College Math Teaching

October 31, 2011

Differentiation Under the Integral Sign

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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3 Comments »

  1. […] I’ve been doing: I posted this article; I’ve decided to write up “stuff I always wished I had learned the first time” as a student. This topic, called “differentiation under the integral sign”, is one of those that we […]

    Pingback by Zero energy « blueollie — November 1, 2011 @ 10:24 am

  2. […] does this work? This is where “differentiation under the integral sign” comes into play. So we write . Then […]

    Pingback by Finding a Particular solution: the Convolution Method « College Math Teaching — November 3, 2011 @ 4:09 pm

  3. […] Method 2: The method in the video This uses “differentiation under the integral sign”, which we talk about here. […]

    Pingback by Fourier Transform of the “almost Gaussian” function with a residue integral | College Math Teaching — August 25, 2014 @ 10:00 pm


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