フェルマーの最終定理の証明 (981ﾚｽ)
上下前次1-新
抽出解除 必死ﾁｪｯｶｰ(本家) (べ) 自ID ﾚｽ栞 あぼーん

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ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

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945: 09/29(月)22:06 ID:tsfKlyQm(1/8) AA×

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946: 09/29(月)22:07 ID:tsfKlyQm(2/8) AA×

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947: 09/29(月)22:09 ID:tsfKlyQm(3/8) AA×

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948: 09/29(月)22:10 ID:tsfKlyQm(4/8) AA×

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949: 09/29(月)22:12 ID:tsfKlyQm(5/8) AA×

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950: 09/29(月)22:13 ID:tsfKlyQm(6/8) AAS
∫[0→∞] (sin(x)/x)^2 dx
　　∫[0→∞] (sin(x)/x)^2 dx
　= ∫[0→∞] (1/x^2) sin(x)^2 dx
　= ∫[0→∞] (- 1/x)'sin(x)^2 dx
　= [(- 1/x) sin(x)^2][0→∞] - ∫[0→∞] (- 1/x) (sin(x)^2)'dx

　　[(- 1/x) sin(x)^2][0→∞]
　= [- sin(x)^2/x][0→∞]
　= lim[x→∞](- sin(x)^2/x) - lim[x→0](- sin(x)^2/x)
　= lim[x→∞](- sin(x)^2/x) - lim[x→0](- (sin(x)/x)^2 x)
　= (- 0) - (- 1･0)
　= 0

　　∫[0→∞] (- 1/x) (sin(x)^2)'dx
　= ∫[0→∞] (- 1/x) 2 sin(x) cos(x) dx
　= ∫[0→∞] (- 1/x) sin(2x) dx
　= -∫[0→∞] sin(2x)/x dx
　= -∫[0→∞] (sin(2x)/(2x))･2 dx
　= -∫[0→∞] (sin(2x)/(2x)) (d (2x)/dx) dx
　2x = t
　　∫[0→∞] (- 1/x) (sin(x)^2)'dx
　= -∫[0→∞] (sin(t)/t) (dt/dx) dx
　= -∫[0→∞] (sin(t)/t) dt
　= -π/2

　∫[0→∞] (sin(x)/x)^2 dx
= 0 - (- π/2)
= π/2

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951: 09/29(月)22:15 ID:tsfKlyQm(7/8) AAS
I=∫[0→∞] sin2x cosx (1/x2)dx = (1/2)∫[0→∞] (sin2x sinx)(1/x2)dx
=(1/2){ [sin2x sinx(-1/x)][∞,0]
- ∫[0→∞] (2cos2x sinx + sin2x cosx) (-1/x)dx }
=(1/2)[ 0+∫[0→∞] (1/2){(sin3x-sinx) + (sin3x+sinx)}/x dx ]
=(1/4)∫[0→∞] (2sin3x)/x dx = (1/2)∫[0→∞] (sin3x)/(3x) d(3x)
=π/4・・・・・?
∫[0→∞] sin3x (1/x3)dx
=[sin3x (-1/2x2)][∞,0] - ∫[0→∞] 3sin2x cosx (-1/2x2)dx
　　(sin3x)/x2=sinx(sinx/x)2 → 0・1=0 (x→0) )
　=-0+0+(3/2)∫[0→∞] (sin2x cosx)/x2 dx
　=(3/2)∫[0→∞] (sin2x cosx)/x2 dx
　= (3/2)I = 3π/8　(?から)

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952: 09/29(月)22:16 ID:tsfKlyQm(8/8) AAS
　?[0→∞]x^2/(x^2+1)^3dx＄
　f(z) = z^2/(z^2+1)^3
　　　 Res[f(z),i]
　　 = (1/2)lim_{z→i}{z^2/(z+i)^3}"
　　 = (1/2)lim_{z→i}[2{z/(z+i)^3}'-3{z^2/(z+i)^4}']
　　 = (1/2)lim_{z→i}[2{1/(z+i)^3-3z/(z+i)^4}-3{2z/(z+i)^4-4z^2/(z+i)^5}]
　　 = (1/2)lim_{z→i}[2/(z+i)^3-12z/(z+i)^4+12z^2/(z+i)^5]
　　 = lim_{z→i}[(z+i)^2-6z(z+i)+6z^2]/(z+i)^5
　　 = lim_{z→i}(z^2-4iz-1)/(z+i)^5
　　 = -i/16
　　∴∫_{C}f(z)dz
　　 = i2πRes[f(z),i]
　　 = π/8.
　　∫_{-R〜R}f(z)dz+∫_{Γ}f(z)dz = π/8
　　lim_{R→∞}∫_{Γ}f(z)dz
　= lim_{R→∞}∫_{0〜π}[ie^(3it)/{Re^(2it)+1/R}^3]dt = 0

　　∴∫_{-∞〜∞}x^2/(x^2+1)^3dx = π/8.
　　∫_{0〜∞}x^2/(x^2+1)^3dx = π/16.

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上下前次1-新書関写板覧索設栞歴

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