Why can't I apply Leibniz' rule in the following way?

$$\frac{d}{ds} g(s)\int_0^\infty f(s,x,u) \, du = \int_0^\infty \frac{d}{ds}g(s)f(s,x,u)\,du,$$

assuming $gf$ and $(gf)'$ are continuous on $[0,+\infty]\times [s_0,s_1]$ for some $s_0<s_1\in\mathbb{R}$.

  • 1
    $\begingroup$ It is unclear if there is any relationship between $s$, $x$ and $u$. Are $x$ and $u$ functions of $s$? $\endgroup$
    – Siminore
    Aug 4 '12 at 16:51
  • 1
    $\begingroup$ @Siminore, well, the $u$ is a dummy variable, so... $\endgroup$ Aug 4 '12 at 16:57
  • $\begingroup$ As written, the domain of $f$ seems to be a subset of $\mathbf R^2$, so the expression you've written doesn't make sense (saying that it is continuous in something two-dimensional). Unless you mean that for all $x$, $f(s,x,u)$ is continuous as a function of $s,u$ in the specified range. But in this case the usage of $x$ as a bound of the range is very confusing. $\endgroup$
    – tomasz
    Aug 4 '12 at 20:13

The problem I was originally having was in thinking that $g(s)$ was a constant coefficient of the definite integral. However, since we wish to differentiate w.r.t. $s$ it seems we can not think of $g$ as a constant, which seems obvious now. Instead, we must treat $g\int$ as a product and apply the product rule of differentiation, e.g.

$$\frac{\partial}{\partial s}g(s)\int_0^\infty f(s,x,u)du = g(s)\frac{\partial}{\partial s}\int_0^\infty f(s,x,u)du + \left(\frac{d}{d s}g(s)\right)\int_0^\infty f(s,x,u)du.$$

We can then take the $\partial/\partial s$ inside the integral as required.

  • 1
    $\begingroup$ Indeed, as long as you take a partial derivative, the other variables are as good as constants. I recommend using the standard notation \partial for partial derivatives. $\endgroup$
    – user31373
    Aug 8 '12 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.