Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mu$ be a finite, non-negative Borel measure on $[0,\infty)$ and let

$$f(r) = \int_0^{\infty} e^{-rt} d\mu(t).$$

A book I am reading states that "differentiation under the integral sign is justified because the measure is finite". That is,

$$-\frac{df}{dr} = -\frac{d}{dr} \int_0^{\infty} e^{-rt} d\mu(t) dt = \int_0^{\infty} t e^{-rt} d\mu(t)$$

I'm not exactly sure how this follows. I have the following theorem from my real analysis textbook by Folland: if $\frac{\partial }{\partial r}e^{-rt}$ exists and if there is a $g \in L^1([0,\infty)$ such that

$$\left|\frac{\partial }{\partial r}e^{-rt}\right| \leq g(t)$$

for all $r$ and $t$, then we may interchange derivatives. However, for fixed $r$, I can do some differentiation and find a relative extrema at $t = \frac{1}{r}$ where the derivative of the integrand is $\frac{1}{re}$ at this point. As $r$ becomes very small, this blows up to infinity, so finding a $g(t)$ that bounds $te^{-rt}$ (the derivative of the integrand) seems difficult to me, as $g$ cannot change with $r$. I don't seem to be using the finiteness of the measure, so I know I'm not on the right track. Any suggestions would be greatly appreciated.

share|cite|improve this question
up vote 1 down vote accepted

Have another look at your theorem. You only want to differentiate $f$ at an arbitrary, but fixed point $r_0 \geq 0$. So you can restrict the domain of $r$ to some neighborhood of $r_0$. For all $r$ in this neighbourhood you can choose an integrable function $g$ which dominates the partial derivative with respect to $r$.

share|cite|improve this answer
Thanks! That makes way more sense. – user35959 Dec 13 '12 at 17:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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