Improper integral Convergence theorem

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  • #1
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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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Answers and Replies

  • #2
lanedance
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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
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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
lanedance
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ok good point, missed the positive - must be too late ;)
 
  • #5
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This things happens sometimes, take another glance at my proof =)
Thank you!
 
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  • #6
Dick
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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
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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
Dick
Science Advisor
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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
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[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!
 

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