Evaluating Integrals Divisions for Physics Homework

[emol1414]

Homework Statement

It's a physics problem, where i have to evaluate the root-mean-square radius defined by the expression below. (First for a constant \rho, then for a "(r)dependent" \rho).

Homework Equations

(\int{_0}{^\infty} \rho r^4 dr / \int{_0}{^\infty} \rho r^2 dr) ^(1/2)

The Attempt at a Solution

So... i know that each of this two integrals goes to infinity. But I wonder about their division... I'm not sure what to do.

 

[emol1414]

Homework Equations

(\int{_0}{^\infty} \rho r^4 dr / \int{_0}{^\infty} \rho r^2 dr) ^(1/2)

I'm really trying to edit it and make the expression look nice, but i can't figure out how to do it.
Anywayz... it's the first integral (from 0 to infinity) divided by the second integral (also 0 to infinity). And then the square root of this.

 

[Mark44]

Fixed your LaTeX.

Homework Statement

It's a physics problem, where i have to evaluate the root-mean-square radius defined by the expression below. (First for a constant \rho, then for a "(r)dependent" \rho).

Homework Equations

\sqrt{\frac{\int_0^{\infty} \rho r^4 dr }{\int_0^{\infty} \rho r^2 dr}}

The Attempt at a Solution

So... i know that each of this two integrals goes to infinity. But I wonder about their division... I'm not sure what to do.

 

[emol1414]

Thank you, Mark! ^^

Now... with the correct latex code =P Could anyone give me a enlightenment here, how to work with this?

I thought... if i have \infty both "sides" up and down... i could use L'Hopital. But I keep thinking this would be just 'too easy' Oo Idk why, but doesn't sound right to me.
Any ideias?

 

[emol1414]

Hm... even using L'Hopital (e.g, for a constante \rho) i would get a \infty root-mean-square radius anyway. It doesn't make sense, i think

 

[HallsofIvy]

Of course, the constant \rho cancels. Now, recall that
\int_a^\infty f(x)dx= \lim_{b\to\infty}\int_a^b f(x)dx

\int_0^b r^4 dr= \frac{1}{5}b^5
and
\int_0^c r^2 dr= \frac{1}{5}c^3
and taking the limit as c and b go to infinity independently gives no integral.