两个奇怪的积分
第一个积分的解答:
第二个积分的另一种求法:
来自:http://math.stackexchange.com/questions/324647/integrate-int-01x-x1-xx-1-sin-pi-xdx
http://math.stackexchange.com/questions/958624/prove-that-int-01-sin-pi-xxx1-x1-x-dx-frac-pi-e24
一道杂志征解题的解答
这道题来自MAA的杂志The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), pp. 500-507,可以参考链接http://www.jstor.org/stable/10.4169/amer.math.monthly.122.5.500?seq=1#page_scan_tab_contents
求\[\int_0^\infty {\frac{1}{x}dx} \int_0^x {\frac{{\cos \left( {x - y} \right) - \cos x}}{y}dy} .\]
解.(翻译而来)令$f\left( {x,y} \right) = \frac{{\cos \left( {x - y} \right) - \cos x}}{y}$.对$x>0$,我们有\[\int_0^x {f\left( {x,y} \right)dy} = \int_0^1 {\frac{{\cos \left( {1 - t} \right)x - \cos x}}{y}dt} = x\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\, dudt} .\]
因而对$R>0$,
\[\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx} = \int_0^R {\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\,dudtdx} } .\]
而$|\sin ux|\leq1$,该三重积分是绝对收敛的.由Fubini定理可知积分能交换次序
曲线积分的计算
张元博问了这么一道题：计算曲线积分\[\oint_C {\left( {{{\left( {x + 1} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right)ds} ,\]其中$C$表示曲面$x^2+y^2+z^2=1$与$x+y+z=1$的交线.
解:我是利用常规的三角换元解决的.联立方程有
二重积分难题荟萃
第一个是09年西北大学的考研题,也在史济怀老师的数分书上找得到.这里的解答就是史老爷子自己的.
计算积分\[\iint_D {\frac{1}{{xy\left( {\ln^2 x + {{\ln }^2}y} \right)}}} dxdy,D = \left\{ {\left( {x,y} \right)\left| {x + y \ge 1,{x^2} + {y^2} \le 1} \right.} \right\}.\]
解.作变换$x=e^{r\cos\theta},y=e^{r\sin\theta}$,则
\[\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {r,\theta } \right)}} = \left| {\begin{array}{*{20}{c}}{\cos \theta {e^{r\cos \theta }}}&{ - r\sin \theta {e^{r\cos \theta }}}\\{\sin \theta {e^{r\sin \theta }}}&{r\cos \theta {e^{r\sin \theta }}}\end{array}} \right| = r{e^{r\sin \theta }}{e^{r\cos \theta }}.\]
原积分变为\[I = \iint_\Delta {\frac{{drd\theta }}{r}} .\]
这里的$\Delta$是变换以后的积分区域.注意$x+y=1$和$x^2+y^2=1$分别被变为
\[\left\{ \begin{array}{l}{e^{r\cos \theta }} + {e^{r\sin \theta }} = 1,\\{e^{2r\cos \theta }} + {e^{2r\sin \theta }} = 1.\end{array} \right.\]
现在来分析由上述两条曲线所围成的$(r,\theta)$平面上的区域是什么形状.从第一个式子可以看出$\theta$的变化范围必须使$\cos \theta$和$\sin \theta$都取负值,故$\theta$只能在$\left[\pi,\frac32\pi \right]$中取值.
假设由第一个式子确定的函数为$r=r(\theta)$,则由第二个式子确定的函数便为$r=\frac12 r(\theta)$.因此
\[I = \iint_\Delta {\frac{{drd\theta }}{r}} = \int_\pi ^{\frac{3}{2}\pi } {d\theta } \int_{\frac{1}{2}r\left( \theta \right)}^{r\left( \theta \right)} {\frac{{dr}}{r}} = \frac{\pi }{2}\ln 2.\]
苹果公司总部
多重积分计算的一些题
（1）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且\[\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=x^2+y^2,\]求\[\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]
解：(Hansschwarzkopf)根据Gauss公式
\[\iint\limits_D\left(x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]
解：(Hansschwarzkopf)根据Gauss公式
一个很好的积分题
背景是这个：
然后郝XX跟我说了下他的求法，还是挺有意思的！
求
\[\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} .\]
由Euler公式,我们有
\begin{align*}&{e^{y\cos x}}\cos \left( {y\sin x} \right) + i{e^{y\cos x}}\sin \left( {y\sin x} \right) = {e^{y\cos x}} \cdot {e^{iy\sin x}}\\=& {e^{y\cos x + iy\sin x}} = {e^{y\left( {\cos x + i\sin x} \right)}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} \\=& \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}{{\left( {\cos x + i\sin x} \right)}^n}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}}}{{n!}}\left( {\cos nx + i\sin nx} \right)} \\= &\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} + i\sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\end{align*}
因此有
\[{e^{y\cos x}}\cos \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\cos nx}}{{n!}}} ,{e^{y\cos x}}\sin \left( {y\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{{y^n}\sin nx}}{{n!}}} .\]
在第一个式子中令$y=1$,我们有
\[{e^{\cos x}}\cos \left( {\sin x} \right) = \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} .\]
因此我们有
\begin{align*}&\int_0^{2\pi } {{e^{\cos x}}\cos \left( {\sin x} \right)\cos nxdx} = \int_0^{2\pi } {\left( {\cos nx \cdot \sum\limits_{n = 0}^{ + \infty } {\frac{{\cos nx}}{{n!}}} } \right)dx} \\=& \int_0^{2\pi } {\left( {\cos nx \cdot \frac{{\cos nx}}{{n!}}} \right)dx} = \left\{ \begin{array}{l}\frac{\pi }{{n!}},n \ge 1\\2\pi ,n = 0\end{array} \right..\end{align*}
而对于无穷乘积\[\prod\limits_{n = 3}^\infty {\cos \frac{\pi }{{n!}}} \approx 0.858314.\]也就是管理员一分钟都不能禁他！！！
事实上,我们有
这几天碰到的一些好题
-
求\[\iint\limits_D {{e^x}\cos ydxdy} ,\]其中$D: x^2+y^2\leq 1$.
先证明\[\int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {r\sin \theta } \right)d\theta } = 2\pi .\]事实上,在积分号下求导,得$F'\left( r \right) = \int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {\theta + r\sin \theta } \right)d\theta } $.由归纳法,可知
\[{F^{\left( n \right)}}\left( r \right) = \int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {n\theta + r\sin \theta } \right)d\theta } .\tag{1}\]从而导出${F^{\left( n \right)}}\left( 0 \right) = 0\left( {n = 1,2, \cdots } \right)$.因此根据Taylor公式,我们有
\[F\left( r \right) - F\left( 0 \right) = \sum\limits_{k = 1}^{n - 1} {\frac{{{F^{\left( k \right)}}\left( 0 \right)}}{{k!}}{r^k}} + \frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n} = \frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n}\left( {0 < {\theta _1} < 1} \right).\]
注意到由(1)可得$\left| {{F^{\left( n \right)}}\left( {{\theta _1}r} \right)} \right| \le 2\pi {e^r}$,故知
\[\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{F^{\left( n \right)}}\left( {{\theta _1}r} \right)}}{{n!}}{r^n}} \right| \le \mathop {\lim }\limits_{n \to \infty } \frac{{2\pi {e^r}}}{{n!}}{r^n} = 0,F\left( r \right) \equiv F\left( 0 \right) = 2\pi .\]
\begin{align*}\iint\limits_D {{e^x}\cos ydxdy} &= \int_0^1 {rdr\int_0^{2\pi } {{e^{r\cos \theta }}\cos \left( {r\sin \theta } \right)d\theta } } \\&= 2\pi \int_0^1 {rdr} = \pi.\end{align*}
-
证明:
\[\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \frac{\pi }{{\sqrt 2 }}.\]
证法一: 利用椭圆函数的一些性质.\begin{align*}&\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^{\frac{\pi }{2} - a} {\frac{1}{{\sqrt {\cos x - \cos \left( {\frac{\pi }{2} - a} \right)} }}dx} \\= &\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \sqrt 2 K\left( {\sin \frac{1}{2}\left( {\frac{\pi }{2} - a} \right)} \right) = \sqrt 2 K\left( 0 \right) = \frac{\pi }{{\sqrt 2 }},\end{align*}
其中$K(x)$为Complete Elliptic Integral of the First Kind,以及参考Elliptic Integral of the First Kind.
证法二: 根据
\begin{align*}&\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}} }}dx} \\= &\frac{1}{{\sqrt 2 }}\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\cos \frac{{x + a}}{2}}}{{\frac{\pi }{2} - \frac{{x + a}}{2}}}\frac{{\sin \frac{{x - a}}{2}}}{{\frac{{x - a}}{2}}}} \sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} \\= &\frac{1}{{\sqrt 2 }}\sqrt {\frac{{\frac{\pi }{2} - \frac{{\xi + a}}{2}}}{{\cos \frac{{\xi + a}}{2}}}} \cdot \sqrt {\frac{{\frac{{\xi - a}}{2}}}{{\sin \frac{{\xi - a}}{2}}}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} \left( {a < \xi < \frac{\pi }{2}} \right).\end{align*}
而
\begin{align*}&\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\left( {\frac{\pi }{2} - \frac{{x + a}}{2}} \right)\frac{{x - a}}{2}} }}dx} = 2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\pi \left( {x - a} \right) - {x^2} + {a^2}} }}dx} = 2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {{{\left( {a - \frac{\pi }{2}} \right)}^2} - {{\left( {x - \frac{\pi }{2}} \right)}^2}} }}dx} \\= &2\int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {1 - {{\left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)}^2}} }}d\left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)} = \left. {2\arcsin \left( {\frac{{2x - \pi }}{{\pi - 2a}}} \right)} \right|_a^{\frac{\pi }{2}} = 2 \times \frac{\pi }{2} = \pi .\end{align*}
证法三: 由于
\begin{align*}&\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_a^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\sin x - \sin a} }}dx} = \mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^1 {\frac{{\frac{\pi }{2} - a}}{{\sqrt {\sin \left( {a + \left( {\frac{\pi }{2} - a} \right)t} \right) - \sin a} }}dt} \\= &\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \int_0^1 {\frac{{\frac{\pi }{2} - a}}{{\sqrt {2\sin \left[ {\frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]\cos \left[ {a + \frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]} }}dt} \\= &\int_0^1 {\mathop {\lim }\limits_{a \to \frac{\pi }{2}} \frac{{\frac{\pi }{2} - a}}{{\sqrt {2\sin \left[ {\frac{t}{2}\left( {\frac{\pi }{2} - a} \right)} \right]\sin \left[ {\left( {1 - \frac{t}{2}} \right)\left( {\frac{\pi }{2} - a} \right)} \right]} }}dt} = \int_0^1 {\frac{1}{{\sqrt 2 \cdot \sqrt {\frac{t}{2}\left( {1 - \frac{t}{2}} \right)} }}dt} .\end{align*}
令$\sqrt {\frac{t}{2}} = \sin \theta $,我们有
\[\int_0^1 {\frac{1}{{\sqrt 2 \cdot \sqrt {\frac{t}{2}\left( {1 - \frac{t}{2}} \right)} }}dt} = \int_0^{\frac{\pi }{4}} {2\sqrt 2 d\theta } = \frac{\pi }{{\sqrt 2 }}.\]
证法四: 显然等价于求
\[\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{1}{{\sqrt {\cos x - \cos a} }}dx} = \mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{1}{{\sqrt {2\sin \frac{{a - x}}{2}\sin \frac{{a + x}}{2}} }}dx} .\]
由$0\leq x<a$,则有
\begin{align*}&{\left( {2\sin \frac{{a - x}}{2}\sin \frac{{a + x}}{2}} \right)^{ - \frac{1}{2}}} = {\left[ {2\left( {\frac{{a - x}}{2} + O\left( {{{\left( {a - x} \right)}^3}} \right)} \right)\left( {\frac{{a + x}}{2} + O\left( {{{\left( {a + x} \right)}^3}} \right)} \right)} \right]^{ - \frac{1}{2}}}\\= &{\left( {\frac{{{a^2} - {x^2}}}{2} + O\left( {\left( {{a^2} - {x^2}} \right){{\left( {a + x} \right)}^2}} \right)} \right)^{ - \frac{1}{2}}} = \frac{{\sqrt 2 }}{{\sqrt {{a^2} - {x^2}} }} + O\left( {\frac{{{{\left( {a + x} \right)}^2}}}{{\sqrt {{a^2} - {x^2}} }}} \right).\end{align*}
即有
\begin{align*}\int_0^a {\frac{1}{{\sqrt {\cos x - \cos a} }}dx} &= \int_0^a {\frac{{\sqrt 2 }}{{\sqrt {{a^2} - {x^2}} }}dx} + O\left( {\int_0^a {\frac{{{{\left( {a + x} \right)}^2}}}{{\sqrt {{a^2} - {x^2}} }}dx} } \right)\\&=\frac{\pi }{{\sqrt 2 }} + O\left( {{a^2}} \right) \to \frac{\pi }{{\sqrt 2 }}\left( {a \to 0} \right).\end{align*}
其中在算阶的时候用到了
\[\max \left\{ {\left( {{a^2} - {x^2}} \right){{\left( {a + x} \right)}^2},\left( {{a^2} - {x^2}} \right){{\left( {a - x} \right)}^2},{{\left( {{a^2} - {x^2}} \right)}^3}} \right\} = \left( {{a^2} - {x^2}} \right){\left( {a + x} \right)^2},\]
这里$0\leq x<a$.
推论：设$f(x)$在$[0,a]$具有二阶连续导数,且满足$f'(x)=0,f''(0)<0$,则有
\[\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{{dx}}{{\sqrt {f\left( x \right) - f\left( a \right)} }}} = \frac{\pi }{{\sqrt {2\left| {f''\left( 0 \right)} \right|} }}.\]
\begin{align*}&\mathop {\lim }\limits_{a \to {0^ + }} \int_0^a {\frac{{dx}}{{\sqrt {f\left( x \right) - f\left( a \right)} }}} = \mathop {\lim }\limits_{a \to {0^ + }} \int_0^1 {\frac{{adt}}{{\sqrt {f\left( {at} \right) - f\left( a \right)} }}} \\=& \mathop {\lim }\limits_{a \to {0^ + }} \int_0^1 {\frac{{\sqrt 2 dt}}{{\sqrt {f''\left( {a{\xi _1}} \right){t^2} - f''\left( {a{\xi _2}} \right)} }}} \left( {0 < {\xi _1} < 1,0 < {\xi _2} < 1} \right)\\= &\int_0^1 {\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\sqrt 2 dt}}{{\sqrt {f''\left( {a{\xi _1}} \right){t^2} - f''\left( {a{\xi _2}} \right)} }}} = \int_0^1 {\mathop {\lim }\limits_{a \to {0^ + }} \frac{{\sqrt 2 dt}}{{\sqrt {\left| {f''\left( 0 \right)} \right|\left( {1 - {t^2}} \right)} }}} \\= &\frac{\pi }{{\sqrt {2\left| {f''\left( 0 \right)} \right|} }}.\end{align*}
-
证明:
\[\mathop {\lim }\limits_{n \to \infty } \int_{n\pi }^{\left( {n + 1} \right)\pi } {\frac{x}{{1 + {x^2}{{\sin }^2}x}}dx} = \pi .\]
更一般地,以下极限
\[\mathop {\lim }\limits_{n \to \infty } \int_{n\pi }^{\left( {n + 1} \right)\pi } {\frac{x}{{1 + {x^\alpha }{{\sin }^2}x}}dx} \]
与$n$的关系是什么?
对$x\in [n\pi,(n+1)\pi]$,则有\[\frac{{n\pi }}{{1 + {{\left( {n + 1} \right)}^2}{\pi ^2}{{\sin }^2}x}} < \text{被积函数} < \frac{{\left( {n + 1} \right)\pi }}{{1 + {n^2}{\pi ^2}{{\sin }^2}x}}.\]
-
求\[\int_0^1 {\left( {1 + \ln x} \right)\ln \left( {1 + x} \right)\ln \ln \frac{1}{x}dx} .\]
令$x=e^{-t}$,我们有$$ I = \int_{0}^{+\infty}e^{-t}(1-t)\log(1+e^{-t})\log t\,dt$$
由于
$$ \int_{0}^{+\infty}e^{-(n+1)t}(1-t)\log t\,dt = -\frac{1}{(n+1)^2}\left(1+n\gamma+n\log (n+1)\right),$$
我们有
$$ I = \sum_{n\geq 1}\frac{(-1)^n}{n(n+1)^2}\left(1+n\gamma+n\log (n+1)\right).$$
因此
$$ I = 2-2\log 2-\gamma-\zeta(2)\left(\frac{1}{2}+\log\sqrt{4\pi}\right)+\pi^2\log A,$$
其中 $A$ 是Glaisher-Kinkelin constant.
-
证明:\[I = \int_0^1 {{x^{ - x}}\left( {{{\ln }^2}x - 2} \right)dx} < 0.\]
注意到\[{x^{ - x}} = {e^{ - x\ln x}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k} \cdot \frac{{{x^k}{{\ln }^k}x}}{{k!}}} .\]
故有
\[I = \sum\limits_{k = 0}^\infty {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\int_0^1 {\left( {{x^k}{{\ln }^{k + 2}}x - 2{x^k}{{\ln }^k}x} \right)dx} } \right]} .\]
又
\[\int_0^1 {{x^k}{{\ln }^{k + 2}}xdx} = {\left( { - 1} \right)^k}\int_0^\infty {{e^{ - \left( {k + 1} \right)t}}{t^{k + 2}}dt} = \frac{{{{\left( { - 1} \right)}^k}\left( {k + 2} \right)!}}{{{{\left( {k + 1} \right)}^{k + 3}}}},x = {e^{ - t}}.\]
和\[\int_0^1 {{x^k}{{\ln }^k}xdx} = \frac{{{{\left( { - 1} \right)}^k}k!}}{{{{\left( {k + 1} \right)}^{k + 1}}}}.\]
因此
\[I = \sum\limits_{k = 0}^\infty {\left[ {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}\int_0^1 {\left( {\frac{{{{\left( { - 1} \right)}^k}\left( {k + 2} \right)!}}{{{{\left( {k + 1} \right)}^{k + 3}}}} - 2\frac{{{{\left( { - 1} \right)}^k}k!}}{{{{\left( {k + 1} \right)}^{k + 1}}}}} \right)dx} } \right]} = - \sum\limits_{k = 0}^\infty {\frac{k}{{{{\left( {k + 1} \right)}^{k + 2}}}}} < 0.\]
若干个著名的积分及文献
1.Integrals of the Ising class (D.H. Bailey J.M. Borwein R.E.Crandall) 2.Hypergeometric forms for Ising-class integrals (D.H. Bailey, D. Borwein, J.M. Borwein,R.E. Crandall) 3. Finding General Explicit Formulas for Ising Integral Recursions (D.H. Bailey J.M. Borwein) 4.On Recurrences for Ising Integrals (Johannes Kepler University Linz, Austria)
1.THREE TRIPLE INTEGRALS (G. N. WATSON) 2.WATSON'S THIRD INTEGRAL (Hannah Cairns) 3.ON THE EVALUATION OF GENERALIZED WATSON INTEGRALS (G. S. JOYCE AND I. J. ZUCKER)
1.Box integrals (D.H. Bailey J.M. Borwein R.E. Crandall) 2.Higher-dimensional box integrals (Jonathan M. Borwein O-Yeat Chan y R. E. Crandall) 3.ADVANCES IN THE THEORY OF BOX INTEGRALS (D. H. BAILEY, J. M. BORWEIN, AND R. E. CRANDALL)
1.Euler Sums and Contour Integral Representations (Philippe Flajolet and Bruno Salvy) 2.Experimental evaluation of Euler sums (D.H.Bailey J.M.Borwein andR.Girgensohn) 3.Evaluation of triple euler sums (Jonathan M. Borwein) 4.Harmonic sums,Mellin transforms and Integrals (J.A.M.Vermaseren,NIKHEF)
一系列类似积分相等的证明
前阵子四叶群里有人问道下面这题，虽感觉此题结论优美但无从下手.
一个很火的积分题
求解\[\displaystyle \int_0^1 \frac{\log^2(1-x)\log(x)}{x}dx=-\frac{\pi^4}{180}.\]
解.$For $|z|<1$ we have that
