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Question Number 135821 by mnjuly1970 last updated on 16/Mar/21

$$....nice.....calculus....$$$$provethat::$$$$\mathit{\varphi }={\int }_0^1\left(\frac{ln(1-x)}{1-\sqrt{1-x}}\right)dx=4(1-\zeta \left(2\right))$$$$$$

Answered by mathmax by abdo last updated on 16/Mar/21

$$\mathrm{\Phi }={\int }_0^1\frac{\log (1-\mathrm{x})}{1-\sqrt{1-\mathrm{x}}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{1-\mathrm{x}}=\mathrm{t}\Rightarrow 1-\mathrm{x}={\mathrm{t}}^2\Rightarrow \mathrm{x}=1-{\mathrm{t}}^2$$$$\mathrm{\Phi }=-{\int }_0^1\frac{\log (1-1+{\mathrm{t}}^2)}{1-\mathrm{t}}(-2\mathrm{t})\mathrm{dt}=4{\int }_0^1\frac{\mathrm{tlog}\left(\mathrm{t}\right)}{1-\mathrm{t}}\mathrm{dt}$$$$=4{\int }_0^1\mathrm{tlogt}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}}\mathrm{dt}=4\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^1{\mathrm{t}}^{\mathrm{n}+1}\mathrm{logt}\mathrm{dt}=4\mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}$$$${\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{t}}^{\mathrm{n}+1}\mathrm{logt}\mathrm{dt}={\left[\frac{{\mathrm{t}}^{\mathrm{n}+2}}{\mathrm{n}+2}\mathrm{logt}\right]}_0^1-\frac{1}{\mathrm{n}+2}{\int }_0^1{\mathrm{t}}^{\mathrm{n}+1}\mathrm{dt}$$$$=-\frac{1}{{(\mathrm{n}+2)}^2}\Rightarrow \mathrm{\Phi }=-4\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+2)}^2}$$$$=-4\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}=-4\{\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2}-1\}$$$$=4-4.\frac{{\pi }^2}{6}=4-\frac{2{\pi }^2}{3}$$

Answered by Dwaipayan Shikari last updated on 16/Mar/21

$${\int }_0^1\frac{log(1-x)}{1-\sqrt{1-x}}dx={\int }_0^1\frac{log\left(x\right)}{1-\sqrt{x}}dx$$$$={\int }_0^1\frac{4tlog\left(t\right)}{1-t}dt$$$$=4{\mathrm{\Phi }}^{\prime }\left(2\right)=4(1-\frac{{\pi }^2}{6})$$$$\mathrm{\Phi }\left(\alpha \right)={\int }_0^1\frac{1-x^{\alpha -1}}{1-x}dx=-\gamma +\psi \left(\alpha \right)$$$${\mathrm{\Phi }}^{\prime }\left(\alpha \right)=-{\int }_0^1\frac{x^{\alpha -1}log\left(x\right)}{1-x}dx={\psi }^1\left(\alpha \right)\Rightarrow {\int }_0^1\frac{xlog\left(x\right)}{1-x}=-{\psi }^1\left(2\right)$$$${\mathrm{\Phi }}^{\prime }\left(2\right)=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1)}^2}=-(\frac{{\pi }^2}{6}-1)$$

Commented by mnjuly1970 last updated on 16/Mar/21

$$thanksalotmrpayan...$$

Answered by Ñï= last updated on 16/Mar/21

$${\int }_0^1\frac{ln(1-x)}{1-\sqrt{1-x}}dx={\int }_0^1\frac{lnx}{1-\sqrt{x}}dx={\int }_0^1\frac{(1-\sqrt{x})lnx}{1-x}dx$$$$={\int }_0^1\frac{lnx}{1-x}dx-{\int }_0^1\frac{\sqrt{x}lnx}{1-x}dx$$$$={\int }_0^1\frac{lnx}{1-x}dx-{\int }_0^1\frac{4x^{2}lnx}{1-x^2}dx(\sqrt{x}\to x)$$$$=-{Li}_2\left(1\right)-4{\int }_0^1(1-\frac{1}{1-x^2})lnxdx$$$$=-{Li}_2\left(1\right)-4{\int }_0^{1}lnxdx+4{\int }_0^1\frac{lnx}{(1-x)(1+x)}dx$$$$=-{Li}_2\left(1\right)+4+2{\int }_0^1(\frac{1}{1-x}+\frac{1}{1+x})lnxdx$$$$=-{Li}_2\left(1\right)+4-2{Li}_2\left(1\right)+2{Li}_2(-1)$$$$=-4{Li}_2\left(1\right)+4$$$$=4(1-\zeta \left(2\right))$$

Commented by mnjuly1970 last updated on 16/Mar/21

$$thankyoudomuch...$$

Answered by mnjuly1970 last updated on 16/Mar/21

$$solution:$$$$y^2=1-x$$$$\mathit{\varphi }=2{\int }_0^1\frac{ln\left(\sqrt{1-x}\right)}{1-\sqrt{1-x}}dx$$$$\mathit{\varphi }=4{\int }_0^1\frac{yln\left(y\right)}{1-y}dy$$$$=-4{\int }_0^{1}ln\left(y\right)dy-4{li}_2\left(y\right)$$$$=4-4\zeta \left(2\right)=4(1-\zeta \left(2\right))...$$$$weknowthat:$$$$1:{li}_2\left(x\right)=-{\int }_0^x\frac{ln(1-z)}{z}dz=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{n^2}$$$$1^{\ast }:{li}_2\left(1\right)=\zeta \left(2\right)$$$$1^{\ast \ast }:{\int }_0^{1}ln\left(t\right)dt=-1$$$$$$

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