Integral Convergence and Taylor Series

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Takeover Season
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For this integral: \displaystyle \int_{x=0}^{\frac{\pi }{2}} \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral absolutely converges, since the integrand is non-negative over the range of integration?

Thank you!

For Taylor series, is this correct in the attached image.
In the previous questions, we had n = 1/2 and hence the square root.
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RDKGames
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(Original post by Takeover Season)
For this integral: \displaystyle \int_{x=0}^{\frac{\pi }{2}} \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx

If I was asked to work out the values of n for which the integral converges, then would the answer be the same if I was asked to work out the values of n for which the integral absolutely converges, since the integrand is non-negative over the range of integration?
Yes. If f(x) \geq 0 over some interval I then |f(x)| \equiv f(x).



For that integral, the only bad point is just x=\pi/2.


As x \to \left( \dfrac{\pi}{2} \right)_- we get that \cos x behaves like \dfrac{\pi}{2} - x by simply expanding the cosine function about this point.

So |\cos x| \sim \dfrac{\pi}{2} - x.

Hence the integral behaves like

\displaystyle \int_0^{\pi/2} (\pi/2 - x)^{1-n} dx

For which n does it converge? The substitution u = \pi/2 - x would put it in a more familiar form that you may be used to.
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I am also unsure why you're integrating \dfrac{\cos x}{\sqrt{\pi / 2  - x} } over (0,\pi) since the function doesn't even exist for the upper half of the region.
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(Original post by RDKGames)
Yes. If f(x) \geq 0 over some interval I then |f(x)| \equiv f(x).



For that integral, the only bad point is just x=\pi/2.


As x \to \left( \dfrac{\pi}{2} \right)_- we get that \cos x behaves like \dfrac{\pi}{2} - x by simply expanding the cosine function about this point.

So |\cos x| \sim \dfrac{\pi}{2} - x.

Hence the integral behaves like

\displaystyle \int_0^{\pi/2} (\pi/2 - x)^{1-n} dx

For which n does it converge? The substitution u = \pi/2 - x would put it in a more familiar form that you may be used to.
Perfect! Thank you so much!
Yes, that is one of our standard test functions and so it converges for n - 1 < 1, so n < 2!
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(Original post by RDKGames)
I am also unsure why you're integrating \dfrac{\cos x}{\sqrt{\pi / 2  - x} } over (0,\pi) since the function doesn't even exist for the upper half of the region.
Sorry, yes I just realised, what a mistake! Let's say, I was integrating \dfrac{\cos x}{(\pi / 2  - x)^2} over (0,\pi)
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(Original post by RDKGames)
hi
Also, why can't we say that:

as  cos x \leq 1, then \displaystyle \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx \leq  \frac{1}{(\frac{\pi}{2} - \ x)^n} dx and so as \displaystyle \int_{x=0}^{\frac{\pi }{2}} \frac{1}{(\frac{\pi}{2} - \ x)^n} dx converges for all n < 1, so does our original integral?
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(Original post by Takeover Season)
Also, why can't we say that:

as  cos x \leq 1, then \displaystyle \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx \leq  \frac{1}{(\frac{\pi}{2} - \ x)^n} dx and so as \displaystyle \int_{x=0}^{\frac{\pi }{2}} \frac{1}{(\frac{\pi}{2} - \ x)^n} dx converges for all n < 1, so does our original integral?
You can but it's not as precise.

We have already shown that the integrand converges for n<2 so doing this method instead and showing that it diverges for n<1 misses out the values 1<n<2 for which the integral also converges for.

Approximating in the limit is more precise than just slapping on the global maximum.
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(Original post by RDKGames)
You can but it's not as precise.

We have already shown that the integrand converges for n<2 so doing this method instead and showing that it diverges for n<1 misses out the values 1<n<2 for which the integral also diverges for.

Approximating in the limit is more precise than just slapping on the global maximum.
Oh, so we can show it using the direct comparison test, but it may miss out some of the values for which the integral converges?
Also, do you mean "misses out the values 1<n<2 for which the integral also converges for".

Also, if we use the Taylor's theorem, we can always ensure we get a correct test whether using the limit comparison test or the direct comparison test right as it gives you the perfect bound? Like here we used it y = cos x is bounded above by y = 1.

But, Taylor's expansion would tell us that for x \in [0, \frac{\pi}{2}],  cos x \leq \frac{\pi}{2} - x and then we can use the direct comparison test to show that \displaystyle \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx \leq  \frac{\frac{\pi}{2} - x}{(\frac{\pi}{2} - \ x)^n} = \frac{1}{(\frac{\pi}{2} - \ x)^{n-1}} and then use the direct comparison test for \displaystyle \int _{x\:=0}^{\frac{\pi }{2}}\frac{1}{\left(\frac{\pi }{2}-x\right)^{n-1}}dx\: and show this converges for n < 2.
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(Original post by Takeover Season)
Oh, so we can show it using the direct comparison test, but it may miss out some of the values for which the integral converges?
Also, do you mean "misses out the values 1<n<2 for which the integral also converges for".

Also, if we use the Taylor's theorem, we can always ensure we get a correct test whether using the limit comparison test or the direct comparison test right as it gives you the perfect bound? Like here we used it y = cos x is bounded above by y = 1.

But, Taylor's expansion would tell us that for x \in [0, \frac{\pi}{2}],  cos x \leq \frac{\pi}{2} - x and then we can use the direct comparison test to show that \displaystyle \frac{cos \ x}{(\frac{\pi}{2} - \ x)^n} dx \leq  \frac{\frac{\pi}{2} - x}{(\frac{\pi}{2} - \ x)^n} = \frac{1}{(\frac{\pi}{2} - \ x)^{n-1}} and then use the direct comparison test for \displaystyle \int _{x\:=0}^{\frac{\pi }{2}}\frac{1}{\left(\frac{\pi }{2}-x\right)^{n-1}}dx\: and show this converges for n < 2.
Yes.
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(Original post by RDKGames)
Yes.
Ah thanks, one last query. This just sum up the whole thread! It is actually about the taylor series, forgetting that dodgy integration I had put up.

If I wanted to test for the absolute convergence of an integral because it oscillates and it had an improper integral of the second kind, or it had a problem point either within the interval or at its endpoints. Then, I need to see the Taylor series of the absolute value of the function near the problem point. If I had f(x) was negative over [a,b], how do I find the taylor series of |f(x)| at x = a? Please look at my original post if you wish, I said my explanation there and I was wondering if the approach was correct.

I'd just find the taylor series of -f(x) at x = a and that'd give me the taylor series of |f(x)| at x = a?
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