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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 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,450 Thanks: 2499 Math Focus: Mainly analysis and algebra 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.
May 19th, 2018, 06:01 AM   #3
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Quote:
 Originally Posted by v8archie 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.

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

since $\displaystyle \int_1^\infty \frac{1}{\sqrt{x+2}} \, dx$ diverges

 May 19th, 2018, 10:29 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1103 before you all rush off to try and show this converges to zero... Wolfram|Alpha: Computational Intelligence
 May 19th, 2018, 11:24 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,450 Thanks: 2499 Math Focus: Mainly analysis and algebra The integral clearly doesn't, the integrand does.
 May 19th, 2018, 01:19 PM #6 Senior Member   Joined: Apr 2017 From: New York Posts: 106 Thanks: 6 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.
May 19th, 2018, 04:14 PM   #7
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Quote:
 Originally Posted by Leonardox 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?!
$\displaystyle \int_1^\infty \frac{1}{\sqrt{x^5+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4}} \, dx = \int_1^\infty \frac{1}{x^2} \, dx = 1$

Last edited by skipjack; May 19th, 2018 at 10:13 PM.

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