Section 7.8 : Improper Integrals
10. Determine if the following integral converges or diverges. If the integral converges determine its value.
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\]Show All Steps Hide All Steps
Now there is clearly an infinite limit here, but also notice that there is a discontinuity at \(x = 0\) that we’ll need to deal with.
Based on the material in the notes it should make sense that, provided both integrals converge, we should be able to split up the integral at any point. In this case let’s split the integral up at \(x = - 1\). Doing this gives,
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\]Keep in mind that splitting up the integral like this can only be done if both of the integrals converge! If it turns out that even one of them is divergent then it will turn out that we couldn’t have done this and the original integral will be divergent.
So, not worrying about if this was really possible to do or not let’s proceed with the problem.
Now, we can eliminate the problems as follows,
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \int_{t}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \mathop {\lim }\limits_{s \to {0^ - }} \int_{{ - 1}}^{s}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\] Show Step 2Next, 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 Calc I substitution. Here is the integration work.
\[\int{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = - {{\bf{e}}^{\frac{1}{x}}} + c\]Note that we didn’t do the definite integral here. The limits don’t really affect how we do the integral and the integral for each was the same with only the limits being different so no reason to do the integral twice.
Show Step 3Okay, now let’s take care of the limits on the integral.
\[\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \left. {\left( { - {{\bf{e}}^{\frac{1}{x}}}} \right)} \right|_t^{ - 1} + \mathop {\lim }\limits_{s \to {0^ - }} \left. {\left( { - {{\bf{e}}^{\frac{1}{x}}}} \right)} \right|_{ - 1}^s = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^{\frac{1}{t}}}} \right) + \mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}} + {{\bf{e}}^{ - 1}}} \right)\] Show Step 4We now need to evaluate the limits in our answer from the previous step. Here is the limit work
\[\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^{\frac{1}{t}}}} \right) + \mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}} + {{\bf{e}}^{ - 1}}} \right)\\ & = \,\,\,\,\,\,\,\,\,\left[ { - {{\bf{e}}^{ - 1}} + {{\bf{e}}^0}} \right]\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {0 + {{\bf{e}}^{ - 1}}} \right]\end{align*}\]Note that,
\[\mathop {\lim }\limits_{s \to {0^ - }} \frac{1}{s} = - \infty \]since we are doing a left-hand limit and so \(s\) will be negative. This in turn means that,
\[\mathop {\lim }\limits_{s \to {0^ - }} \left( { - {{\bf{e}}^{\frac{1}{s}}}} \right) = 0\] Show Step 5The final step is to write down the answer!
Now, from the limit work in the previous step we see that,
\[\int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = - {{\bf{e}}^{ - 1}} + 1\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} = {{\bf{e}}^{ - 1}}\]Therefore, each of the integrals are convergent and have the values shown above. This means that we could in fact break up the integral as we did in Step 1. Also, the original integral is now,
\[\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} & = \int_{{ - \infty }}^{{ - 1}}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}} + \int_{{ - 1}}^{0}{{\frac{{{{\bf{e}}^{\frac{1}{x}}}}}{{{x^2}}}\,dx}}\\ & = \,\,\, - {{\bf{e}}^{ - 1}} + 1\,\,\, + \,\,\,\,\,\,\,{{\bf{e}}^{ - 1}}\\ & = 1\end{align*}\]Therefore, the integral converges and its value is 1.