Definite integral doesn't solve with L'Hopital

[Karol]

I want to solve:
$$\int_0^\infty \frac{dx}{\left( x^2+r^2 \right)^{3/2}}=\left[ \frac{x}{r^2\sqrt{x^2+r^2}}\right]_0^\infty$$
I apply L'Hopital's to the denominator:
$$\left(r^2\sqrt{x^2+r^2}\right)'=\frac{xr^2}{\sqrt{x^2+r^2}}$$
I apply again and agin L'Hopital to this but all the time almost the same result.

 

[pwsnafu]

Don't use L'Hopital. Divide the numerator and denominator by x.

 

[Karol]

Only the denominator:
$$\frac{r^2\sqrt{x^2+r^2}}{x}=\frac{r^4\left(x^2+r^2\right)}{x^2}=r^2\left(1+\frac{r^2}{x^2}\right)$$
And the whole integral is:
$$\left[ \frac{x}{r^2\sqrt{x^2+r^2}}\right]_0^\infty=\left[ \frac{1}{r^2\left(1+\frac{r^2}{x^2}\right)}\right]_0^\infty=\frac{1}{r^2}$$
Is that correct?

 

[Mark44]

Only the denominator:
$$\frac{r^2\sqrt{x^2+r^2}}{x}=\frac{r^4\left(x^2+r^2\right)}{x^2}=r^2\left(1+\frac{r^2}{x^2}\right)$$

This isn't valid. You can't just go in and square something to get rid of the radical.

And the whole integral is:
$$\left[ \frac{x}{r^2\sqrt{x^2+r^2}}\right]_0^\infty=\left[ \frac{1}{r^2\left(1+\frac{r^2}{x^2}\right)}\right]_0^\infty=\frac{1}{r^2}$$
Is that correct?

It's the right answer but the work you show is definitely flaky. To evaluate the expression you have on the left, you need to use limits.
$$\frac{x}{r^2\sqrt{x^2 + r^2}} = \frac{x}{r^2x\sqrt{1 + (r^2/x^2)}}$$
Can you find the limit as x → ∞ of the last expression?

 

[Karol]

As x→∞:
$\frac{x}{r^2x\sqrt{1 + (r^2/x^2)}}<\frac{x}{r^2x}=\frac{1}{r^2}$
And on the other hand:
$\frac{x}{r^2x\sqrt{1 + (r^2/x^2)}}>\frac{\frac{x}{\sqrt{1 + (r^2/x^2)}}}{r^2x\sqrt{1 + (r^2/x^2)}}=\frac{x}{\left(1 + r^2/x^2\right)xr^2}=\frac{1}{\left(1 + r^2/x^2\right)r^2}\rightarrow\frac{1}{r^2}$
It's the sandwich rule, am i correct?

 

[Mark44]

As x→∞:
$\frac{x}{r^2x\sqrt{1 + (r^2/x^2)}}<\frac{x}{r^2x}=\frac{1}{r^2}$

For any finite value of x, x/x = 1, so the factors of x in the numerator and denominator cancel. Inside the radical, r²/x² → 0 as x → ∞.

The work would look like this:
$$\lim_{x \to \infty}\frac{x}{x r^2 \sqrt{1 + \frac{r^2}{x^2}}} = \lim_{x \to \infty} \frac{1}{r^2\sqrt{1 + \frac{r^2}{x^2}} } = \frac{1}{r^2}$$
You shouldn't have '<'.

Apparently you're trying to get a lower bound in the work below. Using the sandwich theorem is a lot more work here than is necessary.

And on the other hand:
$\frac{x}{r^2x\sqrt{1 + (r^2/x^2)}}>\frac{\frac{x}{\sqrt{1 + (r^2/x^2)}}}{r^2x\sqrt{1 + (r^2/x^2)}}=\frac{x}{\left(1 + r^2/x^2\right)xr^2}=\frac{1}{\left(1 + r^2/x^2\right)r^2}\rightarrow\frac{1}{r^2}$
It's the sandwich rule, am i correct?

 

[Karol]

Thanks