積分の問題なのですが。。。

質問

次の広義積分を{Dn}を用いて求めよ。

∬(x+1)/(x+y+1)^α dxdy (α＞3) (x≧0,y≧0)
Dn={(x,y):0≦x≦n,0≦y≦n-x}

∬y/(1+x+y^2)^2 dxdy (x≧0,0≦y≦1)
Dn={(x,y):0≦x≦n,0≦y≦1}

∬1/√(x+y)^5 dxdy,( 0≦x≦1, 0≦y≦1, (x,y)≠(0,0) )
Dn={(x,y):0≦x≦1,1/n≦y≦1} U {(x,y):1/n≦x≦1,0≦y≦1/n}

やり方なども教えていただけると助かります。。。

回答No.3

info22_

#1,#2です。

3番目の問題
>∬1/√(x+y)^5 dxdy,( 0≦x≦1, 0≦y≦1, (x,y)≠(0,0) )
>Dn:{(x,y):0≦x≦1,1/n≦y≦1}, U :{(x,y):1/n≦x≦1,0≦y≦1/n}

Dn:{(x,y):0≦x≦1,1/n≦y≦1}, Un: {(x,y):1/n≦x≦1,0≦y≦1/n}

I=∬[D] 1/√(x+y)^5 dxdy D:{(x,y)|0≦x≦1, 0≦y≦1, (x,y)≠(0,0)}
=lim[n->∞] {∬[Dn] 1/√(x+y)^5 dxdy+∬[Un] 1/√(x+y)^5 dxdy}
=lim[n->∞] {∫[0,1] dx∫[1/n,1] 1/√(x+y)^5 dy+∫[1/n,1] dx∫[0,1/n] 1/√(x+y)^5 dy}
=lim[n->∞] {∫[0,1] [(-2/3)(x+y)^(-3/2)] [y:1/n,1] dx+∫[1/n,1] [(-2/3)(x+y)^(-3/2)] [y:0,1/n] dx }
=lim[n->∞] {∫[0,1] (2/3)[(n^(3/2)(nx+1)^(-3/2)-(x+1)^(-3/2)] dx
+∫[1/n,1] (2/3)[x^(-3/2)-n^(3/2)(nx+1)^(-3/2)] dx }
=(2/3)lim[n->∞] {∫[0,1] [(n^(3/2)(nx+1)^(-3/2)-(x+1)^(-3/2)] dx
+∫[1/n,1] [x^(-3/2)-n^(3/2)(nx+1)^(-3/2)] dx }
=(2/3)lim[n->∞] { [-2(n^(1/2))/(nx+1)^(1/2)+2/(x+1)^(1/2)] [0,1]
+[-2/x^(1/2)+2(n^(1/2))/(nx+1)^(1/2)] [1/n,1] }
=(4/3)lim[n->∞] { [-(n^(1/2))/(nx+1)^(1/2)+1/(x+1)^(1/2)] [0,1]
+[-1/x^(1/2)+(n^(1/2))/(nx+1)^(1/2)] [1/n,1] }
=(2/3)lim[n->∞] { -[2√(n(n+1))-2n(√n)-(√2)n+2n-2(√n)-(√2)+2]
+[(2√(n(n+1))+(2-√2)n(√n)-2n+(2-√2)(√n)-2] }/(n+1)
=(2/3)lim[n->∞] {4-(√2) } { (√n)-1}
=∞（発散）

回答No.2

info22_

#１です。

１番目に続いて２番目だけ

I=
>∬[x≧0,0≦y≦1] y/(1+x+y^2)^2 dxdy
=lim[n->∞]∫[0,n] dx ∫[0,1] y/(1+x+y^2)^2 dy
=lim[n->∞]∫[0,n] dx [-1/{2(y^2+x+1)}] [y:0,1]
=lim[n->∞]∫[0,n] [1/(2x+2)-1/(2x+4)] dx
=lim[n->∞] (1/2)[log|x+1|-log|x+2|] [0,n]
=lim[n->∞] (1/2)[log{(n+1)/(n+2)}+log(2)]
=lim[n->∞] (1/2)log{(1+(1/n))/(1+(2/n)}+(1/2)log(2)
=(1/2)log(1)+(1/2)log(2)
=(1/2)log(2)
(ただし対数は自然対数とする。)

お礼

本当に助かりました！
ありがとうございました！！

回答No.1

info22_

とりあえず、まず１番目だけ。
書きづらいのでαをaと書きます。

I=∬(x+1)/(x+y+1)^a dxdy
=lim[n->∞]∫[0,n] (x-1)dx∫[0,n-x] 1/(x+y+1)^a dy
=lim[n->∞]∫[0,n] (x-1)dx [((x-1)(y+x+1)^(1-a))/(1-a)] [y:0,n-x]
=lim[n->∞]∫[0,n] (x-1) [(x-1)(n+1)^(1-a)-(x-1)(x+1)^(1-a)]/(1-a) dx
=lim[n->∞] {1/(1-a)}[{(a-2)x^3+ax^2+(6-a)x-a+4}e^{-alog(x+1)}/(a^2-5a+6)+{(n+1)^(1-a)}{(1/2)x^2-x}] [x:0,n]
=lim[n->∞] -[e^{-alog(n+1)}
*({2(a-4)(n+1)^a+(a-2)(a-3)n^3+a(3-a)n^2+2a(3-a)n-2a+6}e^{alog(n+1)}
+{(n+1)^a}{2(a-2)n^3+2(2a-3)n^2+2an+2})
]/[2{(n+1)^a}(a-1)(a-2)(a-3)]
=-(a-4)/{(a-1)(a-2)(a-3)}

(☆)途中の計算は煩雑なので省略していますが、丹念に計算間違いしないように自身で計算し確認して下さい。