Passing Derivatives Through Integrals

1. The Setup

The basic question here is the following: suppose that $f(x,t)$ is a function of two variables which is integrable on $[a,b]$ as a function of $x$ for each $t$. When is the function

\[ F(t) := \int_a^b f(x,t) dx \]

differentiable as a function of $t$? We already know what the answer should be:

it should be possible to differentiate $F$ by passing a $t$-derivative “through”

the integral, i.e.,

\[ F'(t) = \int_a^b \frac{\partial f}{\partial t}(x,t) dx. \]

(Here the partial derivative is simply the derivative in the $t$ variable only.) When is this justified?

2. A Nonexample

To see that the question is nontrivial, consider the following nonexample (assume througout that $t \in (-\frac{1}{2},\frac{1}{2})$ and $x in [0,1]):

\[ f(x,t) := \begin{cases} t^{-1} x & 0 \leq x \leq t \\ 2 - t^{-1} x & 0 < t < x \leq 2t \\ 0 & 0 < 2t < x \leq 1 \\ 0 & t = 0 \\ - f(x,-t) & t < 0 \end{cases}\]

Below is a plot of this function $f(x,t)$ for $x \in [0,1]$ and values of $t$ given by $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}$, and $-\frac{1}{6}$. Observe that the graph is triangular on $[0,2|t|]$ when $t \neq 0$ with height $1$ for positive $t$ and height $-1$ for $t < 0$. Everywhere else, the function vanishes.

Figure. Plot of the function $f(x,t)$ for $t = \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, -\frac{1}{6}$

Clearly $f(x,t)$ is continuous as a function of $x$ for each $t \in (-\frac{1}{2},\frac{1}{2})$. Using what we know about areas of triangles,

\[ F(t) := \int_0^1 f(x,t) dx = t \]

for all $t \in (-\frac{1}{2},\frac{1}{2})$, so $F$ is differentiable at $t = 0$ and has $F'(0) = 1$. Notice also that for each $x \in [0,1]$, the map $t \mapsto f(x,t)$ is differentiable at $t=0$. This is because for each $x$, $f(x,t) = 0$ for all sufficiently small $t$. (If $x = 0$, $f(x,t)$ is identically zero, and otherwise $f(x,t) = 0$ when $|t| \leq x/2$.) Thus

\[ \frac{\partial f}{\partial t}f(x,0) := \lim_{h \rightarrow 0} \frac{f(x,h)-f(x,0)}{h}\]

exists and equals zero for each $x \in [0,1]$. But clearly our computations also show that

\[ 1 = F'(0) \neq \int_0^1 \frac{\partial f}{\partial t}(x,0) dx = 0. \]

3. A Positive Result

One way to achieve a positive result is to require a stronger form of convergence of difference quotients (think about the example of passing derivatives through infinite series). A sufficient condition is to require that the difference quotients converge uniformly:

Theorem (Passing Derivatives Through Integrals)

Suppose that for each $t \in (t_0-\delta,t_0+\delta)$, $f(x,t)$ is

a continuous function of $x \in [a,b]$ and that

\[ \frac{\partial f}{\partial t}(x,t_0) := \lim_{h \rightarrow 0} \frac{f(x,t_0+h) - f(x,t_0)}{h}\]

exists in the uniform sense for all $x \in [a,b]$, i.e., that

\[ \sup_{x \in [a,b]} \left| \frac{f(x,t_0+h) - f(x,t_0)}{h} - \frac{\partial f}{\partial t}(x,t_0)\right| \rightarrow 0 \]

as $h \rightarrow 0$. Then $\frac{\partial f}{\partial t}(x,t_0)$ is continuous (and thus Riemann integrable) on $[a,b]$ and the function

\[ F(t) := \int_a^b f(x,t) dx \]

is differentiable at $t = t_0$ with

\[ F'(t_0) = \int_a^b \frac{\partial f}{\partial t}(x,t_0) dx. \]

Proof

The first thing to observe is that uniform convergence of the difference quotients

\[ \frac{f(x,t_0+h) - f(x,t_0)}{h}\]

implies that $\frac{\partial f}{\partial t}(x,t_0)$ is continuous as a function of $x$ since the difference quotients are themselves continuous as functions of $x$ and since uniform convergence preserves continuity.

Next, write the difference quotients for $F'(t_0)$ in terms of the definition of $F$:

\[{}\frac{F(t_0+h) - F(t_0)}{h}{}\]

\[{}= \int_a^b \frac{f(x,t_0+h)-f(x,t_0)}{h} dx{}\]

Now, because the integrand converges uniformly to $\frac{\partial f}{\partial t}(x,t_0)$, we know that the integral must converge to $\int_a^b \frac{\partial f}{\partial t}(x,t_0) dx$, completing the proof.

It can be a bit tedious to verify uniform convergence of the difference quotient, so it is useful to have a less general but simpler hypothesis. In this case, we will see later that it suffices to assume that all of the following conditions hold:

The functions $x \mapsto f(x,t)$ are all continuous in $x$ on $[a,b]$ for each $t \in (t_0-\delta,t_0+\delta)$,
The functions $t \mapsto f(x,t)$ are differentiable in $t$ for each $t \in (t_0-\delta,t_0+\delta)$ and each $x \in [a,b]$, and
The function $\frac{\partial f}{\partial t}(x,t)$ is continuous as a function of two variables.

This is possible to prove using the Mean Value Theorem in the $t$ variable and the fact that continuity of functions of two variables on a compact set implies uniform continuity. It will be an exercise to fill in these details when the appropriate time comes.