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Can I apply L'Hopital's rule to this integral expression?

  1. Oct 9, 2013 #1
    Hi,

    Suppose I have

    [tex]\lim_{r\to 0} \left\{\int_0^{\pi} \frac{f(r,t)}{r^2}dt - \int_0^{\pi} \frac{g(r,t)}{r} dt\right\}[/tex]

    and both integrals tend to infinity. So I combine them:

    [tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt[/tex]

    now at this point, the numerator in the integrand does not go to zero as r goes to zero but rather [itex]\cos(t)[/itex] so it does go to zero at one point in the interval of integration. Can I apply L'Hospital's rule and conclude:

    [tex]\lim_{r\to 0} \int_0^{\pi} \frac{f(r,t)-r g(r,t)}{r^2} dt\overset{?}{=}\lim_{r\to 0} \int_0^{\pi} \frac{\frac{d}{dr}\left[f(r,t)-r g(r,t)\right]}{\frac{d}{dr}r^2}dt[/tex]

    Ok thanks,
    Jack
     
  2. jcsd
  3. Oct 9, 2013 #2

    SteamKing

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    I don't see your purpose here. You state that both integrals are non-convergent. That should be end of story.
    However, if you are going to apply l'Hopital, I think the correct approach is to find the antiderivative first, assuming this is possible, and then apply l'Hopital when evaluating the endpoints, also assuming that the antiderivative is well behaved in between. Taking derivatives of the integrand before integrating I don't think is appropriate, and I don't see how integrating such an expression could produce a true equality in your last equation.
     
  4. Oct 9, 2013 #3
    If in $$

    \lim_{r \to 0} \int\limits_a^b \frac {f(r, t) } {g(r, t)} dt

    $$ both ##f(0, t) = 0 ## and ## g(0, t) = 0 ##, then by repeating the proof of l'Hôpital's rule, one can show, perhaps with some additional conditions, that it is equal to $$

    \lim_{r \to 0} \int\limits_a^b \frac {f_r(r, t) } {g_r(r, t)} dt

    $$ where the r subscript denotes differentiation with respect to r.
     
  5. Oct 9, 2013 #4

    Borek

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    Staff: Mentor

    Sorry for OT, but I don't understand. My math is so rusty everything above multiplication table is a challenge, but I don't see why difference of two functions tending to infinity can't converge on some finite number. How is it different from

    [tex]\lim_{x \to \infty} x - x = 0[/tex]

    Or am I completely missing the point of your post?
     
  6. Oct 9, 2013 #5
    Hi guys. Sorry, I should have just posted the actual integrals:

    [tex]\lim_{r\to 0}\int_{\pi}^0 \frac{e^{-r \sin(t)} \cos[2 t-r \cos(t)]-r\sin(t)}{r^2} dt[/tex]

    Now, if I separate them and numerically integrate both with increasingly smaller r, the individual integrals are diverging but the difference appears to be approaching [itex]\pi/2[/itex]. And this is the correct answer through a proof by contradition: Everything else with the problem is correct and we know the answer to the problem is [itex]\pi/2[/itex] and this is the only part left so it must be [itex]\pi/2[/itex].

    However, surprisingly Mathematica can integrate it directly. I then take the limit as r goes to zero, and Mathematica reports the answer is [itex]-\pi/2[/itex]. This appears to be a problem.

    Finally, if I attempt to apply L'Hopital blindly without justification, two times, I end up with a final result of:

    [tex]-1/2 \int_{\pi}^0 dt=\pi/2[/tex]

    This is part of a homework thread:

    https://www.physicsforums.com/showthread.php?t=715171

    involving the limit:

    [tex]\lim_{\rho\to 0}\int_{\pi}^0 \frac{e^{i \rho e^{it}}-i\rho e^{it}}{\rho^2 e^{2 i t}}dt[/tex]

    and the answer to the question is [itex]\pi/2[/itex] and this integral is all that is left so must converge to that answer unless some logic error has been made but I checked it quite throughly and could find none.

    . . . ugh, you guys wouldn't happen to know how to evaluate the original limit more directly would you?
     
    Last edited: Oct 9, 2013
  7. Oct 9, 2013 #6
    Let ##f(r, t)## be the function in the numerator. Then ## f(0, t) = \cos 2t ##. It is not zero, yet it has this property: $$

    \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f(0, t)} {r^2} dt = \lim_{r \to 0} \int\limits_{\pi}^0 \frac {\cos 2t} {r^2} dt = 0

    $$ So $$

    \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f(r, t)} {r^2} dt = \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f(r, t)} {r^2} dt - \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f(0, t)} {r^2} dt = \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f(r, t) - f(0, t)} {r^2} dt

    $$ By Cauchy's mean value theorem, the latter is equal to $$

    \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f_r(\rho, t)} {2\rho} dt
    $$ where ## \rho \in [0, r] ##. But as ## r \to 0##, so does ##\rho \to 0 ##, so the latter limit is $$

    \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f_r(r, t)} {2r} dt
    $$

    So at least one invocation of l'Hôpital's rule is justified. For the second, we would need to evaluate ##f_r(r, t)##, which seems rather messy.
     
  8. Oct 9, 2013 #7
    So I cheated and used Wolfram Alpha. ## f_{r}(0, t) = 0 ##, which means l'Hôpital's rule can be applied again, yielding

    $$ \lim_{r \to 0} \int\limits_{\pi}^0 \frac {f_{rr}(r, t)} {2} dt $$

    Using WA again, I found that ## f_{rr}(0, t) = -1 ##, resulting in ##\pi/2## overall.
     
  9. Oct 9, 2013 #8
    Outstanding voko! I'll have to spend more time with it though to understand your analysis, and I'll cross-reference your work at the original HW question for completeness.
     
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