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Homework Help: Definite integral doesn't solve with L'Hopital

  1. Nov 15, 2014 #1
    • Originally posted in a technical math forum, so is missing the homework template
    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:
    I apply again and agin L'Hopital to this but all the time almost the same result.
  2. jcsd
  3. Nov 15, 2014 #2


    User Avatar
    Science Advisor

    Don't use L'Hopital. Divide the numerator and denominator by x.
  4. Nov 15, 2014 #3
    Only the denominator:
    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?
  5. Nov 15, 2014 #4


    Staff: Mentor

    This isn't valid. You can't just go in and square something to get rid of the radical.
    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?
  6. Nov 16, 2014 #5
    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?
  7. Nov 17, 2014 #6


    Staff: Mentor

    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.
  8. Nov 18, 2014 #7
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