Improper Integrals

Up to this point, we have always integrated over a finite interval $ a\leq x \leq b $ . However, there are some occassions when we want to integrate over an unbounded region $ a \leq x < \infty $ . For instance, if we have a radioactive substance, individual particles can decay at any time. That means if we want to compute the average time for a particle to decay, we will have to integrate over an unbounded time interval.

To take care of this problem, we will integrate over some finite interval $ a \leq x \leq b $ and then consider what happens as b becomes very large. In particular,

\[ \int_a^\infty f(x)~dx = \lim_{b\to\infty} \int_a^b f(x)~dx \]

As we'll see, this limit may exist in which case we say that the integral converges . However, the limit may not exist and we then say that the integral diverges.


Examples

  1. Here we'll evaluate the improper integral $ \int_0^\infty e^{-x}~dx $ whose graph is shown to the right.

    \begin{eqnarray*} \int_0^\infty e^{-x}~dx & = & \lim_{b\to\infty} \int_0^b e^{-x}~dx \\ & = & \lim_{b\to\infty} -e^{-x}|_0^b \\ & = & \lim_{b\to\infty} (1-e^{-b}) \\ & = & 1 \end{eqnarray*}

    In this case, the integral converges and we may think of the area under the curve as being finite.

  2. Our next integral will be divergent.

    \begin{eqnarray*} \int_0^\infty 1~dx & = & \lim_{b\to\infty} \int_0^b 1~dx \\ & = & \lim_{b\to\infty} x|_0^b \\ & = & \lim_{b\to\infty} b \\ & = & \infty \end{eqnarray*}

    In this case, the integral diverges. This is not too surprising since we don't expect there to be a finite amount of area under the curve.

  3. We'll now work out an important class of examples which illustrate the property of convergence. Consider

    \begin{eqnarray*} \int_1^\infty \frac{1}{x^p}~dx & = & \lim_{b\to\infty} \int_1^b \frac{1}{x^p}~dx \\ & = & \lim_{b\to\infty} \frac{1}{1-p}x^{1-p}|_1^b \\ & = & \lim_{b\to\infty} \frac{1}{1-p}(b^{1-p}-1) \end{eqnarray*}

    If the exponent is positive, which is the case when $ p < 1 $ , then the integral will diverge. However, if the exponent is negative, which is the case when $ p > 1 $ , then the integral converges. This can be demonstrated below.

    The blue curve represents the graph $ y = x^p $ . When $ p > 1 $ , this function grows very rapidly which means that the graph is steep. When we consider the reciprocal $ \frac{1}{x^p} $ , it thus approaches the x axis very quickly and so the area bounded by this graph is finite.

    However, when $ p < 1 $ , the function $ y = x^p $ grows very slowly meaning that its graph is relatively flat. This means that the graph $ y = \frac{1}{x^p} $ approaches the x axis very slowly and the area bounded by the graph is infinite.

    In fact, this demonstrates the phenomenon that the convergence of the integral is determined by how fast the graph approaches the x axis---in other words, how fast the function approaches 0.

  4. We can use these examples to understand the convergence of other improper integrals. For example, if we are interested in something more complicated like:

    \[ \int_1^\infty \frac{1}{x^3+x}~dx \]

    it would be time-consuming to try and evaluate this by antidifferentiation. However, we can determine whether the integral converges or not in a simpler way. Notice that, for positive x , we have $ x > 0 $ which means that $ x^3 + x > x^3 $ . This means that $ \frac{1}{x^3+x} < \frac 1{x^3} $ . Consequently,

    \[ \int_1^\infty \frac{1}{x^3+x}~dx < \int_1^\infty \frac 1{x^3}~dx = \frac 12 \]

    because the integral $ \int_1^\infty \frac1{x^3}~dx $ was computed in the example above. This means that the integral we are interested in, $ \int_1^\infty \frac{1}{x^3+x}~dx $ , must converge as well. This can be represented in the following picture:

    We are interested in the orange area which we know is contained within the yellow area. Since the yellow area is finite, we can be sure that the orange area is as well.

  5. $ \int_1^\infty \frac{\sqrt{x^2+x}}{x^2}~dx $

    To investigate this, we might say that $ x^2 $ is much larger than $ x $ for large values of x . This means that $ x^2 + x \approx x^2 $ for large x and so the integrand

    \[ \frac{\sqrt{x^2+x}}{x^2} \approx \frac{\sqrt{x^2}}{x^2} = \frac 1x \]

    Since the integral $ \int_1^\infty \frac 1x ~dx $ does not converge, we can't expect our integral to converge either. To see this more carefully, notice that $ x^2 + x > x^2 $ which means that

    \[ \int_1^\infty \frac{\sqrt{x^2+x}}{x^2}~dx > \int_1^\infty \frac 1x =\lim_{b\to\infty} \int_1^b \frac 1x~dx = \lim_{b\to\infty} \ln b =\infty \]

    Again, we understand that the orange area, which is contained within the yellow area, is infinite. This must mean that the yellow area is infinite and so our integral diverges.

  6. Gabriel's Horn: a mathematical curiosity

    We'll consider the graph $ y = \frac 1x $ for $ x \geq 1 $ and rotate this about the x axis. This forms a surface known as Gabriel's horn.

    First, we'll compute the volume enclosed by this surface. It is given by

    \[ V = \int_1^\infty \pi \frac 1{x^2}~dx= -\pi \frac 1x|_1^\infty = \pi \]

    Here we see that the integral converges and so the volume enclosed by the horn is finite (and in fact equals $ \pi $ .)

    However, when we compute the surface area, we find that

    \begin{eqnarray*} S & = & 2\pi\int_1^\infty \frac 1x \sqrt{1+\frac 1{x^4}}~dx \\ & > & 2\pi \int_1^\infty \frac 1x = \infty \end{eqnarray*}

    Here, we have used the comparison that $ \sqrt{1 + \frac{1}{x^4}} > 1 $ to show that the surface area of the horn is infinite. This may seem like something of a paradox: the volume is finite so we could fill it up with paint. However, the surface area is infinite so we could never finishing painting the outside of the horn. How would you resolve this seeming paradox?

  7. Let's consider a radioactive substance whose half-life is 1000 years and compute the average time for an indvidual particle to decay. To do this, we'll first have to build the probability density function for the decay.

    Suppose that initially there is a mass $ M_0 $ . We know that after t years, the mass left is $ M(t) = M_0e^{-kt} $ where $ k =\frac{\ln 2}{1000} $ . This means that the amount of mass which has decayed is $ M_0 - M_0e^{-kt} $ and so the fraction of the mass $ M_0 $ which has decayed after t years is

    \[ D(t) = \frac{M_0-M_0e^{-kt}}{M_0} = 1-e^{-kt} \]

    Since this is the fraction of the mass which has decayed, $ D(t) $ represents the cumulative distribution function for this process. This means that the probability density function is $ p(t) = D^\prime(t) = ke^{-kt} $ .

    We can now find the average time for decay:

    \begin{eqnarray*} \mu & = & \int_0^\infty tp(t)~dt = k\int_0^\infty te^{-kt}~dt \\ & = & -te^{-kt}|_0^\infty + \int_0^\infty e^{-kt}~dt \\ & = & -\frac 1ke^{-kt}|_0^\infty \\ & = & \frac 1k = 1450 \mbox{ years} \end{eqnarray*}

    We have concluded that the average time for a particle to decay is 1450 years. This can be compared to the half-life, which is the median time for decay, of 1000 years.