Improper Integrals


The fundamental theorem of Calculus states that we can evaluate a definite integral by
F(b) – F(a) -- the difference between the antiderivative values at the limits of
integration IF the integrand function is continuous on the closed interval [ a, b ].
This means that the limits of integration must be well defined Real numbers, and
the integrand function can't have discontinuities over that interval.

Improper Integrals are those which do not satisfy one or both of these conditions.
Remember that the definite integral represents the area under the curve over the
interval of integration. If one or both of the limits of integration is infinity,
the question becomes can we find a number value for the area under the curve.
If the integrand function has a discontinuity somewhere on the interval of
integration, again, we have to find ways to evaluate the area under the curve.


Note: If the integral can be evaluated, we say it converges or it is convergent.

If the integral becomes infinite or can't be evaluated, we say it diverges or it is divergent.


Note: We may have to use l'Hopital's Rule or other limit techniques to find the final limit.


There are 2 types of improper integrals

Type 1/ Integrals with Infinite Limits of Integration

In this type, one (or both) of the limits of integration is infinity.
We replace the infinity with a parameter (usually t, s or r),
we integrate over a well defined interval [ a, t ], then we limit the parameter t to infinity.
Let's look at pictures and the integral set up to understand the approach.

i) ii)

** f (x) must be continuous over the interval of integration.
Set a = 0 where possible for efficiency.



1/ Assign an area if possible to the region between y = e x and the x-axis, left of x = 1.


We need to integrate from to x = 1, -- so it's the lower limit that is undefined.

The area is:

As t approaches , e t approaches 0. So the integral converges to e square units.


2/ Sketch the graph of and evaluate if possible the integral from to .


Area =

This integral converges to give an area of o square units.


3/ Evaluate if possible the integral of y = x – 3/4 , from x = 1 to .

We want to evaluate

This integral diverges. We cannot assign a value to it.



1) Always replace the infinite limit(s) with a parameter(s), then, at the end, limit it to infinity.

2) Draw the graph if you can since in many cases, it's obvious if the area converges or diverges.

Type 2/ Integrals with Discontinuous Integrands:

Like the first type, there are 3 possible situations:

Here are the images and the parameter set up:

i) If f (x) is discontinuous at b , then:

ii) If f (x) is discontinuous at a , then:

iii) If f (x) is discontinuous at c where a < c < b, then:

If either of the 2 integrals diverges, the whole thing diverges.

Example: (type iii: discontinuity on the interval).

This integral has a vertical asymptote at x = 1, so c = 1.

It becomes

We antidifferentiate to get

The 1st integral becomes . The integral diverges.


Watch out for ones with both kinds of discontinuities!!

For example:

has an infinite limit of integration and an asymptote at x = 1 on the interval.

We break this guy up into 3 integrals.

The setup is:

from t to 0 from 0 to s from s to 3
or or or


1/ Determine if these improper integrals converge or diverge:

a) b)

c) d)

e) f)

g) h)


a) b)

The integral diverges.

The integral converges to 1.

The volume of the solid created by

revolving about the x-axis

from x = 2 to is (1 = answer to a).

The surface area of this solid is
(the answer to b).

This solid has finite volume but infinite area.

If it were a paint can, it could never hold enough paint to cover the outside of the can.


This integral converges to e 0 = 1.


now substitute u = 3 – x and du = – dx

to get:

This integral converges to .


The integral diverges.


There is an asymptote at x = 3.

When we antidifferentiate the 1st integral we get:

The integral diverges.


g) We'll split it at x = 0 but first let's simplify the integrand with a substitution.

If u = – x², then du = – 2x dx so we can replace the integrand with – ½ e u du (easy to integrate)

The antiderivative is e u but u = – x² so we get:

Since , we get:

– ½ [ (1 – 0) + (0 – 1) ] = 0

The integral converges to 0.

h) Let's deal with the integrand's discontinuities first.

x² – 3x + 2 = (x – 2)(x – 1) so the discontinuities are at x = 1 and x = 2: not on the interval.

This means we need only deal with the infinite limit of integration.

So we get:

Now, let's do partial fraction decomposition on the integrand.

Setting x = 2 and x = 1, we find A = 1 and B = – 1.

Applying the rule of logs that says a difference of 2 logs is the log of the quotient,
we divide (x – 2) by (x – 1) to get:

and now we evaluate the limit.

It becomes ln 2 – ln 1 = ln 2 – 0 = ln 2. The integral converges.


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