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April 28th, 2013, 06:59 AM   #1
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Integrals

Judge if those integrals converge:









Thank you.
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April 28th, 2013, 11:48 AM   #2
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Re: Integrals

[color=#000000]I think it is easy to prove that the last two converge,

c. It is obvious that we have a problem at point 0, so knowing that close to zero for a small ?, , so the integral converges.

d. , so again this integral converges.[/color]
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April 28th, 2013, 12:44 PM   #3
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Re: Integrals

1) converges. Transform by y^2 = x and get an integrand 2(y^35)exp(-y)
2) diverges. Concatonate pairs of intervals of length ? (starting at ?/2)
and you have an integrand ~ 1/x^(2/3) as x -> ?.
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April 28th, 2013, 01:03 PM   #4
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Re: Integrals

Quote:
2) diverges. Concatenate pairs of intervals of length ? (starting at ?/2)
[color=#000000]Since I don't see why the same one starting from 1, should diverge. [/color]
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April 29th, 2013, 05:29 AM   #5
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Re: Integrals

[color=#000000]It was easier than I expected.

b.



look at the boxed number, this is the case which converges for p>1.

For the first integral you could say that it converges since the expotential in the denominator is always greater that the every power of x and (note that here we have a problem at infinity since that limit at 0 is 0).

So you could say that for a very big number M, for all

so all the given integrals a,b,c,d converge.
[/color]
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April 29th, 2013, 01:11 PM   #6
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Re: Integrals

Quote:
Originally Posted by ZardoZ
Quote:
2) diverges. Concatenate pairs of intervals of length ? (starting at ?/2)
[color=#000000]Since I don't see why the same one starting from 1, should diverge. [/color]
The formula for ?(2/3) puzzles me.
The usual definition I am familiar with has an integrand of exp(-x)/?x.
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April 29th, 2013, 01:29 PM   #7
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Re: Integrals

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Originally Posted by mathman
The formula for ?(2/3) puzzles me. The usual definition I am familiar with has an integrand of exp(-x)/?x.
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April 30th, 2013, 08:37 AM   #8
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Re: Integrals

Quote:
Originally Posted by Z

Hi Z , could you illustrate what you have done here?
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May 4th, 2013, 12:05 PM   #9
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Re: Integrals

Quote:
Originally Posted by zaidalyafey
Quote:
Originally Posted by Z

Hi Z , could you illustrate what you have done here?
What about a hint ?
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May 4th, 2013, 01:14 PM   #10
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Re: Integrals

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Originally Posted by zaidalyafey
Quote:
Originally Posted by zaidalyafey
What about a hint ?
[color=#000000]
See here. I didn't respond earlier because I needed time to do the computations. I used mathematica to get the result I posted earlier, so I didn't have an answer by that time.[/color]
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