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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.
ምሳሌ · integrate e^(-x) dx from 0 to oo
እርምጃ በደረጃ
- \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.
መልሱን አሳይ
አሁን ችግርን ይምረጡ፣ ወይም ጻፉ ወይም በራሳችሁ ቅርጽ አቅርቡ። እርምጃው ሁሉ፣ ምስል፣ መልሱ እስከምትጠይቅ ድረስ የተደበቀ ነው።
ፊደላት
ለሙሉ መግለጫ፣ ምስል፣ እና በውስጡ ያለውን ፊደል ትርጓሜ ለማየት ማንኛውንም ምልክት ይጫኑ።
እንዴት: 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.
ጥያቄዎች
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.