Takeover Season
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So, it seems like whenever we have an improper integral of the first kind with an integrand with sine/cosine in it without a power of x in the denominator, we are sort of in trouble, because we can't use Taylor's series because we don't know the behaviour of sin x or cos x at  x = \infty. So, either we have absolute convergence and we can show the function converges, or we don't have absolute convergence and so we may have conditional convergence, provided the original integral does converge.

So, can we say that if we have any function of the form \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{sin\:x}{x^n}\:dx or \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{cos\:x}{x^n}\:dx for  A &gt; 0, then we can sort of show that the integral converges by using absolute convergence trick provided n > 1. If n < 1, then we have that \displaystyle \frac{|sin\:x|}{x^n}\: , \frac{|cos\:x|}{x^n}\: \leq \frac{1}{x^n}\: and then by the Direct Comparison Test, as \displaystyle \int _{x=A}^{\infty }\:\frac{1}{x^n}\:dx diverges and so the test fails. If n > 1, we have absolute convergence and we can use the Direct Comparison Test because our original integrand is now non-negative because of the absolute value.

I said A > 0 just to avoid having two problem points. If A = 0, I know that we can split it into 0 to B and B to inf, where for the first integral, we can always say by Taylor's series, sin x behaves like x when x tends to 0 and ... finish it off.

So, is my intuition correct in thinking that if:
n > 1, we can show this integral converges by absolute convergence.
If n <= 1, then for whatever values of n the integral converges (if it does), then because we can't use Taylor's theorem to inspect the behaviour of sin x / cos x at tex] x = \infty[/tex], we have to use the trick of integration by parts shown in the attachment, which of course eventually does use absolute convergence, so as said earlier to use it, we need at least a power of x greater than 1 in the denominator and so our original power of x i.e. n must be greater than 0, so when \displaystyle \frac{1}{x^n} is differentiated, it gives a power of at least 1 to use absolute convergence trick. So, I can guess it converges for  n &gt; 0.

Again, I am more into whether my intuition is correct over how to approach these problems using this trick taught in my lectures, rather than if I guessed the values of n for which it converges correctly.
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RDKGames
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(Original post by Takeover Season)
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So, it seems like whenever we have an improper integral of the first kind with an integrand with sine/cosine in it without a power of x in the denominator, we are sort of in trouble, because we can't use Taylor's series because we don't know the behaviour of sin x or cos x at  x = \infty. So, either we have absolute convergence and we can show the function converges, or we don't have absolute convergence and so we may have conditional convergence, provided the original integral does converge.

So, can we say that if we have any function of the form \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{sin\:x}{x^n}\:dx or \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{cos\:x}{x^n}\:dx for  A &gt; 0, then we can sort of show that the integral converges by using absolute convergence trick provided n > 1. If n < 1, then we have that \displaystyle \frac{|sin\:x|}{x^n}\: , \frac{|cos\:x|}{x^n}\: \leq \frac{1}{x^n}\: and then by the Direct Comparison Test, as \displaystyle \int _{x=A}^{\infty }\:\frac{1}{x^n}\:dx diverges and so the test fails. If n > 1, we have absolute convergence and we can use the Direct Comparison Test because our original integrand is now non-negative because of the absolute value.

I said A > 0 just to avoid having two problem points. If A = 0, I know that we can split it into 0 to B and B to inf, where for the first integral, we can always say by Taylor's series, sin x behaves like x when x tends to 0 and ... finish it off.

So, is my intuition correct in thinking that if:
n > 1, we can show this integral converges by absolute convergence.
If n <= 1, then for whatever values of n the integral converges (if it does), then because we can't use Taylor's theorem to inspect the behaviour of sin x / cos x at tex] x = \infty[/tex], we have to use the trick of integration by parts shown in the attachment, which of course eventually does use absolute convergence, so as said earlier to use it, we need at least a power of x greater than 1 in the denominator and so our original power of x i.e. n must be greater than 0, so when \displaystyle \frac{1}{x^n} is differentiated, it gives a power of at least 1 to use absolute convergence trick. So, I can guess it converges for  n &gt; 0.

Again, I am more into whether my intuition is correct over how to approach these problems using this trick taught in my lectures, rather than if I guessed the values of n for which it converges correctly.
Sounds good.

sin(x)/x over an infinite domain is a good example of a convergent, but not absolutely convergent integral.

It is a good exercise for you to show that it is not absolutely convergent.

Although indeed, in determining that it is convergent, you use absolute convergence after IBP. But this only shows that the leftover integral converges absolutely, and not the original integral.
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Takeover Season
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(Original post by RDKGames)
Sounds good.

sin(x)/x over an infinite domain is a good example of a convergent, but not absolutely convergent integral.

It is a good exercise for you to show that it is not absolutely convergent.

Although indeed, in determining that it is convergent, you use absolute convergence after IBP. But this only shows that the leftover integral converges absolutely, and not the original integral.
Yes, that is great. This was the example in the lectures too, he showed that sin x / x is not absolutely convergent and but it is convergent and so it is conditionally convergent.

Yes that is true, we still have to evaluate the other [box] from integration by parts i.e. the other term with the leftover integral and for this we always use Sandwich Theorem, so that will really only converge if n > 0. So, the [box] converges always for n > 0 and the leftover integral converges absolutely for 0 < n <= 1, depending on what power of n we have to start with.
Hence in all cases, in the limiting case, we have n > 0 which is what this integral converges for.

How does that sound?
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RDKGames
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(Original post by Takeover Season)
Yes, that is great. This was the example in the lectures too, he showed that sin x / x is not absolutely convergent and but it is convergent and so it is conditionally convergent.

Yes that is true, we still have to evaluate the other [box] from integration by parts i.e. the other term with the leftover integral and for this we always use Sandwich Theorem, so that will really only converge if n > 0. So, the [box] converges always for n > 0 and the leftover integral converges absolutely for 0 < n <= 1, depending on what power of n we have to start with.
Hence in all cases, in the limiting case, we have n > 0 which is what this integral converges for.

How does that sound?
You don't need the Sandwich theorem to see that \displaystyle \lim_{t \to \infty} \dfrac{\cos t}{t^n} = 0 for all n &gt; 0. Think about it; the numerator just oscillates between -1 and 1 but the denominator gets large so it dominates and brings the entire quotient to zero.

And I am a bit lost, where your n's coming from and what do they represent? What integrals precisely are you talking about?
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Takeover Season
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(Original post by RDKGames)
You don't need the Sandwich theorem to see that \displaystyle \lim_{t \to \infty} \dfrac{\cos t}{t^n} = 0 for all n &gt; 0. Think about it; the numerator just oscillates between -1 and 1 but the denominator gets large so it dominates and brings the entire quotient to zero.

And I am a bit lost, where your n's coming from and what do they represent? What integrals precisely are you talking about?
Hi, I understand your intuition above, thank you. I realise that too, but I just said it because I would state the Sandwich theorem in the exam.

With regards to the n's and all, I have attached an image explaining.


Name:  WhatsApp Image 2020-04-26 at 07.12.33.jpeg
Views: 12
Size:  157.5 KB

Basically, to show this integral \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{sin\:x}{x^n}\:dx converges, if n > 1, I'd show it directly through absolute convergence. If 0 < n <= 1, I'd use IBP trick and then show the leftover integral converges absolutely and show the (box) also converges, so my whole integral converges. The (box) converges for n > 0 and the integral converges absolutely for all n > 0. The limiting case is when n tends to zero and for n <= 0, neither the box nor the integral converges i.e. the limit of the box doesn't exist and the integral does not converge absolutely.
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RDKGames
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(Original post by Takeover Season)
Hi, I understand your intuition above, thank you. I realise that too, but I just said it because I would state the Sandwich theorem in the exam.

With regards to the n's and all, I have attached an image explaining.


Name:  WhatsApp Image 2020-04-26 at 07.12.33.jpeg
Views: 12
Size:  157.5 KB

Basically, to show this integral \displaystyle \int _{x\:=\:A}^{\infty }\:\frac{sin\:x}{x^n}\:dx converges, if n > 1, I'd show it directly through absolute convergence. If 0 < n <= 1, I'd use IBP trick and then show the leftover integral converges absolutely and show the (box) also converges, so my whole integral converges. The (box) converges for n > 0 and the integral converges absolutely for all n > 0. The limiting case is when n tends to zero and for n <= 0, neither the box nor the integral converges i.e. the limit of the box doesn't exist and the integral does not converge absolutely.
Sounds good.
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Takeover Season
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#7
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(Original post by RDKGames)
Sounds good.
Thanks, so is the 'limit' part in the picture correct i.e. I cannot say \displaystyle \frac{\sqrt{x}\left|sin\:x\right  |}{x+100} behaves like \displaystyle \frac{\sqrt{x}}{x} as x \rightarrow \infty because \lim _{x\to \infty }\displaystyle \left(\frac{\left|sin\:x\right|}  {1+\:\frac{100}{x}}\right) \neq 1? Just confirming it is 'all good'.
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