1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiation under integral sign

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    I have to evaluate the numerical value of the derivative of the following integral for x=1

    [tex]\int_{0}^{\ ln\ x}\ e^{\ -\ x\ (\ t^2\ -\ 2)}\ dt[/tex]

    2. Relevant equations

    The formula for differentitation under integral sign.

    3. The attempt at a solution

    The upper limit term is straightforward:it is

    [tex]\frac{\ 1}{\ x}\ e^{\ -\ x[\ (\ ln\ x)^{\ 2}\ -\ 2]}[/tex]

    The other part is

    [tex]\int_0^{\ ln\ x}\frac{\partial}{\partial\ x}\ e^{\ -\ x(\ t^2\ -2)}\ dt\ =\ -\ e^{\ -\ 2\ x}\ [\int_0^{\ ln\ x}\ t^2\ e^{\ -\ x\ t^2}\ dt\ -\ 2\int_0^{\ ln\ x}\ e^{\ -\ x\ t^2}\ dt\ ][/tex]

    The later can be evaluated and I got the following:

    [tex]\ -\ e^{\ -\ 2\ x}\ [\frac{\ -(\ ln\ x)\ e^{\ -\ x(\ ln\ x)^2}}{\ 2\ x}\ +\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{4x\sqrt{ux}}\ du\ -\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{\sqrt{ux}}\ du}][/tex]

    I found the result as above.However,the two integrals neither cancel with each other nor can be evaluated.Can anyone please check and tell what should be done further.

    Last edited: Jan 17, 2010
  2. jcsd
  3. Jan 17, 2010 #2
    [tex]F(x) = \int_{0}^x e^{-e^x (t^2 - 2)} \text{ d}t[/tex]
    so you want to find the derivative of F(ln(x)) which you can do using the chain rule.
  4. Jan 18, 2010 #3
    Does not help;it ultimately reduces to what I have got...

    The thing lies in putting the limits without explicitly solving the final two inntegrals.They give zero.
  5. Jan 18, 2010 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    I helped neelakash with this problem on another forum (http://www.sciforums.com/showthread.php?t=99010). As he arrived at the correct answer there, I have no qualms posting the solution here for future reference by others.

    neelakash did use the appropriate technique for differentiating under the integral sign, the Leibniz Integral Rule:

    [tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt =
    \int_{a(x)}^{b(x)} \frac{\partial} {\partial x} f(t,x)\,dt +
    f(b(x),x)\frac{db(x)}{dx} - f(a(x),x)\frac{da(x)}{dx}[/tex]

    In this particular problem,

    [tex]f(t,x) = \exp\left(-x(t^2-2)\right),\quad a(x)=0, \quad b(x)=\ln x[/tex]

    The partial derivative of f(t,x) wrt x is

    [tex]\frac{\partial}{\partial x}f(t,x) = -(t^2-2) \exp\left(-x(t^2-2)\right)[/tex]

    Applying the Leibniz Integral Rule,

    [tex]\frac{d}{dx}\left(\int_0^{\ln x} \exp\left(-x(t^2-2)\right)\,dt\right) =
    -\left(\int_0^{\ln x} (t^2-2) \exp\left(-x\bigl(t^2-2)\right) \,dt\right) +

    There is a sign error in the original post (that exp(-2x) should be an exp(2x)). Additionally, neelakash carried the integration a step too far. That integral on the right-hand side is evaluable in terms of the error function erf(x).

    However, there is no reason to do this. neelakash finally saw the "Oh, SNAP!" light that makes this problem particularly easy. From that other forum,
  6. Jan 18, 2010 #5
    You're right, I'm sorry for the wrong suggestion. Anyway try substituting x=1 in the terms you found. The integrals should disappear due to ln(x)=0 and the upper limit term should become e^2.
  7. Jan 18, 2010 #6


    User Avatar
    Science Advisor

    When x= 1, ln(x)= ln(1)= 0 so x ln(x)= 1(0)= 0. Both integrals are from 0 to 0 and so are qual to 0.
  8. Jan 18, 2010 #7
    Yes,we need not carry out the integral explicitly as the answer comes from observation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook