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Integral , uniform convergence

  • Thread starter Madou
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  • #1
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I shall prove that this integral is uniformly convergent or not:

wj6x3t.jpg


and if it is convergent, i must describe its uniform convergence
 

Answers and Replies

  • #2
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please, say something!
 
  • #3
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Ask a question first.
 
  • #4
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okay
 
  • #5
1,838
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Clearly it is not uniformly convergent. You can compute the integral for alpha>0 by substituting X = t/sqrt(alpha). Then you find that the integral as a function of alpha, I(alpha) = constant.

If the integral were to converge uniformly, then that implies that
I(alpha) would be a continuous function. Then it follows that, in particular, the limit for alpha to zero would be given by I(0). But
I(0) does not exist.
 
  • #6
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So, one more time, this is what i'm doing - i'm inspecting for uniform convergence:
[tex] \int_{0}^{\infty}{\frac{\sqrt{\alpha}}{\sqrt{{\alpha}^{2}x^{4}+1}}dx} [/tex] where [tex]\alpha \in \ [0,1][/tex].
 
Last edited:
  • #7
1,838
7
So, one more time, this is what i'm doing - i'm inspecting for uniform convergence:
[tex] $\int_{0}^{\infty}{\frac{\sqrt{\alpha}}{\sqrt{{\alpha}^{2}x^{4}+1}}dx}$, where $\alpha \in {[0,1]}$
[/tex]
There was a mistake in my previous reply, I(0) is of course zero.

So, what you can do is show that for alpha not equal to zero, I(alpha) = constant and that this constant is not equal to zero. Then, if the integral were uniformly convergent, the limit of the integral would be the integral of the limit.

If you take the limit of alpha to zero inside the integral you get zero, if you first evaluate the integral and then take the limit you find a different result. This then proves that the integral does not converge uniformly.
 
  • #8
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Well, first i'm trying to clarify the definition of the uniform convergence of the improper integral.
 
  • #9
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No-no, first the definition of the uniform convergence of the funсtional series.
 
  • #10
1,838
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The improper integral is the limit of the integral from zero to R for R to infinity. Let's call the integral I(R, alpha) and the limit for fixed alpha of I(R, alpha) for R to infinity I(alpha). This then means that for every epsilon there exists a Q such that for all R > Q we have:

|I(R, alpha) - I(alpha)| < epsilon

Uniform convergence means that Q can be chosen independent of alpha. You can then prove that if I(R, alpha) for fixed R is continuous as a function of alpha, then I(alpha) will also be a continuous function.
 
  • #11
1,838
7
So, if we define the function [itex]I(\alpha)[/itex] as

[tex]\lim_{R\rightarrow \infty}I(R,\alpha) = I(\alpha)[/tex]

and it is the case that for all R the function [itex]|I(R, \alpha)[/itex] is continuous, we want to prove that the limit function [itex]|I(\alpha)[/itex] is continuous as well, if the limit for R to infinity converges uniformly. So, we want to prove that:

[tex]\lim_{\alpha\rightarrow\beta}I(\alpha) = I(\beta)[/tex]

This means that for every [itex]\epsilon>0[/itex] there should exists a [itex]\delta[/itex], such that we have

[tex]|I(\alpha)-I(\beta)|<\epsilon[/tex]

if we choose [itex]\alpha[/itex] such that

[tex]|\alpha-\beta|<\delta[/tex]

Now, uniform convergence for the limit of R to infinity implies that we can always find an [itex]R_{0}[/itex] for which

[tex]|I(R_{0},\alpha)-I(\alpha)|<\frac{\epsilon}{3}[/tex]

is true for all [itex]\alpha[/itex].

Then because [itex]I(R_{0},\alpha)[/itex] is continuous as a function of [itex]\alpha[/itex], i.e. we have that

[tex]\lim_{\alpha\rightarrow\beta}I(R_{0}, \alpha)=I(R_{0}, \beta)[/tex]

we can thus be sure that there exists a [itex]\delta[/itex] such that:

[tex]|I(R_{0},\alpha)-I(R_{0},\beta)|<\frac{\epsilon}{3}[/tex]

is true for [itex]\alpha[/itex] in the interval

[tex]|\alpha-\beta|<\delta[/tex]

For such [itex]\alpha[/itex] we have that

[tex]|I(\alpha)-I(\beta)| <\epsilon[/tex]

because

[tex]
\begin{align*}
|I(\alpha)-I(\beta)| &= |I(\alpha) - I(R_{0},\alpha) +
I(R_{0},\alpha)-I(R_{0},\beta) + I(R_{0},\beta) - I(\beta)|\\
& \leq
|I(\alpha) - I(R_{0},\alpha)| + |I(R_{0},\alpha)-I(R_{0},\beta)| +
| I(R_{0},\beta) - I(\beta)|<\epsilon
\end{align*}
[/tex]
 
  • #12
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You say
if you first evaluate the integral and then take the limit you find a different result.
And how do i first evaluate the integral?
 
  • #13
42
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alright, thank you, anyway - i think i have nearly solved the problem. I'll tell everyone, if i finish.
 
  • #14
1,838
7
Call the integral (as I did above) [itex]I(\alpha)[/itex]. Then if [itex]\alpha\neq 0[/itex] we can substitute [itex]x = \frac{t}{\sqrt{\alpha}}[/itex]. We then have:

[tex]\frac{1}{1+\alpha^{2}x^4}=\frac{1}{1+t^2}[/tex]

and

[tex]dx = \frac{dt}{\sqrt{\alpha}}[/tex]

So, the integral becomes:

[tex]I(\alpha)= \int_{0}^{\infty}\frac{\sqrt{\alpha}dx}{1+\alpha^{2}x^{4}}=\int_{0}^{\infty}\frac{dt}{1+t^4}[/tex]

So, we see that [itex]I(\alpha)[/itex] actually does not depend on[itex]\alpha[/itex]. It is a constant function. However, this is only the case for nonzero [itex]\alpha[/itex]. If [itex]\alpha=0[/itex] we have:

[tex]I(0)=\int_{0}^{\infty}\frac{0 dx}{1} = 0 [/tex]

Since

[tex]\int_{0}^{\infty}\frac{dt}{1+t^4}[/tex]

is clearly nonzero, we see that [itex]I(\alpha)[/itex] is a discontinuous function. This then contradicts uniform convergence, because of the theorem that relates uniform convergence to the limit function being continuous (see my previous posting).
 

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