- #1

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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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- Thread starter Song
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- #1

- 47

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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.

- #2

- 13,109

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

Daniel.

- #3

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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.

[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

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

Daniel.

- #6

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Please see the comparison test..

But I don't know how to solve the problem by using that.

- #7

shmoe

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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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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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