#Improper Integrals

A definite integral is considered "improper" if the interval of integration is unbounded, the function being integrated is unbounded on the interval of integration, or both. In such cases, the usual definition of the definite integral does not apply.

###Example 1

Consider the function $\displaystyle f(x)=\frac{1}{x^2}$. Suppose we wanted the area under this curve for $x\ge1$; in other words, we want $\displaystyle \int_1^{\infty} f(x)\,dx$. This is an infinite region, so you might assume that it has infinite area. However, in this case the area is actually finite.

To see this, consider the integral $\displaystyle\int_1^t \frac{1}{x^2}\,dx$ for any $t>1$. This is a normal definite integral, and the answer is $\displaystyle 1-\frac{1}{t}$.

What happens as $t\to\infty$? We have $\displaystyle 1-\frac{1}{t}\to 1$.

So it make sense to say $\displaystyle\int_1^{\infty}f(x)\,dx=1$.

###Infinite Intervals

The previous example falls under the first type of improper integral, when one or both of the limits of integration is $\pm\infty$. In this case, the region under the curve is infinite in the horizontal direction.

Here is the definition:

If $\displaystyle\int_a^tf(x)\,dx$ exists for every $t\ge a$, then we define $\displaystyle\int_a^{\infty}f(x)\,dx=\lim_{t\to \infty}\int_a^tf(x)\,dx$, provided this limit exists.

If the limit exists, we say the improper integral converges (or is convergent). Otherwise, we say it diverges (or is divergent).

Similarly, if $\displaystyle\int_t^bf(x)\,dx$ exists for every $t\le b$, then we define $\displaystyle\int_{-\infty}^bf(x)\,dx=\lim_{t\to -\infty}\int_t^bf(x)\,dx$, provided this limit exists.

Also, if there is a number $a$ such that $\displaystyle\int_{-\infty}^af(x)\,dx$ and $\displaystyle\int_a^{\infty}f(x)\,dx$ both converge, then we define $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^af(x)\,dx+\int_a^{\infty}f(x)\,dx$.

Note: If these integrals converge for one value of $a$, then they converge for any value of $a$. The answer does not depend on the choice of $a$.

It is interesting that $\displaystyle\lim_{t\to\infty}\int_{-t}^tf(x)\,dx$ may not equal $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx$.

###Example 2

Let's explore integrals of the form $\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx$.

We already saw what happens when $p=2$ above, so let's try $p=1,\ 3,\ \frac{1}{2},\ \frac{3}{2},$ etc.

###Unbounded Integrands

The second type of improper integral is when the integrand (the function being integrated) is unbounded on the interval of integration. In this case, the region under the curve is infinite in the vertical direction.

###Example 3

Consider $\displaystyle\int_0^1 \ln(x)\, dx$.

We have finite limits of integration, but notice that $\ln(x)$ is not bounded on the interval $[0,1]$, since $\displaystyle\lim_{x\to 0^+}\ln(x)=-\infty$.

Notice that for $0 < t < 1$ we have $\displaystyle\int_t^1\ln(x)\,dx=t-t\ln(t)-1$.

If we try to compute this integral in Sage, it will ask us for more information.

Sage needs some information about t. We can give it this information using the assume command.

What happens as $t\to0$?

So it makes sense to define $\displaystyle\int_0^1\ln(x)\,dx=-1$.

Here's the definition in general:

If $f(x)$ is continuous on the interval $(a,b]$ and is unbounded near $a$, then we define $\displaystyle\int_a^bf(x)\,dx=\lim_{t\to a^+}\int_t^bf(x)\,dx$, provided this limit exists.

Similarly, if $f(x)$ is continuous on the interval $[a,b)$ and is unbounded near $b$, then we define $\displaystyle\int_a^bf(x)\,dx=\lim_{t\to b^-}\int_a^tf(x)\,dx$, provided this limit exists.

Also, if $f(x)$ is continuous on the intervals $[a,c)$ and $(c,b]$ and unbounded near $c$, then we define $\displaystyle\int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$, provide both of these converge.

###Example 4

Let's look at a graph of the integrand. Notice that it is unbounded near 0.

Thus, this is an improper integral, so $$\int_0^3\frac{1}{x\sqrt{x}}\,dx=\lim_{t\to 0^+}\int_t^3\frac{1}{x\sqrt{x}}\,dx=\lim_{t\to 0^+}-\frac{2\sqrt{3}}{3}+\frac{2}{\sqrt{t}}=\infty$$

Therefore, this integral diverges.

Note that Sage can handle improper integrals. It informs us that this integral is divergent (see the last line of the error output).

Example 5

What if our improper integral has its unbounded value in the interior of the integration interval? We split into two differnt integrals! Consider the integral,

For this integral, there is a vertical asymptote at $x=1$

To evaluate this improper integral, we split up our integral into a left and right portion,

Both of these integrals are improper so we replace the offending limit of integration with a limit:

Evaluating, we find,

As one (actually both) of these integrals diverge, the origional integral does not converge. It is wrong to say that the integral is $\infty+(-\infty)=0$.

###Example 6

When we look at the graph, we see that the integrand is unbounded near 9.

So we have $$\int_1^9\frac{1}{\sqrt{9-x}}\,dx=\lim_{t\to 9^-}\int_1^t\frac{1}{\sqrt{9-x}}\,dx=\lim_{t\to 9^-}(4\sqrt{2}-2\sqrt{9-t})=4\sqrt{2}=2^{5/2}$$

Here is the direct computation:

###Example 7

Notice that the interval of integration is unbounded, and the integrand is unbounded near 0. So we need to split this up into two integrals.

So we have

Note: The choice of 1 as the other limit of integration was arbitrary.

Here's the direct calculation:

Note: If you get an error that says "Assumption is redundant" or "Assumption is inconsistent," then find a blank line, type forget(), and hit run.

#Improper Integrals Assignment

For each integral below:

• Graph the integrand (the function inside the integral).

• State why the integral is improper (there may be more than one reason).

• Set up an appropriate limit to compute the value of the integral and calculate the limit.

• Evaluate the integral, and compare the answer with the limit above.

###Question 1

$\displaystyle\int_0^{\infty} \frac{1}{x^2+2x+4}\,dx$

###Question 2

$\displaystyle\int_0^1\frac{1}{\sqrt{x}}\,dx$

###Question 3

$\displaystyle\int_0^{10}\frac{1}{ (x-5)^{2}}\,dx$