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Double integral of Logarithmic function

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Post Sat Oct 08, 2016 11:51 pm

Posts: 98
By using mathematica tool, we can obtain
\[\int\limits_0^1 {\frac{{\ln x}}{{1 - \frac{x}{2}}}\left( {\int\limits_0^{\frac{x}{2}} {\frac{1}{t}dt} } \right)} dx = \frac{7}{2}\zeta \left( 3 \right) - \zeta \left( 2 \right)\ln 2 - \frac{1}{3}{\ln ^3}2,\]
\[\int\limits_0^1 {\frac{{\ln x}}{{1 - \frac{x}{2}}}\left( {\int\limits_0^{\frac{x}{2}} {\frac{{\ln \left( {1 - t} \right)}}{t}dt} } \right)} dx = \frac{1}{8}\zeta \left( 4 \right) - \frac{1}{2}\zeta \left( 2 \right){\ln ^2}2 + \frac{1}{{12}}{\ln ^4}2 + \frac{1}{2}\zeta \left( 3 \right)\ln 2.\]
In general, the following integral of logarthmic fuction is wether or not can be expressed in terms of a linear rational combination of $ln2$, zeta values and polylogarithms?
\[\int\limits_0^1 {\frac{{{{\ln }^k}x}}{{1 - \frac{x}{2}}}\left( {\int\limits_0^{\frac{x}{2}} {\frac{{{{\ln }^m}\left( {1 - t} \right)}}{t}dt} } \right)} dx = ?\]
Here $m$ and $k$ are positive integers.


Posts: 8
$$\int\limits_0^1 {\frac{{\ln x}}{{1 - \frac{x}{2}}}\left( {\int\limits_0^{\frac{x}{2}} {\frac{1}{t}dt} } \right)} dx$$

fails to converge as the inner integral would be $\ln(x)-\ln(2)-\ln(0)$ but

$$I=\int\limits_0^1 {\frac{{\ln x}}{{1 - \frac{x}{2}}}\left( {\int\limits_1^{\frac{x}{2}} {\frac{1}{t}dt} } \right)} dx$$

does converge. We have

$$\begin{align*}
I&=\int_0^1\frac{\ln(x)}{1-\frac{x}{2}}\left (\int_1^{\frac{x}{2}}\frac{1}{t}dt \right )dx \\
&=\int_0^1\frac{\ln(x)\ln(x/2)}{1-\frac{x}{2}}dx \\
&=\int_0^1\frac{\ln^2(x)}{1-\frac{x}{2}}dx-\ln(2)\int_0^1\frac{\ln(x)}{1-\frac{x}{2}}dx \\
&=\sum_{n=0}^\infty\int_0^1\left ( \frac{x}{2} \right )^n\ln^2(x)dx-\ln(2)\sum_{n=0}^\infty\int_0^1\left ( \frac{x}{2} \right )^n\ln(x)dx \\
&=\sum_{n=0}^\infty\frac{2}{2^n(n+1)^3}+\ln(2)\sum_{n=0}^\infty\frac{1}{2^n(n+1)^2} \\
&=4Li_3\left ( \frac{1}{2} \right )+2\ln(2)Li_2\left(\frac{1}{2} \right ) \\
&= \frac{7}{3}\zeta(3)-\frac{1}{3}\ln^3(2)-\frac{\pi^2}{6}\ln(2).
\end{align*}$$

I assume it was just a typo as this matches your result.

I'll update this post if I get anything else.
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