Improper integral Convergence theorem

In summary, we cannot prove that if the series of integrals is convergent, then the integral itself is convergent without using the fact that f(x) is positive. Therefore, the statement is false and a counterexample is provided by lanedance's example.
  • #1
estro
241
0

Homework Statement



f(x) is a continuous and positive function when [tex] x\in[0,\infty)[/tex]. (#1)

[tex] x_n [/tex] is a monotonic increasing sequence, [tex]x_0=0[/tex] [tex],x_n \rightarrow \infty[/tex]. (#2)

Prove or contradict:

[tex] \mbox{If } \sum_{n=0}^\infty \int_{x_n}^{x_(n+1)} f(x)dx \mbox{ is convergent (#3) then } \int_{0}^\infty f(x)dx \mbox{ is also convergent.}[/tex]3. The attempt

[tex](*3)\ and\ by\ the\ Cauchy\ Criterion\ \Longrightarrow\ \forall\ \epsilon>0\ \exists\ N_1>0,\ so\ \forall\ m>k>N_1[/tex]

[tex]\left \int_{x_(k+1)}^{x_(m+1)} f(x)dx=(*1)=| \sum_{n=0}^\infty \int_{x_n}^{x_(n+1)} f(x)dx|<\epsilon\mbox{ (*4)} \right [/tex][tex](*2)\ \Longrightarrow\ \forall\ N_1>0\ \exists\ N>0\ so\ \forall\ n>N_1,\ x_n>N\ \ \ \ (*5) } [/tex]

[tex](*4)\ and\ (*5)\ \Longrightarrow\ \forall\ m>k>N\ \int_{x_(k+1)}^{x_(m+1)} f(x)dx<\epsilon\\\Longrightarrow\ Cauchy\ Criterion\ \int_{0}^\infty f(x)dx\ is\ convergent. [/tex]

It seems right to me, but I'm not sure...
I think i also have vice versa proof.
 
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  • #2
how about this as a counter example...

let
[tex]f(x) = cos(2 \pi x) [/tex]
[tex]x_n = n \in \mathds{N} [/tex]

Then
[tex] \int_{x_n}^{x_{n+1}} f(x)dx = \int_{x_n}^{x_{n+1}} cos(x)dx
= 0 [/tex]

But
[tex] \int_{0}^{x}cos(2 \pi x)dx = -\frac{1}{\pi} sin(x) [/tex]

though i guess it likely changes if it is true for any series...
 
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  • #3
But f(x) supposed to be positive and continuous when [tex]x \in [0,\infty)[/tex]

I'm actually taking about integral inside a series

And if I am wrong where is my blunder in the proof?
 
  • #4
ok good point, missed the positive - must be too late ;)
 
  • #5
This things happens sometimes, take another glance at my proof =)
Thank you!
 
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  • #6
Put an upper bound of N on the sum on the left side instead of infinity. Then you know the left side approaches a limit L as N->infinity. Put a definite upper bound of M on the right side integral. Then you want to show the integral also approaches L as M->infinity. You know for any integer N the sum is equal to the integral with an upper limit of x_N+1. So define F(M)=integral from 0 to M of f(x). You know lim F(x_n)->L as n->infinity. You just want to show F(M)->L as M->infinity. I.e. you just have to worry about the points M that are in between the x_i.
 
  • #7
I wasn't able to fully understand all the points.
(The limit idea)

---------------------------------------------

Homework Statement



f(x) is a continuous and positive function when [tex] x\in[0,\infty)[/tex]. (*1)

[tex] x_n [/tex] is a monotonic increasing sequence, [tex]x_0=0[/tex] [tex],x_n \rightarrow \infty[/tex]. (*2)

Prove or contradict:

[tex] \mbox{If } \sum_{n=0}^\infty \int_{x_n}^{x_(n+1)} f(x)dx \mbox{ is convergent (*3) then } \int_{0}^\infty f(x)dx \mbox{ is also convergent.}[/tex]
-------------------------------------------------------------

[tex]From\ (*3)\ \Rightarrow\ \forall\ \epsilon\ > 0\ \exists\ N_1>0\ so\ \forall\ m>k>N_1 [/tex]

[tex]|\sum_{n=k+1}^{m} \int_{x_n}^{x_{n+1}} f(x)dx|=|\int_{x_{k+1}}^{x_{m+1}} f(x)dx|<\epsilon\ \ \ \ \ (*4) [/tex]

[tex]From\ (*2)\ \Rightarrow\ \exists\ N_1>0\ so\ \forall\ n>N\ x_n>N\ \ \ \ \ (*5)[/tex]

[tex]From\ (*4)\ and\ (*5)\ \Rightarrow\ \forall\ \epsilon>0\ and\ \forall\ s>t>N[/tex]

[tex]|\int_{t}^{s}f(x)dx|<\epsilon [/tex]

[tex]So\ by\ the\ Cauchy\ Criterion\ \int_{N}^\infty f(x)dx\ is\ convergent\ \ \ \ \ (*6) [/tex]

[tex]From\ (*1)\ and\ (*6)\ \int_{0}^{\infty} f(x)dx\ is\ convergent [/tex]
 
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  • #8
estro said:
I wasn't able to fully understand all the points.
(The limit idea)

---------------------------------------------

Homework Statement



f(x) is a continuous and positive function when [tex] x\in[0,\infty)[/tex]. (*1)

[tex] x_n [/tex] is a monotonic increasing sequence, [tex]x_0=0[/tex] [tex],x_n \rightarrow \infty[/tex]. (*2)

Prove or contradict:

[tex] \mbox{If } \sum_{n=0}^\infty \int_{x_n}^{x_(n+1)} f(x)dx \mbox{ is convergent (*3) then } \int_{0}^\infty f(x)dx \mbox{ is also convergent.}[/tex]
-------------------------------------------------------------

[tex]From\ (*3)\ \Rightarrow\ \forall\ \epsilon\ > 0\ \exists\ N_1>0\ so\ \forall\ m>k>N_1 [/tex]

[tex]|\sum_{n=(k+1)}^{m} \int_{x_n}^{x_{n+1}} f(x)dx|=|\int_{x_{k+1}}^{x_{m+1}} f(x)dx|<\epsilon\ \ \ \ \ (*4) [/tex]

[tex]From\ (*2)\ \Rightarrow\ \exists\ N_1>0\ so\ \forall\ n>N\ x_n>N\ \ \ \ \ (*5)[/tex]

[tex]From\ (*4)\ and\ (*5)\ \Rightarrow\ \forall\ \epsilon>0\ and\ \forall\ s>t>N[/tex]

[tex]|\int_{t}^{s}f(x)dx|<\epsilon [/tex]

[tex]So\ by\ the\ Cauchy\ Criterion\ \int_{N}^\infty f(x)dx\ is\ convergent\ \ \ \ \ (*6) [/tex]

[tex]From\ (*1)\ and\ (*6)\ \int_{0}^{\infty} f(x)dx\ is\ convergent [/tex]

Did you understand any of the points? You 'proof' completely misses the point. Where did you use that f(x) is positive? You need to use that. Otherwise lanedance's example shows it's false.
 
  • #9
[tex]From\ (*3)\ and\ (*2)\ and\ (*1)\ \forall\ m:\ \ \ \ \ 0\leq S_m=\int_{0}^{m} f(x)dx\ \leq L[/tex]

I understand now why my proof is wrong and all your points, thank you!
 

1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite, or the integrand itself is infinite at one or more points in the interval of integration.

2. What is the convergence theorem for improper integrals?

The convergence theorem for improper integrals states that if the integral converges, then the limit of the integral as the upper or lower limit of integration approaches infinity is equal to the improper integral.

3. How do you determine if an improper integral converges?

To determine if an improper integral converges, you can use the comparison test, limit comparison test, or the ratio test. These tests compare the integrand to a known function and determine if it has similar convergence properties.

4. What is the significance of the convergence theorem for improper integrals?

The convergence theorem for improper integrals allows us to evaluate integrals that would otherwise be impossible to calculate. It also helps us determine the convergence or divergence of integrals and understand the behavior of functions at infinity.

5. Can an improper integral converge at both limits of integration?

Yes, an improper integral can converge at both limits of integration. This is known as a convergent improper integral. However, it is also possible for an improper integral to converge at one limit and diverge at the other, in which case it is called a semi-convergent improper integral.

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