西西11年送的几个CMC培训题整理

1.若存在正整数$a$，使得连续函数$f:[0,+\infty)\to[0,+\infty)$满足$f[f(x)]=x^a$，对任意的$x\in[0,+\infty)$，求证：$\int_0^1{[f(x)]}^2 dx\geq\frac{2a-1}{a^2+6a-3}$.

2.\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{{\left( {2n - 1} \right)}^{2011}}}}\sum\limits_{k = 0}^{n - 1} {\int_{2k\pi }^{\left( {2k + 1} \right)\pi } {{x^{2010}}{{\sin }^3}x{{\cos }^2}xdx} } .\]

3.\[\int_0^\infty  {\frac{{\left[ {\frac{x}{\pi }} \right]}}{{{e^x} + 1}}dx} ,\]其中$[a]$为取整函数.

4.\[\int_1^\infty  {\frac{{dx}}{{2\left[ x \right] + 3{{\left[ x \right]}^2} + {{\left[ x \right]}^3}}}} .\]

5.\[\sum\limits_{n = 2}^\infty  {\ln \left( {1 - \frac{1}{{n\left( {n - 1} \right)}}} \right)} .\]

6.求证：\[\mathop {\lim }\limits_{n \to \infty } \frac{{\ln C_{np}^0 + \ln C_{np}^p + \ln C_{np}^{2p} +  \cdots  + \ln C_{np}^{np}}}{{{n^2}}} = \frac{p}{2},\frac{p}{2} \in {N^ * }.\]

7.求极限\[\mathop {\lim }\limits_{x \to 0} \int_0^{ + \infty } {\frac{{{y^3}\left( {y - 3} \right){e^{ - y}}}}{{1 - {e^{ - xy}}}}dy} .\]

8.设$f_1(x)=x,f_2(x)=x^x,\ldots,f(x)={{x^x}^{\scalebox{-1}[1]{$\ddots$}}}^x$，求\[\mathop {\lim }\limits_{x \to 1}\frac{{{f_n}\left( x \right) - {f_{n - 1}}\left( x \right)}}{{{{\left( {1 - x} \right)}^n}}}.\]

9.求\[\int_0^1 {\left( {\frac{{{x^{A - 1}}\cos \left( {B\ln x} \right) - {x^{CA - 1}}\cos \left( {BC\ln x} \right)}}{{\ln x}}\sum\limits_{k = 0}^\infty  {{x^{D{C^k}}}} } \right)dx} \]的值，其中$A,B,D>0$，且$C>1$（后面的$DC^k$单指$C$的$k$次方）.

解答区：（未经特别说明均为来自百度贴吧网友的解答）

1.（_伽罗华_）易得$f$为单射且严格单调递增，$f(0)=0,f(1)=1$，从而\[\int_0^1 {f\left( x \right)dx}  + \int_0^1 {{f^{ - 1}}\left( x \right)dx}  = 1.\]

因此\begin{align*}1 = \int_0^1 {f\left( x \right)dx}  + \int_0^1 {f\left( {{x^{\frac{1}{a}}}} \right)dx}  = \int_0^1 {f\left( x \right)dx}  + \int_0^1 {f\left( x \right) \cdot a{x^{a - 1}}dx} \\= \int_0^1 {f\left( x \right)\left( {a{x^{a - 1}} + 1} \right)dx}  \le \sqrt {\int_0^1 {{f^2}\left( x \right)dx} }  \cdot \sqrt {\int_0^1 {{{\left( {a{x^{a - 1}} + 1} \right)}^2}dx} }. \end{align*}化简即可.

2.

3.

4.

5.（tian27546西西）因为\[\cos x = \prod\limits_{n = 1}^\infty  {\left( {1 - \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}{\pi ^2}}}} \right)} .\]令$x=\frac{\sqrt{5}}{2}\pi$得到\[\cos \frac{{\sqrt 5 }}{2}\pi  = \prod\limits_{n = 1}^\infty  {\frac{{4{n^2} - 4n - 4}}{{{{\left( {2n - 1} \right)}^2}}}} .\]又\[\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{2n!!}}{{\left( {2n - 1} \right)!!}}} \right)^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.\]两式作除整理得：\[\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 2}^n {\frac{{{i^2} - i - 1}}{{i\left( {i - 1} \right)}}}  = \mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 2}^n {\left( {1 - \frac{1}{{i\left( {i - 1} \right)}}} \right)}  =  - \frac{{\cos \left( {\frac{{\sqrt 5 }}{2}\pi } \right)}}{\pi }.\]故\[\sum\limits_{n = 2}^\infty  {\ln \left( {1 - \frac{1}{{n\left( {n - 1} \right)}}} \right)}  = \ln \left( { - \frac{{\cos \left( {\frac{{\sqrt 5 }}{2}\pi } \right)}}{\pi }} \right).\]

另一昵称混乱网友的解答：

\begin{align*}\prod\limits_{n = 2}^\infty  {\frac{{{n^2} - n - 1}}{{n\left( {n - 1} \right)}}}  = \prod\limits_{n = 2}^\infty  {\frac{{\left( {n - \alpha } \right)\left( {n - \beta } \right)}}{{n\left( {n - 1} \right)}}} \\= \frac{1}{{\Gamma \left( { - \alpha } \right)}} \cdot \frac{1}{{\Gamma \left( { - \beta } \right)}} \cdot \frac{{{n^{ - \alpha }}}}{{ - \alpha \left( {1 - \alpha } \right)}} \cdot \frac{{{n^{ - \beta }}}}{{ - \beta \left( {1 - \beta } \right)}}\\= \frac{1}{{\Gamma \left( { - \alpha } \right)}} \cdot \frac{1}{{\Gamma \left( { - \beta } \right)}} = \frac{{ - 1}}{{\Gamma \left( {1 - \alpha } \right)\Gamma \left( {1 - \beta } \right)}}\\= \frac{{ - 1}}{{\frac{\pi }{{\sin \left[ {\left( {1 - \frac{{1 + \sqrt 5 }}{2}} \right)\pi } \right]}}}} =  - \frac{{\cos \frac{{\sqrt 5 }}{2}\pi }}{\pi },\end{align*}其中$\alpha  = \frac{{1 + \sqrt 5 }}{2},\beta  = \frac{{1 - \sqrt 5 }}{2},\alpha  + \beta  = 1,\alpha \beta  =  - 1.$

6.

博士哥？（creasson）利用Binet公式

7.（博士哥）\begin{align*}&\int_0^\infty  {\frac{{{y^3}\left( {y - 3} \right){e^{ - y}}}}{{1 - {e^{ - xy}}}}dy}  = \int_0^\infty  {{y^3}\left( {y - 3} \right){e^{ - y}}\sum\limits_{n = 0}^\infty  {{e^{ - ny}}} dy} \\&= \sum\limits_{n = 0}^\infty  {\frac{{6 - 18nx}}{{{{\left( {1 + nx} \right)}^5}}}}  =  - \frac{{3x\mathrm{PolyGamma}\left[ {3,\frac{1}{x}} \right] + \mathrm{PolyGamma}\left[ {4,\frac{1}{x}} \right]}}{{{x^5}}}.\end{align*}可以利用$\mathrm{PolyGamma\left[0,\frac1x\right]}$的渐近表示式求出该极限3.

(tian27546西西)我们先证明几个引理：

引理1：证明$t>0$时，\[\frac{1}{t} + \frac{n}{2} < \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}}  < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^n}}} + n\left( {n = 1,2,3, \cdots } \right).\]

证明.理设$f(x)=\frac{n}{{1+xt}^{n+1}}$，因为$f''\left( x \right) = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right){t^2}}}{{{{\left( {1 + xt} \right)}^{n + 3}}}} > 0$，可见函数$y=f(x)$的曲线是下凹曲线.

连接$(0,f(0)),(1,f(1)),(2,f(2)),\ldots,(k,f(k)),\ldots$.在这折线下的梯形面积之和为

\[\sum\limits_{k = 0}^\infty  {\frac{{f\left( k \right) + f\left( {k + 1} \right)}}{2}}  =  - \frac{{f\left( 0 \right)}}{2} + \sum\limits_{k = 0}^\infty  {f\left( k \right)}  =  - \frac{n}{2} + \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} .\]因为$y=f(x)$曲线下凹，所以折线下的梯形面积大于$[0,\infty)$上曲线$y=f(x)$下的曲边梯形面积，有

\begin{align*}&- \frac{n}{2} + \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}}  > \int_0^\infty  {f\left( x \right)dx}  = \int_0^\infty  {\frac{n}{{{{\left( {1 + xt} \right)}^{n + 1}}}}dx}  =  - \frac{1}{{t{{\left( {1 + xt} \right)}^n}}}\left| {_0^\infty } \right. = \frac{1}{t}\\&\Rightarrow \frac{1}{t} + \frac{n}{2} < \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} .\end{align*}

过$(1,f(1)),(2,f(2)),\ldots,(k,f(k)),\ldots$点分别作曲线$y=f(x)$的切线，从$x=\frac12$到$x\to\infty$切线下的梯形面积之和为\[\sum\limits_{k = 1}^\infty  {f\left( k \right)}  =  - f\left( 0 \right) + \sum\limits_{k = 0}^\infty  {f\left( k \right)}  =  - n + \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}} .\]

因为$y=f(x)$曲线下凹，所以切线下的梯形面积小于$\left[\frac12,+\infty\right)$上曲线$y=f(x)$下的曲边梯形面积，有

\[ - n + \sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}}  < \int_{\frac{1}{2}}^\infty  {f\left( x \right)dx}  = \int_{\frac{1}{2}}^\infty  {\frac{n}{{{{\left( {1 + xt} \right)}^{n + 1}}}}dx}  =  - \frac{1}{{t{{\left( {1 + xt} \right)}^n}}}\left| {_{\frac{1}{2}}^\infty } \right. = \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^n}}}\]

得到\[\sum\limits_{k = 0}^\infty  {\frac{n}{{{{\left( {1 + kt} \right)}^{n + 1}}}}}  < n + \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^n}}}.\]

引理2：证明\[\mathop {\lim }\limits_{t \to 0} \sum\limits_{k = 0}^\infty  {\frac{{1 - 3kt}}{{{{\left( {1 + kt} \right)}^5}}}}  = \frac{1}{2}.\]

证明.

\[\sum\limits_{k = 0}^\infty  {\frac{{1 - 3kt}}{{{{\left( {1 + kt} \right)}^5}}}}  = \sum\limits_{k = 0}^\infty  {\frac{{4 - 3\left( {1 + kt} \right)}}{{{{\left( {1 + kt} \right)}^5}}}}  = \sum\limits_{k = 0}^\infty  {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}}  - \sum\limits_{k = 0}^\infty  {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}} .\]

由上面引理1可知

\[\frac{1}{t} + 2 < \sum\limits_{k = 0}^\infty  {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}}  < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} + 4,\frac{1}{t} + \frac{3}{2} < \sum\limits_{k = 0}^\infty  {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}}  < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} + 3.\]

所以有

\[\frac{1}{t} + 2 - \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} - 3 < \sum\limits_{k = 0}^\infty  {\frac{4}{{{{\left( {1 + kt} \right)}^5}}}}  - \sum\limits_{k = 0}^\infty  {\frac{3}{{{{\left( {1 + kt} \right)}^4}}}}  < \frac{1}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}} + 4 - \frac{1}{t} - \frac{3}{2}.\]

即有\[\frac{{{{\left( {1 + \frac{t}{2}} \right)}^3} - 1}}{{t{{\left( {1 + \frac{t}{2}} \right)}^3}}} - 1 < \sum\limits_{k = 0}^\infty  {\frac{{1 - 3kt}}{{{{\left( {1 + kt} \right)}^5}}}}  < \frac{5}{2} - \frac{{{{\left( {1 + \frac{t}{2}} \right)}^4} - 1}}{{t{{\left( {1 + \frac{t}{2}} \right)}^4}}}.\]

又因为

8.（博士哥）$x\to1$时，可以得到$f_n(x)=1-(1-x)f_{n-1}(x)+O((1-x)^2)$，所以$x\to1$时\[{f_n}\left( x \right) \approx {\left( {x - 1} \right)^{n - 1}}\left( {x - \frac{1}{{2 - x}}} \right) + \frac{1}{{2 - x}}.\]因此\[\mathop {\lim }\limits_{x \to 1} \frac{{{f_n}\left( x \right) - {f_{n - 1}}\left( x \right)}}{{{{\left( {1 - x} \right)}^n}}} = \mathop {\lim }\limits_{x \to 1} \frac{{{{\left( {x - 1} \right)}^{n - 1}}\left( {x - \frac{1}{{2 - x}}} \right) - {{\left( {x - 1} \right)}^{n - 2}}\left( {x - \frac{1}{{2 - x}}} \right)}}{{{{\left( {1 - x} \right)}^n}}} = 1.\]

9.\begin{align*}\int_0^1 {\left( {\frac{{{x^{A - 1}}\cos \left( {B\ln x} \right) - {x^{CA - 1}}\cos \left( {BC\ln x} \right)}}{{\ln x}}\sum\limits_{k = 0}^\infty  {{x^{D{C^k}}}} } \right)dx} \\= \int_0^\infty  {\left( {\frac{{{e^{ - Ax}}\cos \left( {Bx} \right) - {e^{ - ACx}}\cos \left( {BCx} \right)}}{x}\sum\limits_{k = 0}^\infty  {{e^{ - D{C^k}x}}} } \right)dx} .\end{align*}

利用Laplace变换

\begin{align*}&\mathcal{L}\left( {\left( {{e^{ - Ax}}\cos \left( {Bx} \right) - {e^{ - ACx}}\cos \left( {BCx} \right)} \right)\sum\limits_{k = 0}^\infty  {{e^{ - D{C^k}x}}} } \right)\\&= \sum\limits_{k = 0}^\infty  {\frac{{A + D{C^k} + p}}{{{{\left( {A + D{C^k} + p} \right)}^2} + {B^2}}} - \frac{{AC + D{C^k} + p}}{{{{\left( {AC + D{C^k} + p} \right)}^2} + {{\left( {BC} \right)}^2}}}}. \end{align*}

\[\mathcal{L}^{-1}\left(\frac 1p\right)=1.\]所以原积分

另外：罗账号提供了一个副产品：

证明：\[\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{{2^{n - 1}}\left( {n - 1} \right)!}}\sum\limits_{k = 0}^{n - 1} {\frac{1}{{{2^k}k!}}\frac{{\Gamma \left( {n + k} \right)}}{{\Gamma \left( {n - k} \right)}}}  = \frac{e}{{\sqrt \pi  }}.\]

参考来源：

[1]http://tieba.baidu.com/p/1225531522?pn=2

[2]http://tieba.baidu.com/p/1845244643?share=9105&qq-pf-to=pcqq.group