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### Section 7.8 : Improper Integrals

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

$\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}$

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Hint : Don’t forget that we can’t do the integral as long as there is an infinity in one of the limits! Also, don’t forget that infinities in both limits need as extra step to get set up.
Start Solution

First, we need to recall that we can’t do the integral as long as there is an infinity in one of the limits. Note as well that in this case we have infinities in both limits and so we’ll need to split up the integral.

The integral can be split up at any point in this case and $$w = 0$$ seems like a good point to use for the split point. Splitting up the integral gives,

$\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = \int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}$

Remember as well, that we can only break up the integral like this provided both of the new integrals are convergent! 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 infinities as follows,

$\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \int_{t}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \mathop {\lim }\limits_{s \to \infty } \int_{0}^{s}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}$ 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 Calc I substitution. Here is the integration work.

$\int{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = - \frac{3}{2}\frac{1}{{{w^4} + 1}} + c$

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 3

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

\begin{align*}\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left. {\left( { - \frac{3}{2}\frac{1}{{{w^4} + 1}}} \right)} \right|_t^0 + \mathop {\lim }\limits_{s \to \infty } \left. {\left( { - \frac{3}{2}\frac{1}{{{w^4} + 1}}} \right)} \right|_0^s\\ & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) + \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right)\end{align*} Show Step 4

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

\begin{align*}\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) + \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right)\\ & = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ { - \frac{3}{2}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\frac{3}{2}} \right]\end{align*}

Note that we put the answers for each limit in brackets to make it clear what each limit was. This will be important for the next step.

Show Step 5

The final step is to write down the answer!

Now, from the limit work in the previous step we see that,

\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) = \, - \frac{3}{2}\\ & \\ \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right) = \frac{3}{2}\end{align*}

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 }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\\ & = \,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}\,\\ & = 0\end{align*}

Therefore, the integral converges and its value is 0.