Question Number 49343 by maxmathsup by imad last updated on 05/Dec/18
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}\:. \\ $$
Commented by Abdo msup. last updated on 08/Dec/18
![let give this integral at form of serie I =∫_0 ^1 ((ln(x))/(1+x))dx =∫_0 ^1 ln(x)(Σ_(n=0) ^∞ (−1)^n x^n ) =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n ln(x)dx=Σ_(n=0) ^∞ (−1)^n A_n by parts A_n =∫_0 ^1 x^n ln(x)dx =[(1/(n+1))x^(n+1) ln(x)]_0 ^1 −∫_0 ^1 (1/(n+1)) x^n dx =−(1/((n+1)^2 )) ⇒ I =Σ_(n=0) ^∞ (((−1)^(n+1) )/((n+1)^2 )) =Σ_(n=1) ^∞ (((−1)^n )/n^2 ) but Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4)(π^2 /6) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) but Σ_(n=1) ^∞ (1/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)(π^2 /6) =(π^2 /8) ⇒ I =(π^2 /(24)) −(π^2 /8) =((π^2 −3π^2 )/(24)) =−(π^2 /(12)) .](Q49581.png)
$${let}\:{give}\:{this}\:{integral}\:{at}\:{form}\:{of}\:{serie} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left({x}\right){dx}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{A}_{{n}} \\ $$$${by}\:{parts}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {ln}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${but}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow \\ $$$${I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\frac{\pi^{\mathrm{2}} −\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$$$ \\ $$$$ \\ $$