Differentiation Under the Integral Sign

October 31, 2011

Differentiation Under the Integral Sign

Filed under: advanced mathematics, applied mathematics, calculus, derivatives, differential equations, integrals, Laplace transform — collegemathteaching @ 10:05 pm

Suppose we have $F(s) = \int_a^b f(s,t)dt$ and we’d like to know what $\frac{d}{ds} F$ is.
The answer is $\frac{d}{ds}F(s) = \int_a^b \frac{\partial}{\partial s} f(s,t)dt$ .

This is an important result in applied mathematics; I’ll give some applications (there are many!) in our next post. Both examples are from a first course in differential equations.

First, I should give the conditions on $f(s,t)$ to make this result true: continuity of $f(s,t)$ and $\frac{\partial}{\partial s} f(s,t)$ on some rectangle in $(s,t)$ space which contains all of the points in question (including the interval of integration) is sufficient.

Why is the formula true? The proof isn’t hard at all and it makes use of the Mean Value Theorem and of some basic theorems concerning limits and integrals.

Some facts that we’ll use: if $M = max{|f|}$ on some interval $(a,b)$ , then $|\int_a^b f(t)dt| \leq M |b-a|$ and the Mean Value Theorem.

Now recall from calculus: $\frac{d}{ds} F =lim_{s_0 \rightarrow s} \frac{F(s_0)-F(s)}{s_0 - s} = lim_{s_0 \rightarrow s} \frac{1}{s_0 -s} \int_a^b f(s_0,t)-f(s,t) dt =lim_{s_0 \rightarrow s} \int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s_0} dt$

We now employ one of the most common ~~tricks~~ of mathematics; we guess at the “right answer” and then show that the right answer is what we guessed.

We will examine the integrand (the function being integrated). Does $\frac{f(s_0,t)-f(s,t)}{s_0 - s}$ remind you of anything? Right; this is the fraction from the Mean Value Theorem; that is, there is some $s*$ between $s$ and $s_0$ such that $\frac{f(s_0,t)-f(s,t)}{s_0 - s} = \frac{\partial}{\partial s} f(s*,t)$

Because we are assuming the continuity of the partial derivative, we can say that for $s$ sufficiently close to $s_0$ , $|\frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t)| < \epsilon$

This means that $| \int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t) dt | < \int_a^b |\frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t)| dt < \epsilon (b-a)$

Now realize that $\epsilon$ can be made as small as desired by letting $s_0$ get sufficiently close to $s$ so it follows by the $\epsilon-\delta$ definition of limit that:
$lim_{s_0 \rightarrow s}\int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s} - \frac{\partial}{\partial s} f(s,t) dt=0$ which implies that
$lim_{s_0 \rightarrow s}\int_a^b \frac{f(s_0,t)-f(s,t)}{s_0 - s}dt -\int_a^b \frac{\partial}{\partial s} f(s,t) dt=0$
Therefore $lim_{s_0 \rightarrow s} \frac{F(s_0)-F(s)}{s_0 - s} - \int_a^b \frac{\partial}{\partial s} f(s,t) dt=0$
So the result follows.

Next post: we’ll give a couple of applications of this

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October 31, 2011