4.证明ye-rdr在0≤y≤1非一致收敛. 证明：设=A>0,A0>A和物=e0,1,使得 =m(-e+ewo=eww=e=是>6 所以由非一致收敛的定义，ye-rdz在0≤y≤1非一致收敛 5.设r(a)=0x-e-+dr,a>0,证明∫eed=V 证明：设x2=t,d血=兴，有 装 订 fck-i"cicu-3ave 线 -r- 内 因为函数f(x)=e2是偶函数，所以有 答 rk-2e=2×g= 题 无 效 数学分析（四）试题第7页（共8页） C ¾ ‚ S ‰ K à  ** ** ** ** ** ** ** ** ** ** C ** ** ** ** ** ** ** ** ** ** ¾ ** ** ** ** ** ** ** ** ** ** ‚ ** ** ** ** ** ** ** ** ** 4. y² R +∞ 0 ye−yxdx 3 0 ≤ y ≤ 1 šÂñ. y²:  0 = 1 3 , ∀A > 0, ∃A0 > A Ú y0 = 1 A ∈ [0, 1], ¦     Z +∞ A0 y0e −y0x dx     =     lim p→+∞ Z p A0 e −y0x d(y0x)     =     lim p→+∞ (−e −y0x )    p A0     =     lim p→+∞ (−e −y0p + e −y0A0 )     = e −y0A0 = e − 1 A0 ×A0 = 1 e > 1 3 = . ¤±dšÂñ½Â, R +∞ 0 ye−yxdx 3 0 ≤ y ≤ 1 šÂñ. 5.  Γ(α) = R +∞ 0 x α−1 e −xdx, α > 0, y² R +∞ −∞ e −x 2 dx = √ π. y²:  x 2 = t, dx = dt 2 √ x , k Z +∞ 0 e −x 2 dx = 1 2 Z +∞ 0 t − 1 2 e −t dt = 1 2 Z +∞ 0 t ( 1 2 −1)e −t dt = 1 2 Γ(1 2 ) = 1 2 π. Ϗ¼ê f(x) = e −x 2 ´ó¼ê, ¤±k Z +∞ −∞ e −x 2 dx = 2 Z +∞ 0 e −x 2 dx = 2 × π 2 = π. êÆ©Û(III)ÁK 1 7 £
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