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Question Number 34129 by abdo imad last updated on 01/May/18

$$findthevalueof{\int }_0^{1}ln\left(x\right)ln(1+x)dx.$$

Commented by math khazana by abdo last updated on 02/May/18

$$wehaveprovedthat$$$${\int }_0^{1}ln\left(x\right)ln(1+x)dx=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n{(n+1)}^2}$$$$letput{S}_n=\sum _{k=1}^n\frac{{(-1)}^k}{k{(k+1)}^2}andletdecompose$$$$F\left(x\right)=\frac{1}{x{(x+1)}^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$a={lim}_{x\to 0}xF\left(x\right)=1$$$$c={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=-1\Rightarrow$$$$F\left(x\right)=\frac{1}{x}+\frac{b}{x+1}-\frac{1}{{(x+1)}^2}$$$$F\left(1\right)=\frac{1}{4}=1+\frac{b}{2}-\frac{1}{4}\Rightarrow 1=4+2b-1=3+2b$$$$\Rightarrow 2b=-2\Rightarrow b=-1\Rightarrow F\left(x\right)=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}$$$$S_n=\sum _{k=1}^n\frac{{(-1)}^k}{k}-\sum _{k=1}^n\frac{{(-1)}^k}{k+1}-\sum _{k=1}^n\frac{{(-1)}^k}{{(k+1)}^2}$$$$\sum _{k=1}^n\frac{{(-1)}^k}{k}\to \sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^k}{k}=-ln\left(2\right)$$$$\sum _{k=1}^n\frac{{(-1)}^k}{k+1}=\sum _{k=1}^{n+1}\frac{{(-1)}^{k-1}}{k}-1\to \sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k-1}}{k}-1$$$$=ln\left(2\right)-1$$$$\sum _{k=1}^n\frac{{(-1)}^k}{{(k+1)}^2}=\sum _{k=1}^{n+1}\frac{{(-1)}^{k-1}}{k^2}-1$$$$\to \sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k-1}}{k^2}=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$=\frac{{\pi }^2}{8}-\frac{1}{4}\frac{{\pi }^2}{6}=\frac{3{\pi }^2-{\pi }^2}{24}=\frac{{\pi }^2}{12}$$$$S_n=-ln\left(2\right)-ln\left(2\right)+1-\frac{{\pi }^2}{12}+1$$$$S_n=2-2ln\left(2\right)-\frac{{\pi }^2}{12}.$$

Commented by abdo mathsup 649 cc last updated on 02/May/18

$$S_{\mathrm{\infty }}=2-2ln\left(2\right)-\frac{{\pi }^2}{12}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18

$$fromgraphitisclearthat$$$$areaofthecurveisgreaterthanzero$$$$butlessthan0.5$$$$so0.5>\underset{0}{\stackrel{1}{\int }}lnx.ln(1+x)>0$$$$$$$$$$

Commented by math khazana by abdo last updated on 02/May/18

$$sirtanmaythevalueof{\int }_0^{1}ln\left(x\right)ln(1+x)dxmust$$$$benegativebecauseln\left(x\right)⩽0forx\in ]0,1].$$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/May/18

$$yestruebutseethegraph..areaunder$$$$thecurveisnegativeareaso$$$$\underset{0}{\stackrel{1}{\int }}lnx.ln(1+x)dxistheareaunderthecurve$$$$butnegativearea$$$$so0.5>\mid \underset{0}{\stackrel{1}{\int }}lnx.ln(1+x)dx\mid >0$$$$$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/May/18

$$correctionpls$$$$areaunderthecurveisnegetiveareaso$$$$0.5>\mid \underset{0}{\stackrel{1}{\int }}lnx.ln(1+x)dx\mid >0$$$$$$

Answered by MJS last updated on 02/May/18

$$\int f^{\prime }g=fg-\int {fg}^{\prime }$$$$f^{\prime }=\ln (x+1)\Rightarrow f=(x+1)(\ln (x+1)-1)$$$$g=\ln x\Rightarrow g^{\prime }=\frac{1}{x}$$$$fg=(x+1)(\ln (x+1)-1)\ln x$$$$\int {fg}^{\prime }=\int \frac{x+1}{x}(\ln (x+1)-1)dx=$$$$=\int \frac{\ln (x+1)}{x}dx+\int \ln (x+1)dx-\int \frac{1}{x}dx-\int 1dx=$$$$=\int \frac{\ln (x+1)}{x}dx+(x+1)(\ln (x+1)-1)-\ln x-x=$$$$=\int \frac{\ln (x+1)}{x}dx+(x+1)\ln (x+1)-\ln x-2x$$$$$$$$\underset{0}{\stackrel{1}{\int }}\ln x\ln (x+1)dx=$$$$={[(x+1)(\ln x-1)\ln (x+1)-x(\ln x-2)]}_0^1-\underset{0}{\stackrel{1}{\int }}\frac{\ln (x+1)}{x}dx=$$$$\underset{x\to 0^{+}}{\lim }x\ln x=\underset{x\to 0^{+}}{\lim }\frac{\ln x}{\frac{1}{x}}=\left[{\mathrm{l}}^{\prime }\mathrm{Hopital}\right]=$$$$=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\underset{x\to 0^{+}}{\lim }-x=0\Rightarrow$$$$\Rightarrow \underset{x\to 0^{+}}{\lim }(x+1)(\ln x-1)\ln (x+1)-x(\ln x-2)=0$$$$=2-2\ln 2-\underset{0}{\stackrel{1}{\int }}\frac{\ln (x+1)}{x}dx$$$$$$$$\int \frac{\ln (x+1)}{x}dx=$$$$u=-x\to dx=-du$$$$=-\int -\frac{\ln (1-u)}{u}du=-{\mathrm{Li}}_2\left(u\right)=-{\mathrm{Li}}_2(-x)$$$$$$$$\mathrm{polylogarithm}{\mathrm{Li}}_s\left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n^s}$$$$-({\mathrm{Li}}_2(-1)-{\mathrm{Li}}_2\left(0\right))=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{0^n}{n^2}=$$$$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^2}=\frac{{\pi }^2}{12}$$$$$$$$\underset{0}{\stackrel{1}{\int }}\ln x\ln (x+1)dx=2-2\ln 2-\frac{{\pi }^2}{12}\approx -.208761$$

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