$$I_n=[0,1]^n$$
1.$$int_{I_n}(x_1+x_2+cdots+x_n)mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n=frac{n}{2}$$
2.$$int_{I_n}(x_1^2+x_2^2+cdots+x_n^2)mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n=frac{n}{3}$$
3.$$int_{I_n}(x_1+x_2+cdots+x_n)^2mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n=frac{n}{3}+frac{n(n-1)}{4}$$
4.求$$A=lim_{n to infty}int_{[0,1]^n}f(frac{x_1+x_2+cdots+x_n}{n})mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n=f(frac{1}{2})$$
注：$mu(S_n(a))=frac{a^n}{n!}to 0(nto 0),mu(B_{2n}(a))=frac{a^{2n}pi^n}{n!}to 0,mu(I_n)=1$
记$u_n=frac{x_1+x_2+cdots+x_n}{n},I_n=(lbrace |u_n-frac{1}{2}|leqslantdelta bigcap I_nrbrace)bigcup(lbrace |u_n-frac{1}{2}|>delta bigcap I_nrbrace)$
$$
begin{aligned}
&left | A-f(frac{1}{2})right | \
&=left | int_{I_n}(f(u_n)-f(frac{1}{2})mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_nright | \
&leqslant int_{I_n}left |(f(u_n)-f(frac{1}{2})right |mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n\
&=int_{I_delta}left |(f(u_n)-f(frac{1}{2})right |mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n
&+int_{I_nbackslash I_delta}left |(f(u_n)-f(frac{1}{2})right |mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n
end{aligned}\
forall epsilon >0,exists delta >0,when |x-y|<0,|f(x)-f(y)|<epsilon\
therefore above<epsilonmu(I_delta)+2Mint_{I_nbackslash I_delta}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n\
text{下证} lim_{nto +infty}
int_{I_nbackslash I_delta}1mathop{}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n=0\
begin{aligned}
int_{I_nbackslash I_delta}1mathop{}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n&leqslant int_{I_n
backslash I_delta}frac{(u_n-frac{1}{2})^2}{delta^2}mathop{}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n\
&leqslantfrac{1}{delta^2}int_{I_n}(u_n-frac{1}{2})^2mathop{}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n\
&=frac{1}{delta^2}int_{I_n}left[frac{(x_1+x_2+cdots+x_n)^2}{n^2}-frac{x_1+cdots+x_n}{n}+frac{1}{4}right]mathop{}mathop{}mathrm{d}x_1mathop{}mathrm{d}x_2cdotsmathop{}mathrm{d}x_n\
&=frac{1}{12ndelta^2}to 0
end{aligned}
$$
证明$$iint_{I^2}(xy)^{xy}mathop{}mathrm{d}xmathop{}mathrm{d}y=int_{0}^{1}t^tmathop{}mathrm{d}t$$
$$
leftlbrace
begin{aligned}
&u=xy &(0leqslant uleqslant 1)\
&v=x &(uleqslant xleqslant 1)
end{aligned}
right.\
left=-iint u^ufrac{1}{v}mathop{}mathrm{d}umathop{}mathrm{d}v=-int_{0}^{1}mathop{}mathrm{d}uint_{u}^{1}u^ufrac{1}{v}=-int_{0}^{1}u^uln umathop{}mathrm{d}u\
frac{mathop{}mathrm{d}{y^y}}{mathop{}mathrm{d}{y}}=y^y(ln y+1)\
0=int_{0}^{1}mathop{}mathrm{d}(y^y)=int_{0}^{1}y^ymathop{}mathrm{d}y+int_{0}^{1}y^yln ymathop{}mathrm{d}y\
therefore -int_{0}^{1}u^uln umathop{}mathrm{d}u=int_{0}^{1}u^umathop{}mathrm{d}u\
therefore iint_{I^2}(xy)^{xy}mathop{}mathrm{d}xmathop{}mathrm{d}y=int_{0}^{1}t^tmathop{}mathrm{d}t
$$
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