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Improper integrals

Integrals to infinity or across a singularity, defined as limits.

∫₁^∞ 1/x² dx is a limit of ∫₁^b as b → ∞, and it converges to 1; the same integral of 1/x diverges. The boundary between the two, p = 1, is the same boundary as for the p-series. The integral test is exactly this correspondence.

tusaale · integrate 1/x^2 dx from 1 to oo

Calculus · Definite-Integrals

Integrate x^(-2) from 1 to oo

\int_{1}^{\infty} \frac{1}{x^{2}}\, dx

tallaabo tallaabo

  1. \int_{1}^{\infty} \frac{1}{x^{2}}\, dx

    First find an antiderivative F, then evaluate F(b) − F(a).

  2. \int \frac{1}{x^{2}}\, dx = - \frac{1}{x}

    Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).

  3. F(\infty) - F(1) = \left(0\right) - \left(-1\right)

    Fundamental theorem of calculus: plug in the limits.

  4. = 1

    Simplify.

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1

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Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Sida loo: Improper integrals

  1. First find an antiderivative F, then evaluate F(b) − F(a).
  2. Power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) (n ≠ −1).
  3. Fundamental theorem of calculus: plug in the limits.
  4. Simplify.

Su'aalaha dadku waydiiyaan

What is the ε, δ definition actually saying?

That you can make the output as close to the limit as anyone demands (within ε) by keeping the input close enough (within δ). It replaces "approaches" with a challenge-and-response that can be checked.

Why does the harmonic series diverge when its terms go to zero?

Because the terms shrink too slowly: group them as 1/3 + 1/4 > 1/2, 1/5 + … + 1/8 > 1/2, and so on: infinitely many halves.

In ka badan Real Analysis

Sequences and their limitsConvergence of seriesTaylor approximation and errorFunctions of a complex variableContinuity and differentiability, rigorouslySequences of functions and uniform convergence