Homework Help: Improper Integral with x>0 and a>2

In summary: No, tan^2 t = 1.In summary, the student is trying to find an equation to solve for the homework statement, but is not sure how to do it. They think they might have found a way to do it, but are not sure.
  • #1
springo
126
0

Homework Statement


Find:
[tex]\int_{0}^\infty \frac{\sqrt[3]{x}-\sqrt{x}}{x^b+a^b}dx[/tex]
With x>0 and a>2.

Homework Equations



The Attempt at a Solution


I think it's probably something with Beta, but I'm not sure how to change it into the proper form.
I thought changing it into:
[tex]\int_{0}^\infty \frac{\sqrt[3]{x}-\sqrt{x}}{x^b\left(1+\left(\frac{a}{x}\right)^b\right)}dx[/tex]
And then doing:
[tex]\tan^2 t = \left(\frac{a}{x}\right)^b[/tex]
But it makes things too complicated, so I guess it's not the way to go.

Thanks for your help.
 
Last edited:
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  • #2
You can compute integrals of these forms easily by using this result:

[tex]\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx =\frac{\pi}{\sin(\pi p)}[/tex]

for 0<p<1. This result is very easy to prove using contour integration.

If you start from this, you can do a rescaling to change the 1 in the denominator to some other value. You can change the x in the denominator to u^r.

You can also consider an integrand consisting of a linear combination of terms of the above form such that the individual terms would not converge when integrated separately. Nevertheless, you can still pretend that the divergent integrals are given by the (appropriately generalized version of) the above formula and then take the appropriate linear combination. The validity of such a calculation then follows from an analytical continuation argument.
 
  • #3
Everything looks constant to me with the "dt" in there, or maybe I'm insane.
 
  • #4
chriscolose is right (and not insane). The integral as stated does not exist. Its anti-derivative is just a constant (depending on parameters a and x) times t which goes to infinity as t goes to infinity.

Count Iblis assume the O.P. meant
[tex]\int_{0}^\infty \frac{\sqrt[3]{x}-\sqrt{x}}{x^b+a^b}dx[/tex]
 
  • #5
Sorry about that, I meant dx...

@Count Ibis: I haven't studied contour integration, I have studied line integral/path integrals, but nothing with complex variables... I never heard about that result, just the one with Γ(p)Γ(1-p) which yields the same result.
I think I need to do this with my standard Γ and B results.

I just thought of:
[tex]\tan^2 t = \left(\frac{x}{a}\right)^b[/tex]
Factorizing ab.
Could that work?
 
Last edited:

1. What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite, or the integrand has an infinite discontinuity within the limits of integration.

2. How do you solve an improper integral?

To solve an improper integral, we first need to rewrite it as a limit of a definite integral. Then, we evaluate the integral as usual and take the limit as the upper or lower limit of integration approaches infinity or the point of discontinuity.

3. Why is x>0 and a>2 specified in this homework help?

The condition x>0 and a>2 specifies the region of integration for the given improper integral. This condition ensures that the integrand is well-defined and the integral is convergent.

4. Can an improper integral be divergent?

Yes, an improper integral can be divergent. This means that the limit of the definite integral does not exist and the integral does not have a finite value.

5. Can improper integrals be solved using the fundamental theorem of calculus?

Yes, improper integrals can be solved using the fundamental theorem of calculus. However, we need to make sure that the integral is convergent before applying the theorem.

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