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Section 1-8 : Improper Integrals

3. Determine if the following integral converges or diverges. If the integral converges determine its value.

\[\int_{{ - 5}}^{1}{{\frac{1}{{10 + 2z}}\,dz}}\]

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Hint : Don’t forget that we can’t do the integral as long as there is a division by zero in the integrand at some point in the interval of integration!
Start Solution

First, notice that there is a division by zero issue (and hence a discontinuity) in the integrand at \(z = - 5\) and this is the lower limit of integration. We know that as long as that discontinuity is there we can’t do the integral. Therefore, we’ll need to eliminate the discontinuity first as follows,

\[\int_{{ - 5}}^{1}{{\frac{1}{{10 + 2z}}\,dz}} = \mathop {\lim }\limits_{t \to \, - {5^ + }} \int_{t}^{1}{{\frac{1}{{10 + 2z}}\,dz}}\]

Don’t forget that the limits on these kinds of integrals must be one-sided limits. Because the interval of integration is \(\left[ { - 5,1} \right]\) we are only interested in the values of \(z\) that are greater than -5 and so we must use a right-hand limit to reflect that fact.

Show Step 2

Next, let’s do the integral. We’ll not be putting a lot of explanation/detail into the integration process. By this point it is assumed that your integration skills are getting pretty good. If you find your integration skills are a little rusty you should go back and do some practice problems from the appropriate earlier sections.

In this case we can do a simple Calculus I substitution. Here is the integration work.

\[\int{{\frac{1}{{10 + 2z}}\,dz}} = \frac{1}{2}\ln \left| {10 + 2z} \right| + c\]

Note that we didn’t do the definite integral here. The limits don’t really affect how we do the integral and so we held off dealing with them until the next step.

Show Step 3

Okay, now let’s take care of the limits on the integral.

\[\int_{{ - 5}}^{1}{{\frac{1}{{10 + 2z}}\,dz}} = \mathop {\lim }\limits_{t \to \, - {5^ + }} \left. {\left( {\frac{1}{2}\ln \left| {10 + 2z} \right|} \right)} \right|_t^1 = \mathop {\lim }\limits_{t \to \, - {5^ + }} \left( {\frac{1}{2}\ln \left| {12} \right| - \frac{1}{2}\ln \left| {10 + 2t} \right|} \right)\] Show Step 4

We now need to evaluate the limit in our answer from the previous step. Here is the limit work.

\[\int_{{ - 5}}^{1}{{\frac{1}{{10 + 2z}}\,dz}} = \mathop {\lim }\limits_{t \to \, - {5^ + }} \left( {\frac{1}{2}\ln \left| {12} \right| - \frac{1}{2}\ln \left| {10 + 2t} \right|} \right) = \frac{1}{2}\ln \left| {12} \right| + \infty = \infty \] Show Step 5

The final step is to write down the answer!

In this case, the limit we computed in the previous step existed and but was infinity. Therefore, the integral diverges.