积分题1之来自G.Han的一道积分题

今天，收到G.Han的提问，第一个是计算积分

\[\int_0^{\infty}{\frac{\ln x}{(x^2+1)^n}dx}\]顿时不明觉厉，然后在宝典《Table of Integrals, Series, and Products》上找到一个更一般的结果：

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

其中$\gamma$为Euler-Mascheroni常数，$\psi(x)$为Digamma 函数，有：

\begin{align}\psi\left(n+\frac{1}{2} \right)&=-\gamma-2\ln2+\sum\limits_{k=1}^n{\frac{2}{2k-1}}.\\\psi\left(\frac{1}{2}\right)&=-\gamma-2\ln2\end{align}

解：

\begin{align*}{I_n} &= \int_0^{ + \infty } {\frac{{\ln x}}{{{{\left( {{x^2} + 1} \right)}^n}}}dx}  = \int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^n}}}d\left( {x\ln x - x} \right)}\\&= \left[ {\frac{{x\ln x - x}}{{{{\left( {{x^2} + 1} \right)}^n}}}} \right]_0^{ + \infty } + 2n\int_0^{ + \infty }{\frac{{{x^2}\ln x - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}  \\&= 2n\int_0^{ + \infty } {\frac{{\left( {{x^2} + 1} \right)\ln x - \ln x - {x^2}}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}  \\&= 2n\int_0^{ + \infty } {\frac{{\ln x}}{{{{\left( {{x^2} + 1} \right)}^n}}}dx}  - 2n\int_0^{ + \infty } {\frac{{\ln x}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx} - 2n\int_0^{ + \infty } {\frac{{{x^2}}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}\end{align*}

\begin{align*}\Rightarrow 2n{I_{n + 1}} &= \left( {2n - 1} \right){I_n} - 2n\int_0^{ + \infty } {\frac{{{x^2}}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}  \\&= \left( {2n - 1} \right){I_n} - 2n\int_0^{ + \infty } {\frac{{{x^2} + 1 - 1}}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}  \\&= \left( {2n - 1} \right){I_n} - 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^n}}}dx}  + 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx} .\end{align*}

即

\[2n{I_{n + 1}} = \left( {2n - 1} \right){I_n} - 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1}\right)}^n}}}dx}  + 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx} .\]

先证明一个引理

\begin{align*}\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^n}}}dx} &= \frac{1}{2}\mathbf{B}\left( {n - \frac{1}{2},\frac{1}{2}} \right) = \frac{{\sqrt \pi  }}{2}\frac{{\mathbf{\Gamma} \left( {n - \frac{1}{2}} \right)}}{{\left( {n - 1} \right)!}} \\&= \left\{ \begin{array}{l}\frac{{\sqrt \pi  }}{{2\left( {n - 1} \right)!}}\left[ {\sqrt \pi  .\frac{{\left( {2n - 3} \right)!!}}{{{2^{n - 1}}}}} \right] = \frac{{\left( {2n - 3} \right)!!}}{{{2^n}\left( {n - 1} \right)!}}\pi \qquad n \ge 2\\\frac{\pi }{2} \qquad n = 1\end{array} \right..\end{align*}

引理的证明

\begin{align*}\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^n}}}dx} &\underline{\underline {\text{令}t = \frac{1}{{{x^2} + 1}}}} \int_1^0 {{t^n}d\sqrt {\frac{1}{t} - 1} }  = \frac{1}{2}\int_0^1 {{t^{n - \frac{3}{2}}}{{\left( {1 - t} \right)}^{ - \frac{1}{2}}}dt}  \\&= \frac{1}{2}\mathbf{B}\left( {n - \frac{1}{2},\frac{1}{2}} \right) = \frac{{\sqrt \pi  }}{2}\frac{{\mathbf{\Gamma} \left( {n - \frac{1}{2}} \right)}}{{\left( {n - 1} \right)!}} \\&= \left\{ \begin{array}{l}\frac{{\sqrt \pi  }}{{2\left( {n - 1} \right)!}}\left[ {\sqrt \pi  .\frac{{\left( {2n - 3} \right)!!}}{{{2^{n - 1}}}}} \right] = \frac{{\left( {2n - 3} \right)!!}}{{{2^n}\left( {n - 1} \right)!}}\pi \qquad n \ge 2\\\frac{\pi }{2} \qquad n = 1\end{array} \right..\end{align*}

回到原题

\begin{align*}2n{I_{n + 1}} &= \left( {2n - 1} \right){I_n} - 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^n}}}dx}  + 2n\int_0^{ + \infty } {\frac{1}{{{{\left( {{x^2} + 1} \right)}^{n + 1}}}}dx}  \\&= \left( {2n - 1} \right){I_n} - 2n\frac{{\left( {2n - 3} \right)!!}}{{{2^n}\left( {n - 1} \right)!}}\pi  + 2n\frac{{\left( {2n - 1} \right)!!}}{{{2^{n + 1}}n!}}\pi \\&= \left( {2n - 1} \right){I_n} - \frac{\pi }{{\left( {n - 1} \right)!}}\frac{{\left( {2n - 3} \right)!!}}{{{2^n}}}\\&\Rightarrow \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}{I_{n + 1}} = \frac{{\left( {2n - 2} \right)!!}}{{\left( {2n - 3} \right)!!}}{I_n} - \frac{\pi }{2} \cdot \frac{1}{{2n - 1}} \\&\Rightarrow \frac{{\left( {2n - 2} \right)!!}}{{\left( {2n - 3} \right)!!}}{I_n} = \frac{{2!!}}{{1!!}}{I_2} - \sum\limits_{k = 2}^{n - 1} {\frac{\pi }{{2k - 1}}}  =  - \frac{\pi}{2}\sum\limits_{k = 1}^{n - 1} {\frac{1 }{{2k - 1}}} \\&\Rightarrow {I_n} = \left\{ \begin{array}{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 \ge 2\end{array} .\right.\end{align*}

 

 

第二题是个重要的Steffensen积分不等式

设$f,g\in \mathcal{R}[a,b]$，且$f$在$[a,b]$单减,$0<g(x)\leq1$，求证：

\[\int_{b-\lambda}^b{f(x)dx}\leq\int_a^b{f(x)g(x)dx}\leq\int_a^{a+\lambda}{f(x)dx}.\]

其中\[\lambda=\int_a^b{g(x)dx}.\]

证：先证明右边不等式

\begin{align*}&\int_a^{a + \lambda } {f\left( x \right)dx} - \int_a^b {f\left( x \right)g\left( x \right)dx} = \int_a^{a + \lambda } {f\left( x \right)\left[ {1 - g\left( x \right)} \right]dx} - \int_{a + \lambda }^b {f\left( x \right)g\left( x \right)dx}\\ &\ge f\left( {a + \lambda } \right)\int_a^{a + \lambda } {\left[ {1 - g\left( x \right)} \right]dx} = f\left( {a + \lambda } \right)\left( {\lambda - \int_a^{a + \lambda } {g\left( x \right)dx} } \right) - \int_{a + \lambda }^b {f\left( x \right)g\left( x \right)dx} \\&= f\left( {a + \lambda } \right)\int_{a + \lambda }^b {g\left( x \right)dx} - \int_{a + \lambda }^b {f\left( x \right)g\left( x \right)dx} = \int_{a + \lambda }^b {\left[ {f\left( {a + \lambda } \right) - f\left( x \right)} \right]g\left( x \right)dx} \ge 0\end{align*}

左边不等式同理可证.