Need Help with Improper Integral: Convergent or Divergent?

In summary: The fact that there is a 2x in the integrand suggests you should make the substitution u= x^2+ 1. Then your integral becomesI = \int^{\infty}_{1}\frac{du}{u^{1.2}} = \int^{\infty}_{1}\frac{du}{u^{6/5}}which is convergent.
  • #1
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This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

[tex]I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}[/tex]

which, after carrying out the 1.2, becomes

[tex]\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}[/tex]

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

[tex]\frac{1}{x^{p}}[/tex]

I started off by trying a direct comparison test and compared it to [tex]\frac{2x}{x^{2.4}}[/tex] which can simplify to

[tex]\frac{2}{x^{.4}}[/tex]

So, since the above function has [tex]p=.4<1[/tex], it diverges, by the Direct Comparison Test, [tex]I[/tex] diverges.
 
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  • #2
Be careful! [itex](a + b)^p \neq (a^{p} + b^p)[/itex].

Furthermore, the function [itex]2x/x^{2.4}[/itex] is actually slightly larger than your original integrand, so the fact that the integral over this diverges doesn't actually tell you anything about the convergence/divergence of your original integrand.

The fact that there is a [itex]2x[/itex] in the integrand suggests you should make the substitution [itex]u = x^2 + 1[/itex]. Then your integral becomes

[tex]I = \int^{\infty}_{1}\frac{du}{u^{1.2}} = \int^{\infty}_{1}\frac{du}{u^{6/5}}[/tex]

This is an integral you can do, and you should be able to tell if it diverges or converges.
 
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  • #3
ttiger2k7 said:
This was a problem on one of my previous tests that I got wrong entirely. In preparing for my final, I'm attempting to redo it. I was wondering if someone could check my work.

Determine whether the following improper integral is convergent or divergent. If the integeral is convergent, find its value.

[tex]I = \int^{\infty}_{0}\frac{2x}{(x^{2}+1)^{1.2}}[/tex]

which, after carrying out the 1.2, becomes

[tex]\int^{\infty}_{0}\frac{2x}{(x^{2.4}+1)}[/tex]
No, it doesn't! In general, (x+ y)n is NOT xn+ yn!

That's where I get stuck. I know that the function diverges, but I don't know how to prove it.

I thought about trying to manipulate it somehow so that the function I'm comparing it to is in the form

[tex]\frac{1}{x^{p}}[/tex]

I started off by trying a direct comparison test and compared it to [tex]\frac{2x}{x^{2.4}}[/tex] which can simplify to

[tex]\frac{2}{x^{.4}}[/tex]

So, since the above function has [tex]p=.4<1[/tex], it diverges, by the Direct Comparison Test, [tex]I[/tex] diverges.
Let u= x2+ 1, so that du= 2x dx, in your original integral.
 

1. What is an improper integral?

An improper integral is an integral that does not have both limits of integration or has an integrand with a singularity in the interval of integration. This means that the integral does not meet the standard criteria for convergence and may require special techniques to evaluate.

2. How do you determine if an improper integral is convergent or divergent?

The most common method for determining convergence or divergence of an improper integral is to evaluate the limit as the upper bound approaches infinity. If the limit exists and is finite, then the integral is convergent. If the limit does not exist or is infinite, then the integral is divergent.

3. Can an improper integral have both a finite and an infinite limit?

Yes, an improper integral can have both a finite and an infinite limit. This is known as a semi-convergent integral and requires additional analysis to determine the convergence or divergence of the integral.

4. Are there any other methods for determining convergence or divergence of an improper integral?

Yes, there are other methods for determining convergence or divergence of an improper integral, such as the comparison test, limit comparison test, and the ratio test. These methods involve comparing the given integral to a simpler integral with known convergence properties.

5. Can improper integrals be evaluated using numerical methods?

Yes, improper integrals can be evaluated using numerical methods, such as the trapezoidal rule or Simpson's rule. These methods involve approximating the integral using a series of smaller intervals and can be useful when the integral cannot be evaluated analytically.

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