一个漂亮的积分 ∫(0,+∞)lnx dx／(x^2+1)^n

永远

\[\int_0^\infty  {\frac{{\ln x}}{{{{({x^2} + 1)}^n}}}dx}  = \left\{ {\begin{array}{*{20}{l}}
  {0\qquad n = 1} \\
  { - \frac{\pi }{2}\frac{{\left( {2n - 3} \right)!!}}{{\left( {2n - 2} \right)!!}}\sum\limits_{k = 1}^{n - 1} {\frac{1}{{2k - 1}}} \qquad n \geqslant 2}
\end{array}.} \right.\]

永远

一般的，我们有：


\[\int_0^\infty  {\ln x\frac{{dx}}{{{{\left( {{a^2} + {b^2}{x^2}} \right)}^n}}}}  = \frac{{\Gamma \left( {n - \frac{1}{2}} \right)\sqrt \pi  }}{{4\left( {n - 1} \right)!{a^{2n - 1}}b}}\left[ {2\ln \frac{a}{{2b}} - \gamma  - \psi \left( {n - \frac{1}{2}} \right)} \right]\qquad {\text{a}} > {\text{0}},{\text{b}} > {\text{0}}.\]