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May 18th, 2018, 06:43 PM   #1
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Improper integrals

Hello everyone,
Can anyone help me with the antiderivative of this integtral?

I graphically proved it is convergent to zero but algebraicly I could't find the antiderivative of the integral.

I tried u-sub and then arctan. but couldn't get out of it.
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May 18th, 2018, 07:24 PM   #2
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Are you trying to evaluate the integral of just prove convergence? In the latter case you can use $$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x+2}}$$
on the interval.
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May 19th, 2018, 06:01 AM   #3
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Quote:
Originally Posted by v8archie View Post
Are you trying to evaluate the integral of just prove convergence? In the latter case you can use $$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x+2}}$$
on the interval.
maybe this comparison instead ...

$$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x^4+2}}$$

since $\displaystyle \int_1^\infty \frac{1}{\sqrt{x+2}} \, dx$ diverges
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May 19th, 2018, 10:29 AM   #4
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before you all rush off to try and show this converges to zero...

Wolfram|Alpha: Computational Intelligence
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May 19th, 2018, 11:24 AM   #5
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The integral clearly doesn't, the integrand does.
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May 19th, 2018, 01:19 PM   #6
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smaller function diverges mean larger function (above that line) also diverges
or
larger function converges mean smaller function (line below that) also converges.

How can I prove that original function is diverging by proving a larger function is diverging?

Shouldn't I pick smaller function then that and prove it diverges instead of picking larger one which diverges?!

Last edited by skipjack; May 19th, 2018 at 10:02 PM.
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May 19th, 2018, 04:14 PM   #7
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Originally Posted by Leonardox View Post
smaller function diverges mean larger function (above that line) also diverges
or
larger function converges mean smaller function (line below that) also converges.

How can I prove that original function is diverging by proving a larger function is diverging?

Shouldn't I pick smaller function then that and prove it diverges instead of picking larger one which diverges?!
$\displaystyle \int_1^\infty \frac{1}{\sqrt{x^5+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4}} \, dx = \int_1^\infty \frac{1}{x^2} \, dx = 1$
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Last edited by skipjack; May 19th, 2018 at 10:13 PM.
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