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The Lebesgue integral and convergence theorems
Integrating by slicing the range instead of the domain; monotone and dominated convergence.
Riemann slices the x-axis; Lebesgue slices the y-axis and asks how much of the domain sits at each height. The payoff is the convergence theorems: under mild conditions, the limit of the integrals is the integral of the limit. Picture it: horizontal strips under the curve instead of vertical ones. Think it: L¹ and L² become complete spaces, the Hilbert spaces of functional analysis.
Çözümlü örnek · integrate e^(-x) dx from 0 to oo
Adım adım
- \int_{0}^{\infty} e^{- x}\, dx
First find an antiderivative F, then evaluate F(b) − F(a).
- u = - x,\quad du = -1\, dx
Substitute u = - x.
- \int e^{- x}\, dx = \int - e^{u}\, d_u
Rewrite the integral in terms of u.
- \int - e^{u}\, d_u = -1 \int e^{u}\, d_u
Pull the constant -1 out of the integral.
- \int e^{u}\, d_u = e^{u}
∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- = - e^{- x}
Substitute back u = - x.
- F(\infty) - F(0) = \left(0\right) - \left(-1\right)
Fundamental theorem of calculus: plug in the limits.
- = 1
Simplify.
Cevabı göster
Şimdi sen Bir problem seçin, ya da kendi probleminizi yazın ya da çizin. Her adım, bir resim, cevap sorulana kadar gizli.
Burada kullanılan sembolleri
Tüm tanım, resim ve her harfin anlamı için herhangi bir sembolü tıklayın.
Nasıl: The Lebesgue integral and convergence theorems
- First find an antiderivative F, then evaluate F(b) − F(a).
- Substitute u = - x.
- Rewrite the integral in terms of u.
- Pull the constant -1 out of the integral.
- ∫ aᵘ du = aᵘ / ln a (for eˣ that is just eˣ).
- Substitute back u = - x.
- Fundamental theorem of calculus: plug in the limits.
- Simplify.
Sık sorulan sorular
What is wrong with the Riemann integral?
It fails on functions that oscillate too much, and it does not interact well with limits: the limit of integrable functions need not be integrable. Lebesgue's integral fixes both.