Discontinuous Function Integration
Date: Winter 2011-2012
We know that the area under a continuous function on a certain interval, for instance, [a,b], can be found out by definite integration (taking limit of a sum of areas of rectangles). But when the curve is a discontinuous function, the limit does not exist and is not independent of the choice of the height of rectangle in each delta x. Does it mean that the height in an interval delta x including f(x) of undefined value can be any value so that a limiting value (area under the curve) does not exist?
If I understand your question correctly, here is a way out.
The limits of integration are divided into
sections [x1,y1] to [x2,y2] over which the function is continuous.
Then the next section [x2,y2]
to [x3,y3] over which the function is continuous is added to the first
section. You can carry this on as long as you need to include each of the
separate segments. That is not so hard, but I am not completely sure that is
what you are asking. There are some pathological functions where this
pasting together does not work so well.
Suppose you have a function G(x) with the following properties:
G(0) = 0, but G(x1) =1, where x1 > 0, and G(x2) = 0 where x2 > x1.
So the function G(x) "jumps" from a value of (0) to (1), then back
to (0) for the
next value of (x) -- but the values of (x) are irrational. Then the
integral depends upon the limits of the integral.
Then the value of the integral is recursive. It depends upon the
limits of integration, which in turn, depend upon the value of G(x) chosen.
There are many, if not infinite, number of such pathological
I would not say that the height in the vicinity of the discontinuity
can be ANY value, because you know the value of the function on either
side of the discontinuity. It seems natural choose, as the height of
the rectangle, the value of the function at the x value of the middle
of the rectangle. In this case, the height will change as you decrease
dx, depending on where the x midpoint of the discontinuity-straddling
rectangle happens to lie.
In actual practice, of course, you rarely have to worry about this. The
fact that there is a discontinuity usually means that you do not have a
single functional form that represents the function on both sides of the
discontinuity, so you just split the integral into two integrals and add
Even when the function is discontinuous there may be a well-defined
integral. Take for instance a function f defined in [a,b] which has
discontinuity at a point c, but this discontinuity is of the kind
that the lateral limits exist. Then f is integratable and its
integral equals the integral from a to c plus the integral from c to
b. There is a theorem of Lebesgue that states that a bounded
function f on [a,b] is (Riemann) integratable if and only if the set
D of discontinuity points of f is of measure zero, in the sense that
for any \epsilon>0 there exist denumerably many intervals (a_n,b_n)
such that their union contain [a,b] and the sum of their lengths do
not surpass \epsilon.
However, functions such as Dirichlet's (this function is defined on
[0,1], equals 1 on rationals and 0 on irrationals) are not Riemann
integratable, since its discontinuities are the whole [0,1]. If you
wish to integrate these functions, if have to go for Lebesgue's
theory of integration.
The integration of discontinuous functions is a large subject, too
large to go into detail here, but do a Search on the term "integration of
discontinuous functions". You will find the topic discussed on several sites
at different levels of rigor, depending on how much detail you want to know.
Some discontinuous functions can be integrated by dividing up the function
into pieces so that it is "piece-wise" continuous, but this approach has its
limitations. There are discontinuous functions that cannot be integrated.
The process of rectangles only applies when the limit exists. You must arrange your rectangles so that the limit is defined. This occurs when the discontinuity is held as a border between rectangles. Find the sum of the two limits, the two areas, one to the left of the discontinuity and one to the right. Sometimes, provided you choose you algebra wisely, you can find the integral even when a vertical asymptote exists within the range of the domain of your integral.
Dr. Ken Mellendorf
Illinois Central College
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Update: June 2012