Evaluating Integrals Divisions for Physics Homework

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emol1414
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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


([tex]\int{_0}{^\infty} \rho r^4 dr / \int{_0}{^\infty} \rho r^2 dr[/tex]) ^(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.
 
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emol1414 said:

Homework Equations


([tex]\int{_0}{^\infty} \rho r^4 dr / \int{_0}{^\infty} \rho r^2 dr[/tex]) ^(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.
 
Fixed your LaTeX.
emol1414 said:

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


[tex]\sqrt{\frac{\int_0^{\infty} \rho r^4 dr }{\int_0^{\infty} \rho r^2 dr}}[/tex]


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.
 
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 [tex]\infty[/tex] 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?
 
Hm... even using L'Hopital (e.g, for a constante [tex]\rho[/tex]) i would get a [tex]\infty[/tex] root-mean-square radius anyway. It doesn't make sense, i think
 
Of course, the constant [itex]\rho[/itex] cancels. Now, recall that
[tex]\int_a^\infty f(x)dx= \lim_{b\to\infty}\int_a^b f(x)dx[/tex]

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