Question Number 141623 by qaz last updated on 21/May/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}+{x}}{dx}\right)^{\mathrm{2}} ={ln}\:\mathrm{2} \\ $$
Answered by mindispower last updated on 21/May/21
![(∫_0 ^1 (x^n /(1+x))dx)^2 =∫_0 ^1 ∫_0 ^1 (((xy)^n )/((1+x)(1+y)))dxdy ⇔Σ_(n≥0) ∫∫_0 ^1 (((xy)^n )/((1+x)(1+y)))dxdy =∫∫_0 ^1 (1/((1+x)(1+y)(1−xy)))dxdy S=∫_0 ^1 (1/((1+y)))∫_0 ^1 (dx/((1+x)(1−xy)))dy ∫_0 ^1 (dx/((1+x)(1−xy)))=∫_0 ^1 (1/((1+y))).(1/(1+x))+(y/(1+y)).(1/(1−xy))dx =((ln(2))/(1+y))−(1/(1+y))ln(1−y) S=∫_0 ^1 ((ln(2))/((1+y)^2 ))−(1/((1+y)^2 ))ln(1−y)dy =[−((ln(2))/(1+y))]_0 ^1 +lim_(x→1) ((ln(1−y))/(1+y))]_0 ^x +∫_0 ^x (1/((1+y)(1−y)))dy =((ln(2))/2)+lim_(x→1) ((ln(1−x))/(1+x))+(1/2)ln(((1+x)/(1−x))) =((ln(2))/2)+lim_(x→1) ((ln(1+x))/2)+ln(1−x){(1/(1+x))−(1/2)} =ln(2)+lim_(x→1) (((1−x)ln(1−x))/(2(1+x)))=ln(2) ⇔Σ_(n≥0) (∫_0 ^1 (x^n /(1+x))dx)^2 =ln(2)](https://www.tinkutara.com/question/Q141630.png)
$$\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}+{x}}{dx}\right)^{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({xy}\right)^{{n}} }{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)}{dxdy} \\ $$$$\Leftrightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\int\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({xy}\right)^{{n}} }{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)}{dxdy} \\ $$$$=\int\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{y}\right)\left(\mathrm{1}−{xy}\right)}{dxdy} \\ $$$${S}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{xy}\right)}{dy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{xy}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)}.\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{{y}}{\mathrm{1}+{y}}.\frac{\mathrm{1}}{\mathrm{1}−{xy}}{dx} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{1}+{y}}−\frac{\mathrm{1}}{\mathrm{1}+{y}}{ln}\left(\mathrm{1}−{y}\right) \\ $$$${S}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{2}\right)}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} }{ln}\left(\mathrm{1}−{y}\right){dy} \\ $$$$\left.=\left[−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{1}+{y}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}−{y}\right)}{\mathrm{1}+{y}}\right]_{\mathrm{0}} ^{{x}} +\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}}{\left(\mathrm{1}+{y}\right)\left(\mathrm{1}−{y}\right)}{dy} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{2}}+{ln}\left(\mathrm{1}−{x}\right)\left\{\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$={ln}\left(\mathrm{2}\right)+\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}\left(\mathrm{1}+{x}\right)}={ln}\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}+{x}}{dx}\right)^{\mathrm{2}} ={ln}\left(\mathrm{2}\right) \\ $$
Commented by qaz last updated on 22/May/21
$${thanks}\:{sir}\:{power} \\ $$