# Improper integral

## Homework Statement

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

## 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:
$$\int_{0}^\infty \frac{\sqrt[3]{x}-\sqrt{x}}{x^b\left(1+\left(\frac{a}{x}\right)^b\right)}dx$$
And then doing:
$$\tan^2 t = \left(\frac{a}{x}\right)^b$$
But it makes things too complicated, so I guess it's not the way to go.

Last edited:

You can compute integrals of these forms easily by using this result:

$$\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx =\frac{\pi}{\sin(\pi p)}$$

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.

Everything looks constant to me with the "dt" in there, or maybe I'm insane.

HallsofIvy
Homework Helper
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
$$\int_{0}^\infty \frac{\sqrt[3]{x}-\sqrt{x}}{x^b+a^b}dx$$

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:
$$\tan^2 t = \left(\frac{x}{a}\right)^b$$
Factorizing ab.
Could that work?

Last edited: