Question

求极限lim(t->0+) 1/t^2 ∫(0~t)dx∫(0~t-x)e^(x^2+y^2)dy,高手帮个忙,

Answer 1

f(t²) = ∫[0,t] dx ∫[0,t-x] e^(x²+y²) dy 化为极坐标D:0≤ r ≤ t / (sinθ+cosθ),0≤θ≤ π/2

= ∫[0,π/2] dθ ∫[0,t /(sinθ+cosθ)] e^(r²) r dr

= ∫[0,π/2] (1/2) [ e^ { t² /(sinθ+cosθ)² } ﹣1] dθ

令 u = t²,原式 = lim(u->0+) f (u) / u

f '(u) = ∫[0,π/2] (1/2) e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ

原式 = lim(u->0+) f ' (u) 洛必达法则

= lim(u->0) (1/2) ∫[0,π/2] e^ {u / (sinθ+cosθ)² } * 1/(sinθ+cosθ)² dθ

= (1/2) ∫[0,π/2] 1/(sinθ+cosθ)² dθ 第二积分中值定理

= (1/2) (-1/2) cot(θ+π/4) | [0,π/2]

= 1/2