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∫∞01xdx∫x0cos(x−y)−cosxydy.
解.(翻译而来)令f(x,y)=cos(x−y)−cosxy.对x>0,我们有∫x0f(x,y)dy=∫10cos(1−t)x−cosxydt=x∫101t∫11−tsinuxdudt.
因而对R>0,
∫R01x∫x0f(x,y)dydx=∫R0∫101t∫11−tsinuxdudtdx.
而|sinux|≤1,该三重积分是绝对收敛的.由Fubini定理可知积分能交换次序
∫R01x∫x0f(x,y)dydx=∫10∫11−t1t∫R0sinuxdxdudt=∫101−cosRuu∫11−u1tdtdu=−∫10ln(1−u)udu+∫10ln(1−u)ucosRudu.
我们知|ln(1−u)/u|∈L1([0,1]),由Riemann-Lebesgue引理可知
lim
由于{\sum\limits_{n = 1}^\infty {\frac{{{t^{n - 1}}}}{n}} }一致收敛,故可逐项积分.因此我们有
\begin{align*}&\mathop {\lim }\limits_{R \to \infty } \int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)dydx} = - \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}du} \\= &\int_0^1 {\sum\limits_{n = 1}^\infty {\frac{{{t^{n - 1}}}}{n}} dt} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}.\end{align*}