Improper integrals

IMPROPER INTEGRALS

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. Improper integrals cannot be computed using a normal Riemann integral.

Definition of an Improper Integral of Type 1:

(a) If ₁∫^t f(x) dx exists for every number t ≥ a, then

provided this limit exists (as a finite number).

(b) If _t∫^b f(x) dx exists for every number t ≤ b, then

provided this limit exists (as a finite number).

The improper integrals _a∫^‒∞f(x) dx and _‒∞∫^b f(x) dx

are called convergent if the corresponding limit exists and divergent

if the limit does not exist.

(c) The improper integral

_‒∞∫^∞f(x) dx is defined as

where a is any real number. It is said to converge if both terms converge and diverge if either term diverges.

Example 86. Evaluate ₁∫^∞ 1/x dx if possible.

Solution:

The limit does not exists as a finite number and so the improper integral ₁∫^∞ 1/x dx is divergent.

Example 87. Evaluate ₄∫^∞ 1/√x dx if possible.

Solution:

The limit does not exist as a finite number and so the improper integral ₄∫^∞ 1/√x dx is divergent.

Example 88. Evaluate ₂∫^∞ 1/x² dx if possible.

Solution:

Example 89. : (p‒Test) For what values of p is ₁∫^∞ 1/x^p dx convergent?

Solution: We know that if p = 1, then the integral is divergent,

so let's assume that p ≠ 1. Then

and so the integral converges. On the other hand, if p < 1, then p−1 < 0 and so 1 / t^p‒1 = t^1‒p → ∞ as t→ ∞ and the integral diverges.

So, ₁∫^∞ 1/x^P dx is convergent if p > 1 and divergent if p ≤ 1.

Example 90. Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

Solution:

The integral ₁∫^∞ x dx is divergent by the p‒test, since p = ‒1≤1. Therefore, _‒∞∫^∞ x dx is divergent.

(d) Applying integration by parts, we have

Example 91. Determine if the following integral is convergent or divergent. If it is convergent find its value.

Solution:

There really isn't much to do with these problems once you know how to do them. We'll convert the integral to a limit/integral pair, evaluate the integral and then the limit.

So, the limit is infinite and so this integral is divergent.

Example 92. Determine if the following integral is convergent or divergent. If it is convergent find its value.

Solution:

In this case we've got infinities in both limits. The process we are using to deal with the infinite limits requires only one infinite limit in the integral and so we'll need to split the integral up into two separate integrals. We can split the integral up at any point, so let's choose x = 0 since this will be a convenient point for the evaluation process. The integral is then,

We've now got to look at each of the individual limits.

So, the first integral is convergent. Note that this does NOT mean that the second integral will also be convergent. So, let's take a look at that one.

This integral is convergent and so since they are both convergent the integral we were actually asked to deal with is also convergent and its value is,

Example 93. Determine if the following integral is convergent or divergent. If it is convergent find its value.

 _‒2∫^∞ sin(x) dx

Solution:

First convert to a limit.

This limit doesn't exist and so the integral is divergent

Type 2: Discontinuous Integrands

(1) If f is continuous on [a, b) and is discontinuous at b, then

if this limit exists (as a finite number).

(2) If f is continuous on (a, b] and is discontinuous at a, then

if this limit exists (as a finite number).

The improper integral _a∫^bf(x) dx is called convergent, if the corresponding limit exists and divergent if the limit does not exist.

(3) If f has a discontinuity at c, where a<c<b, then the improper integral _a∫^bf(x)dx is defined as

It is said to converge if both terms converge and diverge if either term diverges.

Example 94. Evaluate ₁∫² [ dx / 1‒x] dx if possible.

Solution: We first note that the given integral is improper because f(x) = 1/(1‒x) has

the vertical asymptote x = 1. We have

 = ‒ ∞ (divergent)

Example 95. Determine if the following integral is convergent or divergent. If it is convergent find its value.

Solution:

We first note that the given integral is improper because f(x) = 1/√(3‒x) has the vertical asymptote x = 3. We have

= 2√3

The limit exists and is finite and so the integral converges and the integral's value is 2√3.

Example 96. Determine if the following integral is convergent or divergent. If it is convergent find its value _‒2∫³ 1/x³ dx.

Solution:

This integrand is not continuous at x = 0 and so we'll need to split the integral up at that point.

Now we need to look at each of these integrals and see if they are convergent.

At this point we're done. One of the integrals is divergent that means the integral that we were asked to look at is divergent. We don't even need to bother with the second integral.

Example 97. Determine if the following integral is convergent or divergent. If it is convergent find its value.

₀∫^∞ 1/x² dx

Solution: This is an integral over an infinite interval that also contains a discontinuous integrand. To do this integral we'll need to split it up into two integrals so each integral contains only one point of discontinuity. It is important to remember that all of the processes we are working with in this section so that each integral only contains one problem point.

We can split it up anywhere but pick a value that will be convenient for evaluation purposes.

In order for the integral in the example to be convergent we will need BOTH of these to be convergent. If one or both are divergent then the whole integral will also be divergent.

We know that the second integral is convergent by the fact given in the infinite interval portion above. So, all we need to do is check the first integral.

So, the first integral is divergent and so the whole integral is divergent.

Example 98. Evaluate ₀∫³ [ dx / √9‒x² ] if possible.

Solution:

We first note that the given integral is improper because f(x) = 1 / √9‒x² has the vertical asymptotes x = 3. We have

Example 99. Evaluate _‒1∫¹ dx/x if possible.

Solution:

We first note that the given integral is improper because f(x) = 1/x has the vertical asymptote x = 0.

We have

it follows that _‒1∫¹ dx/x is divergent.

Example Evaluate  if possible.

Solution: