Takeover Season
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Suppose I want to test for the absolute convergence of \displaystyle \int _{x=0}^1\:\frac{sinx}{x\left(x+1  00\right)}dx.

How would I go about it, does anyone have any ideas?

I was thinking that:

\displaystyle \left|\frac{sin\:x}{x\left(x+100  \right)}\right| = \frac{|sinx|}{x\left(x+100\right  )} and that

 \displaystyle 0 \leq \frac{\left|sinx\right|}{x\left(  x+100\right)} \leq \frac{1}{x\left(x+100\right)} but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since sin\:x\:\approx \:x for x near 0, then can I say that sin\:x\ \leq \:x and so \left|sinx\right|\: \leq \:x for x near 0 and so \displaystyle \frac{\left|sinx\right|}{x\left(  x+100\right)} \leq \frac{x}{x\left(x+100\right)} = \frac{1}{\left(x+100\right)} and as that converges, then by the DCT, so does our integral absolutely?
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RDKGames
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(Original post by Takeover Season)
Suppose I want to test for the absolute convergence of \displaystyle \int _{x=0}^1\:\frac{sinx}{x\left(x+1  00\right)}dx.

How would I go about it, does anyone have any ideas?

I was thinking that:

\displaystyle \left|\frac{sin\:x}{x\left(x+100  \right)}\right| = \frac{|sinx|}{x\left(x+100\right  )} and that

 \displaystyle 0 \leq \frac{\left|sinx\right|}{x\left(  x+100\right)} \leq \frac{1}{x\left(x+100\right)} but this doesn't help because the integral for the LHS converges and the integral for the RHS diverges and so the direct comparison test definitely doesn't help here.

I thought since sin\:x\:\approx \:x for x near 0, then can I say that sin\:x\ \leq \:x and so \left|sinx\right|\: \leq \:x for x near 0 and so \displaystyle \frac{\left|sinx\right|}{x\left(  x+100\right)} \leq \frac{x}{x\left(x+100\right)} = \frac{1}{\left(x+100\right)} and as that converges, then by the DCT, so does our integral absolutely?
Over the interval (0,1) we have |sin(x)| < x so the second approach is correct.
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(Original post by RDKGames)
Over the interval (0,1) we have |sin(x)| < x so the second approach is correct.
Thank you, why did you talk about an open interval rather than (0,1], just wondering.
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RDKGames
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(Original post by Takeover Season)
Thank you, why did you talk about an open interval rather than (0,1], just wondering.
Does it matter?
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(Original post by RDKGames)
Does it matter?
I don't know, that's why i asked.

Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?
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RDKGames
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(Original post by Takeover Season)
I don't know, that's why i asked.

Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?
If I asked you to integrate something over the region (0,1) and then ask you to integrate over [0,1] instead, how is your working out going to be different?
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(Original post by RDKGames)
If I asked you to integrate something over the region (0,1) and then ask you to integrate over [0,1] instead, how is your working out going to be different?
I don't think it will, I'll do it the same way.
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RDKGames
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(Original post by Takeover Season)
I don't think it will, I'll do it the same way.
So it doesnt matter whether I talk about the open interval or the closed one, or the half-open one.
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(Original post by RDKGames)
So it doesnt matter whether I talk about the open interval or the closed one, or the half-open one.
Thank you, would my above approach still work if Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

Would I have to split the integral from x = 0 to x = 1 and use the above method and then say from x = 1 to x = 2pi, the integrand is continuous and well defined so the integral is 'proper' and converges.

So as both integrals converge, the integral from x = 0 to x = 2pi does too?
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(Original post by Takeover Season)
Thank you, would my above approach still work if Also, just wondering if my upper limit of the integral was x = 2 pi, then would this still hold?

Would I have to split the integral from x = 0 to x = 1 and use the above method and then say from x = 1 to x = 2pi, the integrand is continuous and well defined so the integral is 'proper' and converges.

So as both integrals converge, the integral from x = 0 to x = 2pi does too?
Yep
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(Original post by RDKGames)
Yep
thanks, so I'd definitely be wrong in saying that sin x <= x over x in (0,2pi) right?

Just wondering, when i take Taylor series of sin x about x = 0, how do I know that sin x <= x for x in (0,1).

Like how would I know for what values around x = 0 will the inequality hold?

It just seems to hold everytime I have an integral like here from 0 to 1, but from 0 to 5, it'd be different etc.
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RDKGames
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(Original post by Takeover Season)
thanks, so I'd definitely be wrong in saying that sin x <= x over x in (0,2pi) right?
Nope.

< is correct but so is the less strict <=

Just wondering, when i take Taylor series of sin x about x = 0, how do I know that sin x <= x for x in (0,1).
No need to consider Taylor; it just helps you to understand the behaviour of the integrand for small x.

Just sketch the graphs of sin x and x.

x > sin x for +ve x

x < sin x for -ve x
Last edited by RDKGames; 6 months ago
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(Original post by RDKGames)
No need to consider Taylor. Just sketch the graphs.

x > sin x for +ve x

x < sin x for -ve x
oh right, thank you.

i think what i was meant to say is, when i try to find the behaviour of sin x near x = 0, the problem point of the integral.
i use the taylor series to find this of sin x about x = 0 to find this.

then i see that sin x <= x for this.

i guess after knowing this, I'd just draw sin x and x because they're simple and see for what x does the inequality hold.
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(Original post by Takeover Season)
oh right, thank you.

i think what i was meant to say is, when i try to find the behaviour of sin x near x = 0, the problem point of the integral.
i use the taylor series to find this of sin x about x = 0 to find this.

then i see that sin x <= x for this.

i guess after knowing this, I'd just draw sin x and x because they're simple and see for what x does the inequality hold.
I suggest you recall this inequality regarding x and sin x since its can slap it on many integrals to show their convergence.

You may even search the web for its proof since its not too difficult.

As I said, Taylor expansion was just to get you thinking of the behaviour of the integrand near the bad point, and is often more useful in more complicated integrands.
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(Original post by RDKGames)
I suggest you recall this inequality regarding x and sin x since its can slap it on many integrals to show their convergence.

You may even search the web for its proof since its not too difficult.

As I said, Taylor expansion was just to get you thinking of the behaviour of the integrand near the bad point, and is often more useful in more complicated integrands.
Thank you, that is very helpful indeed.
Yes, that makes sense!
I was just wondering that does the Taylor Expansion usefulness diminish when testing for absolute convergence? If I want to find the Taylor Expansion of |f(x)|, do I just find the Taylor Expansion of f(x) and then take the absolute value of both sides?

I was just wondering because if I can't do that, then it would be difficult, especially if f(x) <= h(x) for some interval from the Taylor Expansion, but then the findings doesn't hold when we take |f(x)|.
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