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Question Number 114045 by mnjuly1970 last updated on 17/Sep/20

$$...advancedcalculus...$$$$$$$$i:provethat::$$$${\int }_0^1\frac{ln(1+ln(1-x))}{ln(1-x)}dx\stackrel{?}{=}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+1)}{n^2}$$$$ii:$$$$provethat::$$$$\mathrm{\Omega }={\int }_0^1\frac{ln(1+x)}{x(1+x^2)}dx\stackrel{?}{=}\frac{5{\pi }^2}{48}$$$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$

Answered by Dwaipayan Shikari last updated on 16/Sep/20

$${\int }_0^1\frac{log(1+x)}{x(1+x)}={\int }_0^1\frac{log(1+x)}{x}-\frac{log(1+x)}{(1+x)}dx$$$$={\int }_0^1{(-1)}^n\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n-1}}{n}-{\int }_1^2\frac{logu}{u}du$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{1}{n^2}-\frac{1}{2}{\left[{\left(logu\right)}^2\right]}_1^2=\frac{{\pi }^2}{12}-\frac{1}{2}{\left(log2\right)}^2$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{I}={\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}(1+\mathrm{x})}\mathrm{dx}={\int }_0^1(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1})\ln (1+\mathrm{x})\mathrm{dx}$$$$={\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}-{\int }_0^1\frac{\ln (1+\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}\mathrm{but}$$$${\int }_0^1\frac{\ln (1+\mathrm{x})}{1+\mathrm{x}}\mathrm{dx}{=}_{1+\mathrm{x}=\mathrm{t}}{\int }_1^2\frac{\ln \left(\mathrm{t}\right)}{\mathrm{x}>\mathrm{t}}\mathrm{dt}={\left[\frac{1}{2}{\ln }^2\mathrm{t}\right]}_1^2=\frac{{\ln }^2\left(2\right)}{2}$$$$\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \ln (1+\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \frac{\ln (1+\mathrm{x})}{\mathrm{x}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}-1}}{\mathrm{n}}\Rightarrow$$$${\int }_0^1\frac{\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^2}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}$$$$=-\{2^{1-2}-1\}\xi \left(2\right)=\frac{{\pi }^2}{12}\Rightarrow \mathrm{I}=\frac{1}{2}{\ln }^2\left(2\right)-\frac{{\pi }^2}{12}$$$$$$

Commented by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{sorry}\mathrm{I}=\frac{{\pi }^2}{12}-\frac{1}{2}{\ln }^2\left(2\right)$$

Commented by mnjuly1970 last updated on 17/Sep/20

$$thankyousir$$$$thatwasmymistake$$$$denominatorisx(1+x^2)$$$$ieditedit.$$

Answered by mathmax by abdo last updated on 16/Sep/20

$$\mathrm{A}={\int }_0^1\frac{\ln (1+\ln (1-\mathrm{x}))}{\ln (1-\mathrm{x})}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{tbe}\mathrm{changement}\ln (1-\mathrm{x})=-\mathrm{t}\Rightarrow$$$$1-\mathrm{x}={\mathrm{e}}^{-\mathrm{t}}\Rightarrow \mathrm{x}=1-{\mathrm{e}}^{-\mathrm{t}}\Rightarrow \mathrm{A}={\int }_0^{+\mathrm{\infty }}\frac{\ln (1+\mathrm{t})}{-\mathrm{t}}\left({\mathrm{e}}^{-\mathrm{t}}\right)\mathrm{dt}$$$$=-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}\ln (1+\mathrm{t})}{\mathrm{t}}\mathrm{dt}=-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{t}}^{\mathrm{n}-1}}{\mathrm{n}}\right)\mathrm{dt}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{n}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{\mathrm{\Gamma }(\mathrm{n}-1)}{\mathrm{n}}-1$$

Commented by mnjuly1970 last updated on 17/Sep/20

$$A=-{\int }_0^{\mathrm{\infty }}\frac{ln(1-t)e^{-t}}{t}dtis$$$$correct.pleasecheckit$$$$.thankyousomuchfor$$$$youreffort.$$

Answered by mindispower last updated on 17/Sep/20

$$i){\int }_0^1\frac{ln(1+ln(1-x))}{ln(1-x)}dx={\int }_0^1\frac{ln(1+ln\left(x\right))}{ln\left(x\right)}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{ln(1-x)}{-x}{e}^{-x}dxsomthingwrong$$$$ln(1-x)\in \mathbb{C}x>1$$$$$$$$$$

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