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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.

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1 Answer 1

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 $r$.

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Thanks! That makes way more sense. –  user35959 Dec 13 '12 at 17:19

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