Improper integral

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  • #1
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basically I'm stuck with this problem........

Integral the upper limit is positive infinite and the lower limit is 1
X/(sqrt(1+x^6) dx......

can someone give me an idea on how to start this?...I really don't know. Thanks.
 

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  • #2
dextercioby
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Are you sure it's this...?

[tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx \ = ...? [/tex]

Daniel.
 
  • #3
dextercioby
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If so, then

[tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx = {}_{2}F_{1}\left(\frac{1}{6},\frac{1}{2}; \frac{7}{6}; -1\right) [/tex]

, with F Gauss' hypergeometric function.

Daniel.
 
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  • #4
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I don't know how to do that. How do you use the comparison theorem to do this? Thanks.
 
  • #5
dextercioby
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What comparison theorem are you referring to ...?

A simple substitution [itex] x^{2}=t [/itex] will bring your integral to a simpler form, however that wouldn't help too much.

Daniel.
 
  • #7
shmoe
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It's always a good idea to state your question in full. I suspect yours is

Prove the integral [tex] \int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx[/tex] is convergent.

Specifically you don't care about finding a value for this integral. When x is large, what simple polynomial is the denominator kinda like? This should suggest a function for your comparison test.
 
  • #8
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When X is large, x/(sqrt(1+x^6)) is getting close to the x-axis?
x/(sqrt(1+x^2)) is greater than x/(sqrt(1+x^6))..is my logic right?
 
  • #9
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I'm sorry that I didn't state my question clearly at the beginning.
 
  • #10
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..........................
 
  • #11
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[tex] \frac{x}{\sqrt{1+x^{6}}} [/tex] is about [tex] \frac{x}{x^{3}} = \frac{1}{x^{2}} [/tex]. Since [tex] \frac{1}{x^{2}} [/tex] is convergent ([tex] p > 1 [/tex]), then so is the original integral by the comparison test (the original function is less than the new function)
 
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  • #12
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but why did you choose x/(x^3)?
 
  • #13
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[tex]\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}} = \frac{1}{x^{3}} [/tex]. The 1 doesnt matter for large enough x.
 
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  • #14
shmoe
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courtrigrad said:
[tex]\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}} = \frac{1}{x^{3}} [/tex]. The 1 doesnt matter for large enough x.

I don't consider pointing this out to be picky- you never have [tex]\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}}[/tex].

Much better to write [tex]\frac{1}{\sqrt{1+ x^{6}}}\sim \frac{1}{ \sqrt{x^{6}}}[/tex] with a precise asymptotic meaning behind [tex]\sim[/tex].
 

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