9.7: Kinetic energy and Improper Integrals

Recall example 9.5.3 in which we computed the work required to lift an object from the surface of the earth to some large distance \(D\) away. Since \(F=k/x^2\) we computed $$\int_{r_0}^D {k\over x^2}\,dx=-{k\over D}+{k\over r_0}.$$ We noticed that as \(D\) increases, \(k/D\) decreases to zero so that the amount of work increases to \(k/r_0\). More precisely, $$ \lim_{D\to\infty}\int_{r_0}^D {k\over x^2}\,dx= \lim_{D\to\infty}-{k\over D}+{k\over r_0}={k\over r_0}. $$ We might reasonably describe this calculation as computing the amount of work required to lift the object "to infinity,'' and abbreviate the limit as $$ \lim_{D\to\infty}\int_{r_0}^D {k\over x^2}\,dx= \int_{r_0}^\infty {k\over x^2}\,dx. $$ Such an integral, with a limit of infinity, is called an improper integral. This is a bit unfortunate, since it's not really "improper'' to do this, nor is it really "an integral''---it is an abbreviation for the limit of a particular sort of integral. Nevertheless, we're stuck with the term, and the operation itself is perfectly legitimate. It may at first seem odd that a finite amount of work is sufficient to lift an object to "infinity'', but sometimes surprising things are nevertheless true, and this is such a case. If the value of an improper integral is a finite number, as in this example, we say that the integral converges, and if not we say that the integral diverges.

Here's another way, perhaps even more surprising, to interpret this calculation. We know that one interpretation of $$\int_{1}^D {1\over x^2}\,dx$$ is the area under \(y=1/x^2\) from \(x=1\) to \(x=D\). Of course, as \(D\) increases this area increases. But since $$\int_{1}^D {1\over x^2}\,dx=-{1\over D}+{1\over1},$$ while the area increases, it never exceeds 1, that is $$\int_{1}^\infty {1\over x^2}\,dx= 1.$$ The area of the infinite region under \(y=1/x^2\) from \(x=1\) to infinity is finite.

Consider a slightly different sort of improper integral: \(\int_{-\infty}^\infty xe^{-x^2}\,dx\). There are two ways we might try to compute this. First, we could break it up into two more familiar integrals: $$ \int_{-\infty}^\infty xe^{-x^2}\,dx= \int_{-\infty}^0 xe^{-x^2}\,dx+\int_{0}^\infty xe^{-x^2}\,dx. $$ Now we do these as before: $$ \int_{-\infty}^0 xe^{-x^2}\,dx=\lim_{D\to\infty} \left.-{e^{-x^2}\over2}\right|_D^0=-{1\over2}, $$ and $$ \int_0^\infty xe^{-x^2}\,dx=\lim_{D\to\infty} \left.-{e^{-x^2}\over2}\right|_0^D={1\over2}, $$ so $$ \int_{-\infty}^\infty xe^{-x^2}\,dx=-{1\over2}+{1\over2}=0.$$ Alternately, we might try $$ \int_{-\infty}^\infty xe^{-x^2}\,dx= \lim_{D\to\infty}\int_{-D}^D xe^{-x^2}\,dx= \lim_{D\to\infty}\left.-{e^{-x^2}\over2}\right|_{-D}^D= \lim_{D\to\infty} -{e^{-D^2}\over2}+{e^{-D^2}\over2}=0. $$ So we get the same answer either way. This does not always happen; sometimes the second approach gives a finite number, while the first approach does not; the exercises provide examples. In general, we interpret the integral \(\int_{-\infty}^\infty f(x)\,dx\) according to the first method: both integrals \(\int_{-\infty}^a f(x)\,dx$ and $\int_{a}^\infty f(x)\,dx\) must converge for the original integral to converge. The second approach does turn out to be useful; when \(\lim_{D\to\infty}\int_{-D}^D f(x)\,dx=L\), and \(L\) is finite, then \(L\) is called the Cauchy Principal Value of \(\int_{-\infty}^\infty f(x)\,dx\).

Here's a more concrete application of these ideas. We know that in general $$W=\int_{x_0}^{x_1} F\,dx$$ is the work done against the force \(F\) in moving from \(x_0\) to \( x_1\). In the case that \(F\) is the force of gravity exerted by the earth, it is customary to make \(F < 0\) since the force is "downward.'' This makes the work \(W\) negative when it should be positive, so typically the work in this case is defined as $$W=-\int_{x_0}^{x_1} F\,dx.$$ Also, by Newton's Law, \(F=ma(t)\). This means that $$W=-\int_{x_0}^{x_1} ma(t)\,dx.$$ Unfortunately this integral is a bit problematic: \(a(t)\) is in terms of \(t\), while the limits and the "\(dx\)'' are in terms of \(x\). But \(x\) and \(t\) are certainly related here: \(x=x(t)\) is the function that gives the position of the object at time \(t\), so \(v=v(t)=dx/dt=x'(t)\) is its velocity and \(a(t)=v'(t)=x''(t)\). We can use \(v=x'(t)\) as a substitution to convert the integral from "\(dx\)'' to "\(dv\)'' in the usual way, with a bit of cleverness along the way: $$\eqalign{ dv&=x''(t)\,dt=a(t)\,dt=a(t){dt\over dx}\,dx\cr {dx\over dt}\,dv&=a(t)\,dx\cr v\,dv&=a(t)\,dx.\cr }$$ Substituting in the integral: $$ W=-\int_{x_0}^{x_1} ma(t)\,dx=-\int_{v_0}^{v_1} mv\,dv= -\left.{mv^2\over2}\right|_{v_0}^{v_1}=-{mv_1^2\over2}+{mv_0^2\over2}. $$ You may recall seeing the expression \(mv^2/2\) in a physics course---it is called the kinetic energy of the object. We have shown here that the work done in moving the object from one place to another is the same as the change in kinetic energy.

We know that the work required to move an object from the surface of the earth to infinity is $$W=\int_{r_0}^\infty {k\over r^2}\,dr={k\over r_0}.$$ At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force on an object of mass \(m\) is \(F=9.8m\). The radius of the earth is approximately 6378.1 kilometers or 6378100 meters. Since the force due to gravity obeys an inverse square law, \(F=k/r^2\) and \(9.8m=k/6378100^2\), \(k= 398665564178000m\) and \(W=62505380 m\).

Now suppose that the initial velocity of the object, \(v_0\), is just enough to get it to infinity, that is, just enough so that the object never slows to a stop, but so that its speed decreases to zero, i.e., so that \(v_1=0\). Then $$62505380 m=W=-{mv_1^2\over2}+{mv_0^2\over2}={mv_0^2\over2}$$ so $$v_0=\sqrt{125010760}\approx 11181\quad\hbox{meters per second},$$ or about 40251 kilometers per hour. This speed is called the escape velocity. Notice that the mass of the object, \(m\), canceled out at the last step; the escape velocity is the same for all objects. Of course, it takes considerably more energy to get a large object up to 40251 kph than a small one, so it is certainly more difficult to get a large object into deep space than a small one. Also, note that while we have computed the escape velocity for the earth, this speed would not in fact get an object "to infinity'' because of the large mass in our neighborhood called the sun. Escape velocity for the sun starting at the distance of the earth from the sun is nearly 4 times the escape velocity we have calculated.

Contributors and Attributions

• David Guichard (Whitman College)

• Integrated by Justin Marshall.