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Question Number 94269 by LPM last updated on 17/May/20

Commented by prakash jain last updated on 17/May/20

ln (1+x)=Σ_(n=1) ^∞ (x^n /n)(−1)^(n+1) ((ln (1+x))/x)=Σ_(n=0) ^∞ (x^n /(n+1))(−1)^n ∫_0 ^1 Σ_(n=0) ^∞ (x^n /(n+1))(−1)^n = [x−(x^2 /2^2 )+(x^3 /3^2 )−+...]_0 ^1 =1−(1/2^2 )+(1/3^2 )−+...=(π^2 /(12))

$$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} = \\ $$$$\left[{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{{x}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} }−+...\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−+...=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Commented by mathmax by abdo last updated on 17/May/20

I =∫_0 ^1 ((ln(1+x))/x)dx we have ln^′ (1+x) =(1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ ln(1+x) =Σ_(n=0) ^∞ (((−1)^n x^(n+1) )/(n+1)) +c (c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) x^n )/n) ⇒((ln(1+x))/x) =Σ_(n=1) ^∞ (((−1)^(n−1) x^(n−1) )/n) ⇒ I =∫_0 ^1 (Σ_(n=1) ^∞ (((−1)^n )/n) x^(n−1) )dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^1 x^(n−1) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) let δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) we have proved δ(x) =(2^(1−x) −1)ξ(x) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =δ(2) =(2^(1−2) −1)ξ(2) =−(1/2)×(π^2 /6)=−(π^2 /(12)) ⇒ ★I =(π^2 /(12))★

$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:\:{we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\:\left({c}=\mathrm{0}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{x}^{{n}} }{{n}}\:\Rightarrow\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{x}^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{x}^{{n}−\mathrm{1}} \right){dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:{let}\:\:\delta\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} } \\ $$$${we}\:{have}\:{proved}\:\delta\left({x}\right)\:=\left(\mathrm{2}^{\mathrm{1}−{x}} −\mathrm{1}\right)\xi\left({x}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\delta\left(\mathrm{2}\right)\:=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow \\ $$$$\bigstar{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\bigstar \\ $$

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