=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{2}{n}} }{{n}} \\ $$$${ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{u}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{u}^{{n}} }{{n}}\:{and} \\ $$$${ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{n}} \\ $$$$\Rightarrow\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$=−\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{2}{n}} }{{n}}+\frac{\mathrm{1}}{{x}}\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}} }{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right)\frac{{x}^{\mathrm{2}{n}−\mathrm{1}} }{{n}} \\ $$$$=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:{and} \\ $$$${I}=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}{n}+\mathrm{1}} {dx} \\ $$$$=−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{2}\right)} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({cosx}\right)}{{sinx}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$

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