Integrating
Discontinuous Functions
Tom McNamara
Department of Mathematics, SWOSU
April 11, 2015
Overview
1 Introduction
2 Integration and Smoothing
3 A Discontinuous Function
4 Limit Processes
The “Nice” Calculus Operation
Calculus students have the impression that differentiation is the
“nice” operation, and integration is trouble.
This is understandable. Differentiation is mechanical. Even
2
e x + tan x
f (x) = p
x 4 − x log3 x
can be tackled by following rules in the correct order.
Overview
1 Introduction
2 Integration and Smoothing
3 A Discontinuous Function
4 Limit Processes
Integration and Anti-Derivatives
Students often conflate “integration” with finding an anti-derivative.
The Fundamental Theorem of Calculus is powerful, but there’s
more to integration.
Indeed,
Z
1
2
e x dx
0
2
e x + tan x
appears much simpler than f (x) = p
.
x 4 − x log3 x
Unfortunately, the Fundamental Theorem of Calculus doesn’t help
us evaluate the definite integral.
Integration and Anti-Derivatives
Hence, it is reasonable that students view integration as being more
difficult. There are many elementary functions who do not have
anti-derivatives.
To understand why integration is "nicer," consider the following
function. Let
Z x
f (x) =
|t|dt.
0
Students are not familiar with the anti-derivative of f (t) = |t|.
Some hesitate to work with this new creature.
Integration and Anti-Derivatives
If we consider our new function’s graph we can see that it is
"nicer" than the absolute value function.
Notice our new function has no corners.
Integration and Anti-Derivatives
Our new function’s graph resembles that of the cubing function,
it’s not quite that simple. We see that if x ≥ 0 we have
x
Z x
Z x
1 2
f (x) =
|t|dt =
tdt = 2 t = 12 x 2 .
0
0
0
On the other hand, if x < 0 we get
Z
x
Z
|t|dt = −
f (x) =
0
x
Z
tdt =
0
0
tdt =
x
0
1 2
2t x
= − 21 x 2 .
Integration and Anti-Derivatives
The function
Z
(
x
|t|dt =
f (x) =
0
1 2
2x
− 12 x 2
if x ≥ 0
.
if x < 0
Indeed, since
(
x
|x| =
−x
if x ≥ 0
if x < 0
it is easy to see where our function’s piecewise defined formula
comes from.
Overview
1 Introduction
2 Integration and Smoothing
3 A Discontinuous Function
4 Limit Processes
The Greatest Integer Function
Each part of the piece-wise formula for the integral is an
anti-derivative of the corresponding piece of the absolute value
function.
Now consider the greatest integer function, f (x) = bxc.
For example, we see that b3.75c = 3 and b−4.0067901c = −5.
This function is graphed below.
The Greatest Integer Function’s Graph
This function is discontinuous at every integer.
A Piecewise Definition
It is possible to consider a piece-wise definition for this function.

..


.





1
if 1 ≤ x < 2



0
if 0 ≤ x < 1
.
bxc =

−1
if
−
1
≤
x
<
0





−2 if − 2 ≤ x < −1




...
Integral of the Greatest Integer Function
Rx
Now consider the function F (x) = 0 btcdt, defined by integrating
the greatest integer function from 0 to x. For example, we compute
that
Z
1
Z
btcdt +
F (3.5) =
0
2
Z
btcdt +
1
3
Z
btcdt +
2
3.5
btcdt.
3
We see that the first three integrals are 0, 1, and 2 respectively.
Note that btc is constant on these intervals. For the last, btc = 3
on that interval. Thus, we have 3 × 0.5 for that component.
Integral of the Greatest Integer Function
Combining all the above work together shows
Z
3.5
btcdt = 0 + 1 + 2 + 3 × 0.5 = 4.5.
0
There was nothing special about x = 3.5. Indeed, for any x with
3 ≤ x < 4 we see that
Z
x
Z
btcdt =
F (x) =
0
1
Z
btcdt +
0
2
Z
btcdt +
1
3
Z
btcdt +
2
x
btcdt.
3
Integral of the Greatest Integer Function
Evaluating these gives
Z x
F (x) =
btcdt = 0 + 1 + 2 + (3x − 3 · 3) = 3x − 6
0
for any x with 3 ≤ x < 4. This matches our concrete x = 3.5
above because 3 × 3.5 − 6 = 4.5.
Integral of the Greatest Integer Function
Applying this procedure to all x values gives

..


.





x −1
if 1 ≤ x < 2



0
if 0 ≤ x < 1
.
F (x) =
−x
if − 1 ≤ x < 0




−2x − 1 if − 2 ≤ x < −1




...
Note that each piece is indeed an anti-derivative of the
corresponding piece in the definition of bxc.
Graphing the Integral
Graphing this function gives
So, the integral of the discontinuous function is, in fact, continuous.
Graphing the Integral
Notice that there are corners on this graph. Thus, the function
Z x
F (x) =
btcdt
0
is continuous for all x ∈ R but is not differentiable at the places
where the integrand is discontinuous.
Overview
1 Introduction
2 Integration and Smoothing
3 A Discontinuous Function
4 Limit Processes
Integration and Sequences of Functions
Integration is (relatively) well behaved with respect to limit
operations.
For example, if fn (x) is a sequence of functions, and
fn (x) → f (x)uniformly
then we can show that
Z
lim
n→∞ a
b
Z
fn (x)dx =
b
f (x)dx.
a
Indeed, uniform convergence interacts well with limits, continuity
and integrals.
The Problem Child
Differentiation is the problem calculus operation.
Consider these questions.
1
2
If fn converges uniformly to f , and the functions fn are
differentiable, does that imply f is also differentiable?
Does fn0 converge to f 0 ?
Let’s consider the second question first.
Example 1
Let fn (x) : [0, 2π] → R be given by fn (x) =
1
sin(nx).
n
We see that fn converges uniformly to the zero function, f (x) = 0.
However, fn0 (x) = cos(nx). From here we get that
0
fn (0) − f 0 (0) = 1.
Thus, fn0 does not even converge point-wise to f 0 .
Example 2
r
1
. These functions are differentiable for all
n2
x ∈ [−1, 1]. Further, we see that
Define fn (x) =
x2 +
1
|x| ≤ fn (x) ≤ |x| + .
n
Thus fn (x) converges uniformly to f (x) = |x|. Unfortunately, this
function is not differentiable at 0.
Thus, we see that the uniform limit of differentiable functions need
not be differentiable.
What Can We Do?
Theorem
Let fn : [a, b] → R be a sequence of differentiable functions with fn0
continuous. Suppose that the derivatives fn0 converge uniformly to a
function g . Suppose also that there exists a point c ∈ [a, b] such
that lim fn (c) exists. Then the fn converge uniformly to a function
n→∞
f and f 0 = g .
So, if fn0 converges uniformly, then so does fn and we get
d
d
lim fn (x) = lim
fn (x).
n→∞ dx
dx n→∞
Idea Behind the Proof
Let L = lim fn (c).
n→∞
Since the fn0 were assumed continuous and converge uniformly to g ,
that g must be continuous.
Define
Z
f (x) = L −
c
Z
g (x)dx +
a
x
g (x)dx
a
for x ∈ [a, b].
Show that fn converges uniformly to this f and that f is
differentiable with f 0 = g .