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Question-130203

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Question Number 130203 by mnjuly1970 last updated on 23/Jan/21
Answered by Dwaipayan Shikari last updated on 23/Jan/21
I(a)−I(β)=∫_0 ^1 (x^a /((1+x)logx))−(x^β /((1+x)log(x)))dx I′(a)−I′(β)=Σ_(n=1) ^∞ (−1)^(n+1) (1/(a+n+1))−Σ_(n=1) ^∞ (−1)^(n+1) (1/(β+n+1)) I(a)−I(β)=Σ_(n=1) ^∞ (−1)^(n+1) (log(a+n+1)−log(β+n+1))+C I(a)−I(β)=Σ_(n=1) ^∞ (−1)^(n+1) log(((a+n+1)/(β+n+1))) (C=0 when a=0 , β=0) I(0)−I(1)=Σ_(n=1) ^∞ (−1)^(n+1) log(((n+1)/(n+2)))=Σ_(n=1) ^∞ (−1)^n log(1+(1/(n+1))) I(a)=∫_0 ^1 ((1−x^a )/((1+x)log(x)))dx ⇒I′(a)=∫_0 ^1 ((−x^a )/((1+x)))dx=−∫_0 ^1 ((x^a −x^(a+1) )/(1−x^2 ))dx =∫_0 ^1 (x^(a+1) /(1−x^2 ))−(x^a /(1−x^2 ))dx x^2 =u =(1/2)∫_0 ^1 ((1−u^((a−1)/2) )/(1−u))−((1−u^(a/2) )/(1−u))du=(1/2)(ψ(((a+1)/2))−ψ(((a+2)/2))) I(a)=(1/2)∫ψ(((a+1)/2))−ψ(((a+2)/2)) da =log(((Γ(((a+1)/2)))/(Γ((a/2)+1))))+C a=0 C=−(1/2)log(π) I(a)=log(((Γ(((a+1)/2)))/(Γ((a/2)+1)(√π))))⇒I(1)=log((2/π))
$${I}\left({a}\right)−{I}\left(\beta\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} }{\left(\mathrm{1}+{x}\right){logx}}−\frac{{x}^{\beta} }{\left(\mathrm{1}+{x}\right){log}\left({x}\right)}{dx} \\ $$$${I}'\left({a}\right)−{I}'\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{a}+{n}+\mathrm{1}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{\beta+{n}+\mathrm{1}} \\ $$$${I}\left({a}\right)−{I}\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({log}\left({a}+{n}+\mathrm{1}\right)−{log}\left(\beta+{n}+\mathrm{1}\right)\right)+{C} \\ $$$${I}\left({a}\right)−{I}\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {log}\left(\frac{{a}+{n}+\mathrm{1}}{\beta+{n}+\mathrm{1}}\right)\:\:\:\:\left({C}=\mathrm{0}\:\:{when}\:{a}=\mathrm{0}\:,\:\beta=\mathrm{0}\right) \\ $$$${I}\left(\mathrm{0}\right)−{I}\left(\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {log}\left(\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {log}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{a}} }{\left(\mathrm{1}+{x}\right){log}\left({x}\right)}{dx}\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{{a}} }{\left(\mathrm{1}+{x}\right)}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −{x}^{{a}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{{x}^{{a}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{u}}−\frac{\mathrm{1}−{u}^{\frac{{a}}{\mathrm{2}}} }{\mathrm{1}−{u}}{du}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}+\mathrm{2}}{\mathrm{2}}\right)\right) \\ $$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}+\mathrm{2}}{\mathrm{2}}\right)\:{da}\:={log}\left(\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}\right)+{C} \\ $$$${a}=\mathrm{0}\:\:\:\:{C}=−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\pi\right) \\ $$$${I}\left({a}\right)={log}\left(\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\sqrt{\pi}}\right)\Rightarrow{I}\left(\mathrm{1}\right)={log}\left(\frac{\mathrm{2}}{\pi}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 23/Jan/21
tayeb allah mr dwaypayan nice way...
$${tayeb}\:{allah}\:{mr}\:{dwaypayan} \\ $$$$\:{nice}\:{way}… \\ $$
Commented by Dwaipayan Shikari last updated on 23/Jan/21
With pleasure
$${With}\:{pleasure} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Jan/21
Another way Σ_(n=1) ^∞ (−1)^n log(1+(1/n))=−log((2/1).(2/3).(3/2).(3/4)...)=log((2/π))
$${Another}\:{way} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {log}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=−{log}\left(\frac{\mathrm{2}}{\mathrm{1}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}…\right)={log}\left(\frac{\mathrm{2}}{\pi}\right) \\ $$

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