積分の問題なのですが。。。

質問

f(x,y)=x^2-y^2/(x^2+y^2)^2とするとき、次を確かめよ。
(1) lim ∬ f(x,y)dxdy = 0 (n→∞, 1/n≦x≦1, 1/n≦y≦1)

(2) ∬f(x,y) dxdy = ∞ (0＜x≦1, 0≦y≦x)

回答No.1

info22_

>f(x,y)=x^2-y^2/(x^2+y^2)^2とするとき、

これは括弧をつけた次式が正しいでしょう。
正：f(x,y)=(x^2-y^2)/(x^2+y^2)^2

(1)
I=lim[n→∞] ∬[D] f(x,y)dxdy, D={(x,y)|1/n≦x≦1, 1/n≦y≦1}
=lim[n→∞] ∫[1/n,1] dy∫[1/n,1] (x^2-y^2)/(x^2+y^2)^2 dx
=lim[n→∞] ∫[1/n,1] dy∫[1/n,1] {2x^2-(x^2+y^2)}/(x^2+y^2)^2 dx
=lim[n→∞] ∫[1/n,1] dy [-x/(x^2+y^2)] [1/n,1]
=lim[n→∞] ∫[1/n,1] [-1/(1+y^2)+n/(n^2y^2+1)] dy
=lim[n→∞] [-tan^-1(y)+tan^-1(ny)] [1/n,1]
=lim[n→∞] [tan^-1(1/n)+tan^-1(n)-(π/2)]
公式：tan^-1(A)+tan^-1(1/A)=π/2 より
=lim[n→∞] [(π/2)-(π/2)]
=lim[n→∞] 0
= 0

(2)
I=∬[D] f(x,y) dxdy, D={(x,y)|0＜x≦1, 0≦y≦x}
=lim[n→∞] ∫[1/n,1] dx∫[1/n,x] (x^2-y^2)/(x^2+y^2)^2 dy
=lim[n→∞] ∫[1/n,1] dx∫[1/n,x] {-2y^2+(x^2+y^2)}/(x^2+y^2)^2 dy
=lim[n→∞] ∫[1/n,1] dx [y/(x^2+y^2)] [1/n,x]
=lim[n→∞] ∫[1/n,1] [1/(2x)-n/(n^2x^2+1)] dx
=lim[n→∞] [(ln|x|)/2 -tan^-1(nx)] [1/n,1]
=lim[n→∞] [(ln(n))/2 -tan^-1(n) +(π/4)]
={lim[n→∞] (ln(n))/2} -(π/2) +(π/4)
= ∞　（発散）

お礼

ありがとうございます！！
本当に助かりました！！