# Differentiating 1D Integrals #

*by Shuang Zhao*

In what follows, we discuss the differentiation of a simple Riemann integral `$I(\theta)$`

over some 1D interval `$(a, b) \subseteq \real$`

:

## The Incomplete Solution #

The derivative of the integral in Eq. \eqref{eqn:I} with respect to `$\theta$`

can *sometimes* be obtained by exchanging the ordering of differentiation and integration:

Precisely, the second equality in Eq. \eqref{eqn:dI_0} requires the integrand $f$ to be **continuous**^{1} throughout the interval `$(a, b)$`

.

### Success Example #

We now provide a toy example where Eq. \eqref{eqn:dI_0} holds.
Let `$f(x, \theta) := x^2 \,\theta$`

. Consider the following integral:

Since `$I = \left[ (x^3 \,\theta)/3 \right]_0^1 = \theta/3$`

, we know that

We now try calculating the same derivative `$\D I/\D\theta$`

using Eq. \eqref{eqn:dI_0}:

which matches the manually calculated result above.

### Failure Example #

We now show another toy example for which simply exchanging differentiation and integration outlined in Eq. \eqref{eqn:dI_0} fails. Let

Then, for any `$0 < \theta < 2$`

, it holds that

and

However, since the integrand $f$ is piecewise-constant in this example, we have `$\D f/\D\theta \equiv 0$`

.
Thus, Eq. \eqref{eqn:dI_0} in this example gives

which does **not** match the manually calculated result in Eq. \eqref{eqn:f_step_dI_manual}.

## The General Solution #

### Examining The Previous Examples #

Before presenting the general expression of the derivative `$\D I/\D\theta$`

, we first examine the examples shown above.

#### The Success Example #

We first examine the success example with the integrand `$f(x, \theta) = x^2 \,\theta$`

.
In the following, we show the graph of `$f(x, \theta)$`

for some fixed `$\theta = \theta_0$`

in the following:
By definition, the integral `$I(\theta_0) := \int_0^1 f(x, \theta) \,\D x$`

equals the signed area (marked in light blue) of the region below the graph.
Further, by adding some small `$\Delta\theta > 0$`

to `$\theta_0$`

, we obtain the graph of `$f(x, \theta_0 + \Delta\theta)$`

and the corresponding signed area `$I(\theta_0 + \Delta\theta)$`

, both illustrated in red:

We recall that the derivative of $I$ with respect to `$\theta$`

is given by the rate at which $I$ changes with `$\theta$`

.
To calculate this rate, we examine the difference between `$I(\theta_0 + \Delta\theta)$`

and `$I(\theta_0)$`

:

Geometrically, this difference equals the (signed) area of the orange region illustrated below:

At each fixed `$0 < x < 1$`

, the integrand of Eq. \eqref{eqn:diffI0_0} satisfies that

Base on this relation, we can rewrite the area difference \eqref{eqn:diffI0_0} as:

In both equations above, the equalities become exact at the limit of `$\Delta\theta \to 0$`

.
By dividing both sides by `$\Delta\theta$`

and taking the limit of `$\Delta\theta \to 0$`

, we have

for any `$0 < \theta_0 < 1$`

.
This agrees with the incomplete solution expressed in Eq. \eqref{eqn:dI_0}.

#### The Failure Example #

So what has been the cause for the failure example? To be specific, what has been missing from the incomplete solution \eqref{eqn:dI_0}?

To understand what has been going on, we again examine the integrand `$f(x, \theta)$`

which, for this example, is the piecewise-constant function defined in Eq. \eqref{eqn:f_step}.

The following are the graphs of `$f(x, \theta)$`

for some fixed `$\theta = \theta_0$`

and `$\theta = \theta_0 + \Delta\theta$`

(for some small `$\Delta\theta > 0$`

), respectively:

Further, the difference `$I(\theta_0 + \Delta\theta) - I(\theta_0)$`

between the signed areas below the two graphs is caused by the rectangle illustrated in orange:

Intuitively, in the success example, the change of signed area is caused by **vertical** shifts of the graph—which is captured by the incomplete solution \eqref{eqn:dI*0}.
On the other hand, in this failure example, the change of signed area is caused by horizontal shifts of the graph _at jump discontinuities*—which is

*missing*from the incomplete solution!

We now calculate the signed area of the orange rectangle shown above.
We first observe that the length of the rectangle’s vertical edge equals the difference `$\Delta f \equiv 1 - 1/2 = 1/2$`

of the integrand `$f(x, \theta)$`

across the discontinuity point.

To calculate the length of the rectangle’s horizontal edge, we let `$x(\theta) = \theta/2$`

denote the jump discontinuity point of `$f(x, \theta)$`

defined in Eq. \eqref{eqn:f_step}.
Then, the (signed) length of the horizontal edge is simply `$x(\theta_0 + \Delta\theta) - x(\theta_0)$`

.

Based on the observations above, we know that

Dividing both sides of this equation by `$\Delta t$`

and taking the limit `$\Delta\theta \to 0$`

produce:

Therefore, we know that

matching the hand-derived result in Eq. \eqref{eqn:f_step_dI_manual}.

### The Full Derivative #

Based on the observations above, we now present the general derivative of the 1D integral expressed in Eq. \eqref{eqn:I}:

which comprises:

A

**interior**component obtained by exchanging differentiation and integration operations—identical to Eq. \eqref{eqn:dI_0}.A

**boundary**component involving a sum over all jump discontinuity points`$\{ x_i(\theta) : i = 1, 2, \ldots \}$`

.

#### Remarks #

Precisely, `$\Delta f(x, \theta)$`

in the *boundary* component is defined as

where `$\lim_{u \uparrow x}$`

and `$\lim_{u \downarrow x}$`

denote **one-sided limits** with `$u$`

approaching `$x$`

from *below* (i.e., `$u < x$`

) and *above* (i.e., `$u > x$`

), respectively.
For any fixed $\theta$, $\Delta f(x, \theta)$ is nonzero (and well-defined) if and only if `$x$`

is a jump discontinuity point of `$f(\cdot, \theta)$`

.

Lastly, when the endpoints `$a$`

and `$b$`

of the integral depend on `$\theta$`

, they should be considered as jump discontinuities with `$\Delta f(a, \theta) = -f(a, \theta)$`

and `$\Delta f(b, \theta) = f(b, \theta)$`

.

In the next section, we will present a generalization of Eq. \eqref{eqn:dI} that describes derivatives of Lebesgue integrals.

Unless otherwise stated, we use “continuous” to indicate the

`$C^0$`

class. ↩︎