Improper Integral Proof

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Homework Statement



Prove that [tex] \int_0^{\infty} sin^2(\pi(x + 1/x))dx [/tex] does not exist.


Homework Equations





The Attempt at a Solution




First, we can construct a sequence as follows:

[tex] \int_0^{\infty} f(x)dx = \lim_{n \rightarrow \infty} S_n [/tex], where [tex] S_n =\int_0^{1} f(x)dx, \int_0^{2} f(x)dx, \int_0^{3} f(x)dx, \int_0^{4} f(x)dx, ..., \int_0^{n} f(x)dx [/tex]. Now, the sequence S_n converges as n approaches infinity if and only if [tex] |S_m - S_n| < \epsilon [/tex] for all epsilon > 0, provided that there is an integer N such that m,n > N. We set m = n + 1. We then have [tex] |S_{n+1} - S_n| = | \int_0^{n+1} f(x)dx - \int_0^{n} f(x)dx |[/tex].

Now,

[tex]\int_0^{n+1} f(x)dx - \int_0^{n} f(x)dx = \int_n^{n+1} f(x)dx = f(c) [/tex] for some c inbetween n and n + 1 (By the MVT of Integral Calculus). And so [tex] |S_{n+1} - S_n}| = f(c)[/tex]. Now, if |f(c)| < epsilon for sufficiently large c (remember that c goes to infinity as n goes to infinity), then this means that the function approaches a limit of 0, which it obviously doesn't as it's periodic. Thus the integral doesn't converge. QED.


How does this proof look?
 
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Answers and Replies

  • #2
Dick
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Nah, I don't buy that. The integrand isn't really periodic in the variable x, is it? It is periodic in the variable u=x+1/x and it's bounded in x. Try to exploit that.
 
  • #3
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You don't think the entire proof works, or just the last part where I say the function is periodic? I'll get cracking back on this problem tomorrow, I've gotta catch some sleep now. Goodnight.
 
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  • #4
Dick
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The whole proof. How do you know |f(c)|<epsilon is impossible? Even if it was periodic. It's a pile of junk. Throw it away. Think about the change of variable u=x+1/x for x larger than some number. Now it's periodic and positive in that range. Obviously divergent. But don't forget about accounting for du vs. dx.
 
  • #5
Dick
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You don't think the entire proof works, or just the last part where I say the function is periodic? I'll get cracking back on this problem tomorrow, I've gotta catch some sleep now. Goodnight.

Goodnight. Think about it fresh in the morning.
 
  • #6
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Sorry about the late reply, I've been tied up lately.

So I've tried using u = x + 1/x. Then du = (1 - 1/x^2)dx and so this is difficult to integrate...infact according to my integrator, it needs to use complex numbers and since I haven't learned that yet, I know that I'm not really looking to integrate the function...


Looking at the graph of the function, it seems that since it's continuously "wavy" and positive over all of R, then the area under the graph from 0 to infinity must be infinity, and so it diverges. I'm just having trouble proving this rigorously.
 
  • #7
Dick
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Sorry about the late reply, I've been tied up lately.

So I've tried using u = x + 1/x. Then du = (1 - 1/x^2)dx and so this is difficult to integrate...infact according to my integrator, it needs to use complex numbers and since I haven't learned that yet, I know that I'm not really looking to integrate the function...


Looking at the graph of the function, it seems that since it's continuously "wavy" and positive over all of R, then the area under the graph from 0 to infinity must be infinity, and so it diverges. I'm just having trouble proving this rigorously.

Right. But the point is that for large values of x, du is ALMOST dx. If x>2 then (1-1/x^2)>=3/4. Isn't that enough to prove it diverges?
 
  • #8
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Hmmm...I'm not sure I'm entirely getting what you're saying, but I'll give it a shot:

For x getting larger and larger, du gets closer and closer to dx. I see that (1 - 1/x^2) increases monotonically towards 1 as x tends to infinity, so we could say that for very large values of x, du = dx + r(x), where r(x) is some function approximating the error. Since du gets closer and closer to dx, we know that r(x) goes to 0 as x approaches infinity. So using the substitution u = 1 + 1/x we have

[tex] \int_0^{\infty} sin^2(\pi(x + 1/x))dx = \lim_{\beta \rightarrow \infty} \int_0^{\beta} sin^2(\pi(u))(du - r(x)) [/tex].

Since as beta tends to infinity r(x) must go to 0 and we will have du = dx, we have [tex] \lim_{\beta \rightarrow \infty} \int_0^{\beta} sin^2(\pi(u))(du - r(x)) = \int_0^{\infty} sin^2(\pi(u))du [/tex] which obviously diverges since the integrand is positive and periodic.
 
  • #9
Dick
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You have the right idea. But the expression is pretty awkward. Why not just say that for x>2 and u=x+1/x that the integral of sin^2(pi*(x+1/x))dx=sin^2(pi*u)(1-1/x^2)dx>the integral of sin^2(pi*u)*(4/3)*du. Since the latter diverges, the former must diverge.
 
  • #10
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There we go...thanks for all the help!

Besides practice, practice, and some more practice, do you have any tips on how to recognize how one should start the proof?
 
  • #11
Dick
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Feelings from the gut. I think you had that to begin with. You KNEW it diverged, right? See the big picture. Like, this is 'almost' this. Now learn to put 'almost' into words. That's how I do it.
 
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