11. 记A = ∫∫{D} f(x,y) dxdy, 由条件得f(x,y) = A²xy+1/4.
于是A = ∫∫{D} f(x,y) dxdy
= A²·∫∫{D} xy dxdy + 1/4·∫∫{D} 1 dxdy
= A²·∫{0,1}∫{0,1} xy dxdy +1/4
= A²·(∫{0,1} y dy)(∫{0,1} x dx) +1/4
= A²/4+1/4.
解得A = 2±√3.
代回得f(x,y) = (7+4√3)xy+1/4或f(x,y) = (7-4√3)xy+1/4
12. ∫{0,1} x·ln(x)/(x-1) dx
= ∫{0,1} -(1-t)·ln(1-t)/t dt (换元t = 1-x)
= ∫{0,1} (1-t)/t·∑{1 ≤ n} t^n/n dt (ln(1-t)在t = 0处幂级数展开)
= ∫{0,1} 1/t·∑{1 ≤ n} t^n/n - ∑{1 ≤ n} t^n/n dt
= ∫{0,1} ∑{1 ≤ n} t^(n-1)/n - ∑{1 ≤ n} t^n/n dt
= ∫{0,1} 1+∑{2 ≤ n} t^(n-1)/n - ∑{2 ≤ n} t^(n-1)/(n-1) dt
= ∫{0,1} 1+∑{2 ≤ n} (1/n-1(n-1))t^(n-1) dt
= 1+∑{2 ≤ n} (1/n-1(n-1))·∫{0,1} t^(n-1) dt (由级数在(-1,1)内闭一致收敛, 可逐项积分)
= 1+∑{2 ≤ n} (1/n-1(n-1))·1/n
= 1-∑{2 ≤ n} 1/(n²(n-1)),
即所求证.
11
设∫∫D f(x,y)dxdy=t
那么f(x,y)=t^2xy+1/4
那么∫∫D f(x,y)dxdy=∫∫D (t^2xy+1/4)dxdy=(t^2+1)/4=t
解得t=2±√3
所以f(x,y)=t^2xy+1/4=(7±4√3)xy+1/4
12再想想
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