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Question Number 123977 by mnjuly1970 last updated on 29/Nov/20
Answered by mathmax by abdo last updated on 29/Nov/20
let A =∫_0 ^(π/2) ln(cosx)dx we have A =−(π/2)ln(2) also A =∫_0 ^(π/2) ln(((e^(ix) +e^(−ix) )/2))dx =∫_0 ^(π/2) ln(e^(ix) +e^(−ix) )dx−(π/2)ln(2) ⇒ ∫_0 ^(π/2) ln(e^(ix) +e^(−ix) )dx=0 ⇒∫_0 ^(π/2) ln(e^(ix) (1+e^(−2ix) ))=0 ⇒ [i(x^2 /2)]_0 ^(π/2) +∫_0 ^(π/2) ln(1+e^(−2ix) )dx we have 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) ⇒ ∫_0 ^(π/2) ln(1+e^(−2ix) )=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^(π/2) e^(−2inx) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)×(1/(−2in))[e^(−2inx) ]_0 ^(π/2) =(i/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ){(−1)^n −1} =−iΣ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ i(π^2 /8)−iΣ_(n=0) ^∞ (1/((2n+1)^2 ))=0 ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 ))=(π^2 /8) we have Σ_(n=1) ^∞ (1/n^2 )=Σ_(n=1) ^∞ (1/(4n^2 ))+Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ (3/4)Σ_(n=1) ^∞ (1/n^2 )=(π^2 /8) ⇒Σ_(n=1) ^∞ (1/n^2 )=(4/3)×(π^2 /8) =(π^2 /6)
$$\mathrm{let}\:\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{A}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:\mathrm{also} \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\frac{\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2}}\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{e}^{\mathrm{ix}} \:+\mathrm{e}^{−\mathrm{ix}} \right)\mathrm{dx}−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{−\mathrm{ix}} \right)\mathrm{dx}=\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{e}^{\mathrm{ix}} \left(\mathrm{1}+\mathrm{e}^{−\mathrm{2ix}} \right)\right)=\mathrm{0}\:\Rightarrow \\ $$$$\left[\mathrm{i}\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2ix}} \right)\mathrm{dx}\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{ln}^{'} \left(\mathrm{1}+\mathrm{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}^{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{u}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2ix}} \right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{e}^{−\mathrm{2inx}} \mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}×\frac{\mathrm{1}}{−\mathrm{2in}}\left[\mathrm{e}^{−\mathrm{2inx}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{2}} }\left\{\left(−\mathrm{1}\right)^{\mathrm{n}} −\mathrm{1}\right\}\:=−\mathrm{i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{i}\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\mathrm{i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{4n}^{\mathrm{2}} }+\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}×\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Commented by mnjuly1970 last updated on 29/Nov/20
thank you so much sir max very nice as always...
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{max} \\ $$$${very}\:{nice}\:{as}\:{always}… \\ $$
Commented by mathmax by abdo last updated on 30/Nov/20
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$
Answered by Dwaipayan Shikari last updated on 30/Nov/20
∫_0 ^(π/2) log(2cosx)dx =(π/2)log(2)−(π/2)log(2)=0 Σ^∞ (1/n^2 ) =(1/1^2 )+(1/2^2 )+.... ((sinx)/x)=(1−(x^2 /π^2 ))(1−(x^2 /(4π^2 )))(1−(x^2 /(9π^2 )))... 1−(x^2 /(3!))+...=1−(x^2 /π^2 )−(x^2 /(4π^2 ))+(x^4 /(4π^4 ))−(x^2 /(9π^2 ))+... (x^2 /(3!))=(x^2 /π^2 )+(x^2 /(4π^2 ))+(x^2 /(9π^2 ))+... 1+(1/2^2 )+(1/3^2 )+...=(π^2 /6)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left(\mathrm{2}{cosx}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+…. \\ $$$$\frac{{sinx}}{{x}}=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{9}\pi^{\mathrm{2}} }\right)… \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+…=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }−\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{4}\pi^{\mathrm{4}} }−\frac{{x}^{\mathrm{2}} }{\mathrm{9}\pi^{\mathrm{2}} }+… \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}=\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\mathrm{9}\pi^{\mathrm{2}} }+… \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

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