计算

\[\int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy \]

(tian_275461)

解

\begin{align*}
\int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy &= \int_0^1 \int_0^1 \frac{\ln \left(1+(1-xy) \right)}{1-xy}dx \ dy
\\  &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\int_0^1 \int_0^1 (1-xy)^{n-1}dx \ dy \\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}\int_0^1 \left(\frac{1-(1-y)^n}{y} \right)dy \\  &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}H_n\\
&=\frac{(-1)^{n}}{n}\int_{0}^{1}{(1-x)^{n-1}\ln{x}dx}\\
&=-\int_{0}^{1}{\sum_{n=1}^{\infty}{\frac{(x-1)^{n-1}}{n}\ln{x}}dx}\\
&=\int_{0}^{1}{\frac{\Li_{1}(x-1)\ln{x}}{1-x}dx}\\
&=-\int_{0}^{1}{\frac{\ln{(1+t)}\ln{(1-t)}}{t}dt}
\end{align*}
为了计算上面的式子，我们先算
\[ \int_{0}^{1}{\frac{\ln^{2}{(1+x)}}{x}dx} \]
为此考虑一般的\begin{align*}
I(t) &= \int_0^t \frac{\ln^2(1+x)}{x}dx \\
&= \ln t\ln^2(1+t)-2\int_0^t \frac{\ln(x)\ln(1+x)}{1+x}dx \\
&= \ln^2{t}\ln(1+t)-2\int_1^{1+t}\frac{\ln(y) }{y} \left( \ln y+\ln \left( 1-\frac{1}{y}\right)\right)dy \quad (y=x+1)\\
&= \ln t\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\int_{1}^{1+t}{\frac{\ln{y}}{y}\ln{\left(1-\frac{1}{y}\right)}dy} \quad (z=\frac{1}{y})\\
&=\ln t\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)+2\int_{\frac{1}{1+x}}^{1}{\frac{\ln{z}\ln{(1-z)}}{z}dz}\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\int_{\frac{1}{1+x}}^{1}{\ln{z}d \Li_{2}(z)}\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\left(\ln{z}\Li_{2}(z)\bigg|_{\frac{1}{1+x}}^{1}-\int_{\frac{1}{1+x}}^{1}{\frac{\Li_{2}(z)}{z}dz}\right)\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\ln{(1+t)}\Li_{2}\left(\frac{1}{1+t}\right)-2\Li_{3}\left( \frac{1}{1+t} \right)+2\Li_{3}(1)
 \end{align*}
故
\[ I(1)=-\frac{2}{3}\ln^3{2}-2\ln{2}\Li_{2}\left(\frac{1}{2}\right)-2\Li_{3}\left(\frac{1}{2}\right)+2\Li_{3}(1)\]
由
\[ \Li_{2}\left(\frac{1}{2}\right)=\frac{1}{12}(\pi^{2}-6\ln^2{2}) \]
\[ \Li_{3}\left(\frac{1}{2}\right)=\frac{1}{24}[4\ln^3{2}-2\pi^{2}\ln{2}+21\zeta(3)] \]
\[ I(1)=\frac{\zeta(3)}{4} \]
对已知结论
\[ \int_{0}^{1}{\frac{\ln^2{(1-x)}}{x}dx}=2\zeta(3) \]
作$x=t^{2}$
\[ \Rightarrow \int_{0}^{1}{\frac{\ln^2{(1-t^2)}}{t}dx}=\zeta(3) \]
\[ \Leftrightarrow \int_{0}^{1}{\frac{\ln^2{(1-t)}}{t}dx}+\int_{0}^{1}{\frac{\ln^2{(1+t)}}{t}dx}+2\int_{0}^{1}{\frac{\ln{(1-t)}\ln{(1+t)}}{t}dx}=\zeta(3) \]
\[ \Rightarrow \int_{0}^{1}{\frac{\ln{(1-t)}\ln{(1+t)}}{t}dx}=-\frac{5}{8}\zeta(3) \]
故有
\[ \int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy=\frac{5}{8}\zeta(3) \]

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根据上面的办法，我们可以得到

\[ \int_{0}^{1}{\frac{\ln^{2}(1-t)}{t}dt}=2\zeta(3) \]

\[ \int_{0}^{1}{\frac{\ln^{2}(1+t)}{t}dt}=\frac{1}{4}\zeta(3) \]

\[ \sum_{n=1}^{\infty}{\frac{H_{n}}{n^2}}=2\zeta(3) \]

\[ \int_{0}^{1}\int_{0}^{1}\frac{\ln(1-xy)}{1-xy}dxdy =-\zeta(3) \]

\[ \sum_{n=1}^{\infty}{\frac{H_{n}}{n^3}}=\frac{\pi^4}{72} \]