Applications and improper integrals

A somewhat counter-intuitive result from calculus is that even a function extending infinitely in one direction may confine an area of finite size. In light of this phenomenon, it is important not to jump to any conclusions when dealing with improper integrals; integrals involving infinities.

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    The ancient Greek philosophers, who were also the mathematicians of that time, held rigorous logical reasoning high and refused to accept a claim that could not be backed up by a structured geometrical proof.

    As a consequence, they never found a way around working with infinitesimals: infinitely small numbers, than which any real number is larger.

    The fact that there is no limit to how small a real number can be gives rise to a contradiction, and the concept was abandoned despite its useful properties.

    Since being put to use in the seventeenth century through the development of calculus, this concept has undoubtedly been a crucial part of the human technological advancements. Applications of integrals, for instance, have helped put humans on the moon, and provide us with reliable internet connection.

    Who knows where the world would have been today had not the concept been discredited until then. All this said though, the strive for logical soundness continued and calculus has by now been formalized in a manner that avoids questionable concepts and reasoning.


    As the ancient Greeks had realized, the concept of leads to all sorts of calculation problems, and it requires some alternative methods.

    is not a number, and so we cannot simply plug it in to our equations. Instead, we must analyze what happens as some variable grows larger and larger.

    A function can grow infinitely large, and still give rise to a finite area underneath it

    This is what improper integrals are all about, where either or grows without bound. Despite this ever-increasing behavior, it turns out that the integral, or the area under the curve, may still be a finite number.


    Gabriel's horn is a body formed by rotating the function about the -axis from to .

    Gabriel's horn is called a solid of revolution, since we are forming a solid body by rotating a function about an axis.

    There is a paradox with Gabriel's horn. You can fill it up with a finite volume of paint, but to paint the outside of the horn, you need an infinite amount of paint.

    We will show here that the horn has finite volume. The volume is given by:

    This integral is called improper since it has unbounded limits of integration. The way forward is to evalute the limit:

    Improper integrals

    Example: Water flow

    Imagine that we turn on a faucet and never turn it off again. The flow of water out of the faucet, measured in some unit of volume per unit time, can be given as a function.

    In this case, we let the flow be described by:

    The thing here is, after an infinite amount of time, we won't have an infinite amount of water that have come out of the faucet. And here is why.

    At a small time interval, amount of water has run out of the faucet. If we sum up all of these small amounts over the infinite time interval, we get the integral:

    If we let this faucet run forever, we would only end up with 1 unit of water!

    Improper integrals

    Recall the form of a definite integral:

    where and are the limits of the integration, and is an integrable function on the interval .

    Now is an open interval, and the definite integral will calculate the area under the curve of from to . Should we not need to include the points and ? Not necessarily, but we must be careful, because the end points may make it an improper integral.

    Improper integral

    Let be continuous on the open interval , the integral:

    is said to be an improper integral if it satisfies at least one of two conditions:

    1. , , or both.

    2. as , , or both.

    Integrals fulfilling the first condition are universally regarded as improper integrals of type 1. Likewise, those fulfilling the second one are called improper integrals of type 2.

    Convergence of improper integrals

    It might seem counter intuitive at first, since improper integrals always contain infinities, that the area calculated by this type of integrals can sometimes be a finite number. If it is, we say that the integral converges.

    On the other hand, if the area under the curve obtained by the integral goes to positive or negative infinity, we say that it diverges.

    The term convergence is not unique to improper integrals, as we will see in future chapters, but wherever it appears it is used to describe a quantity that tend to approach some value.

    In the case of improper integrals, this quantity is the area under the curve. As opposed to convergence, divergence means to grow or shrink without bound.

    Evaluating improper integrals

    Considering that we are interested in whether an improper integral approach some finite value or not, it should come as no surprise that the way we evaluate it will be through limits:

    Say the following integral is improper:

    Let's look at how we would treat it depending on what makes it improper.

    Type 1

    If , then:

    If , then:

    Type 2

    If as , then:

    If as , then:

    If the limit we equate the improper integral with exists as a finite number, we say it converges to that number. Otherwise, we say that it diverges.

    In the case that the limit tend toward , we say that the improper integral diverges to positive or negative infinity.

    The p-test for integrals

    As an alternative method to calculating limits, we can in some cases use the -test to determine whether an improper integral converges or diverges.

    Comparing integrals

    Let . Then if at all points on the interval :

    Hence, if we know that converges, must necessarily do so too. Similarly, if diverges, will diverge.

    Now it turns out that the following is true about improper integrals of powers of :


    For :

    The type 1 integral

    converges to

    if , otherwise it diverges to .

    The type 2 integral

    converges to

    if , otherwise it diverges to .

    For example, it seems reasonable when looking at the picture below that converges when integrated around , but not from some to the infinity:

    Conversely, looking at below, it could kind of make sense that integrating from some to infinity, we get a nice convergence, whereas integration around result in divergence.

    In the -test, we make use of these known results to determine whether an improper integral is convergent or divergent, given that we know the function to be greater or smaller than between the limits of integration.

    Piece-wise integration

    This section is about integrals of functions with holes, disobediently deviant points and vertical asymptotes between the integration limits. We'll see that we can integrate them all, with some reservations.

    Improper integrals, type 2

    We have seen when looking at integrability and properties of integrals that we can separate an integral into two without that it changes anything:


    Say now that has a vertical asymptote at . To calculate the integral, we then need to split it in two, and write:

    As an example, let's calculate the integral of on .

    We write:

    Now, the integral is perfectly symmetric around , so we can rewrite it as:

    and from the p-test for integrals we know that the integral converges, as the exponent . Thus, we calculate the integral and get:

    Graphs with point-wise discontinuities

    In front of you stands a function:

    How do you do to integrate on ? Somewhat surprisingly, you can actually integrate as usual without caring about the points of trouble:

    This is valid as long as there is a finite number of points on the finite limits of integration where is either undefined or has a function value which breaks out of the continuous curve.

    So let´s go on, remove as many points as you want from any continuous function, we can still integrate as if nothing happened.

    The difference between these functions and the vertical asymptote example is that there, the whole continuous curve went off towards infinity.

    Here, we only have infinitely small points going there own way. They don't actually contribute with anything to the integrals: the area under one infinitely small point is indeed very small, compared to the total area.

    Arclength parametrization

    Have a look at this funny-looking thing:

    With some values of , mapping to multiple values of , the curve is certainly not a function, and it is important that we do not treat it as such.

    So how do go about treating it instead? The answer is parametrization.

    In mathematics, an arclength is a smooth curve connecting two points. For an arclength in two dimensions, such as the one above, we need two numbers to describe the points along it. We represent a point by , which is not an interval, but coordinates.

    An arclength parametrization relates a parameter to coordinates and

    Instead of the being dependent on according to some function as we are used to, both of these will depend on some other variable, a parameter, usually denoted by .

    Now and can be expressed as functions of , called the parametric equations, independent of each other.

    The arclength parametrization will look like a system of equations, where only the points satisfying both equations will lie on the arc:

    For an arc to arise from these equations, it is necessary that and are continuous functions, so that there are no gaps in the curve.

    Example: Parametrize parabola

    Parametrize the curve that is described by the equation:

    Here, we want to express and as functions of a single paramter , that is:

    This could be done by:

    and the parametrization will look like:

    Example: Parametrize unit circle

    The unit circle is described by the equation:

    Now, we can describe the unit circle using only one parameter

    This can be done by:


    Have a look at Snakey:

    You're a biologist, so you'd like to know how long Snakey is. The issue is that Snakey is sleeping. If you pick him up and stretch him out in front of a ruler, he'll try to bite you. You better find some other way to measure his length.

    Hmmm. How about modeling Snakey's body as a curve? Then you can actually figure out his length. Ain't that clever?

    Let's say Snakey's body is expressed by the curve , where ranges from to, I dunno, . An tiny increase in will induce a change in and . Let's call these changes and . The overall change in length is given by the Pythagorean theorem:

    Now try factoring out , so that:

    If you'd like the total length, you'd add up all those :s, until you've moved from Snakey's head to tail. As we decrease , we wind up with an integral. So Snakey's length can be written as:

    What if Snakey's body corresponds to a function curve? Then we could write:

    Snakey's length turns out to be:

    where and are the relevant integral boundaries.


    A hanging bridge is suspended between two points over a river. The distance between the points is 4 meters and the hanging bridge is described by the formula:

    If we want to calculate the length of the bridge, we would have to calculate the arclength:

    Differentiating to get , our integral becomes:

    Surface- and solid of revolution

    We have a curve in the -plane, what we'll do is to grab the curve and rotate it around the or axis. It may not be revolutionary, but it's pretty. We'll see how we can calculate the surface area and the volume of the object appearing.

    Surface of revolution

    This is some function rotated around the -axis:

    Say we want to know its surface area. We'll build it up step by step. Call an infinite small element of the curve . From the lecture note on curve length, we know that:

    As the circumference of a circle is , with the radius, the surface area of the thin band appearing as we rotate around a line is:

    If the curve is rotated about the -axis, the corresponds to the function value , i.e. . Now, to get the whole surface of revolution, we need to sum all the thin bands.

    This gives the surface of revolution as we rotate the curve of around the -axis:

    If the curve is instead rotated about the -axis, the radius is just and the rest remains the same. So the surface area is:

    Note that like when dealing with curve lengths, we require that the function be continuous.

    Solids of revolution

    Solids of revolution more or less means the volume appearing as we fill in the surface of revolution.

    However, calculating a solid of revolution follows a slightly different, and, maybe surprisingly, easier method.

    We'll build up the volume from thin discs. If we want to know the solid of revolution of between and , we cut up the -axis into small segments of length , depending on . We use the index to refer to the interval .

    For the method to hold, we require to be continuous. Then, by the mean value theorem for integrals, there exists some -value in each interval , so that the volume of one disc is equals:

    Now, if we let the number of discs go to infinity and sum all the discs, we get:

    A side note on vs

    You may wonder why we don't need to bother with for solids of revolution, when we had to for the surfaces. The reason is that when dealing with the surface area, we lose too much precision if we replace by .

    Say we want to calculate the surface area of a cone made up from rotating the curve . Then, each element is . That makes every thin ring that the surface area is built from times bigger using the correct instead of , and that's a big difference.

    But, when approximating the whole volume, the difference between using and is negligible. Making the curve segments infinitely small will eliminate all difference between using or for the volume.

    Example - volume between two curves

    Find the volume between of the shape that is produced by rotating and around the axis, where:

    The first step is to define what the area of intersection will be in our problem. To do this it can be useful to note that in the interval , as we can see below.

    The next step is to define the area that each function encloses. These become and .

    We find the area of intersection to be . We can easily see this in the picture below.

    By multiplying with we get the volume elements . Next, we simply integrate :

    Thus, the volume is .

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