Ryan Fox’s

Investigation #1:

Technology and the Improper Integral

 

 

 

During my first year teaching Advanced Placement Calculus BC, I was going over examples concerning Improper Integrals.  At some point, the class came upon a certain problem along the lines of

After working towards the solution of the problem, we got an answer of 1/9.  One of my most insightful students thought she was onto an idea by suggesting that we could find the answer to the problem simply by looking at the exponent of the denominator.  Believing that I could prove my student wrong, the class went on a block-long exploration that ended up being the greatest lesson I never planned.

 

First Steps: Evaluate Definite Integrals in the form  where p is an integer greater than 1. 

 

            Let’s start by looking at one particular example: . I can use the Graphing Calculator program to look at least some portion of the graph.

 

 

With a little analytical investigation, we can say that the red horizontal line, the x-axis, and the graph of the function colored in purple bound the desired region. 

So now that we know what we are looking at, the big question that the definite integral asks us is “What is the area of this region?”  The version of Graphing Calculator that I am working with does not have the ability to evaluate definite integrals.  At this point I will use the Maple program to evaluate the definite integral.  In fact, the Maple output is given below:

> evalf(Int(1/(x^5),x=1..infinity));

                          0.2500000000

Based upon our decimal answer, we would be tempted to say that the solution to the improper integral is ¼.  This is the point where my students made their first conjecture: the solution to an improper integral in this form will be . 

            As the leader in this wild ride, I pointed out to the students that just because we were correct one instance on our conjecture does not mean we are closing to proving.  More examples would give us a better indication if we are right or not, but all of our examples will not be able to handle the symbolic manipulation needed to prove our conjecture.

which contains the region represented in the following graph

and our Maple output gives us the following:

> evalf(Int(1/(x^10),x=1..infinity));

                          0.1111111111

            Using a calculator we could see that the decimal approximation for 1/9 is indeed .111111 (which would be .1 repeating). 

 

 

            If we take advantage of the Graphing Calculator software program, we can see the graphs of the various rational expressions. By following the link to this particular Graphing Calculator software, we can see various graphs based upon the various exponents in . 

            Maple’s ability to carry out the calculations necessary to validate our suspicions begins to falter when the exponent is greater than 10,000.

 

            However, what should be noted at this point, is that our conjecture for our evaluation of the improper integral is . 

 

Next Steps: Investigate a Lower Limit of Integration other than 1

 

            One of the ways we have been able to verify our solutions quickly is that the answers from Maple are in a somewhat recognizable form: the answers are reciprocals of the natural numbers.  However, what happens if we were to move the lower limit away from 1 and to some other value, whether it be greater than 1 or less than 1. 

 

            First, let us look at what happens when our lower limit of integration is increased.  This will be represented graphically by moving our red line from the above Graphing Calculator output to the right.  An example of such a situation would be to evaluate the improper integral .  If we look at the output from the Graphing Calculator program and make several adjustments (zooming in and moving the graph rightward), we would have the following picture.

From this graph, it is quite possible that the resulting evaluation of our improper integral will give us a very small value.  For example, according to this picture, the functional value at x = 2 is less than 0.01.  We could overestimate the area under the curve by creating rectangles of width 1 and choosing the largest functional value from any point within that one unit interval.  For [2, 3], we could make an even greater overestimation by choosing 0.01 as the largest function value.  This would tell us that the area of this rectangle is 0.01, and the largest possible rectangle we could draw in this region and maintain some sort of a plausible estimate.

 

            Instead of focusing on the estimate, let us begin the process of using other programs to pinpoint a potential exact value of the improper integral.   Our earlier conjecture says that any improper integral in our desired form with a lower limit of integration of 1 would have a solution in the form 1/p, where p is our exponent in the denominator.  Clearly, this conjecture would not hold if the lower limit of integration were greater than 1, since some of the area of the corresponding region has been removed.  The question then becomes how can we determine this new area.  One possible approach would be to take our answer from previous work and subtract out the area from 1 to the new lower limit of integration.  If we followed through with this approach, we would have the following solution.

While this is a perfectly reasonable method to find the value of the improper integral, we might hope to find the solution using just the lower limit of integration and the value of the exponent in the denominator.  If we employ Maple to evaluate integral we have now, we will get the following result:

> evalf(Int(1/(x^10),x=2..infinity));

                        0.0002170138889

The important question is there a fraction that can be written in terms of the lower limit of integration and the value of the exponent in the denominator that would give the decimal approximation as the evaluation of the improper integral.  Let us test a couple of conjectures.  The first conjecture would be to see if has the same decimal approximation as 1/(2^9), where we have taken the lower limit of integration as raised to the value of one less than the value of the exponent in the denominator, modeling our conjecture from the previous part.  If we calculate the decimal approximation of this fraction, we would get the following Maple output:

> 1/(2.^9);

                         0.001953125000

Clearly this approximation does not match the approximation from the improper integral.  We will reject this conjecture, but will put forth a new conjecture: .  If we calculate the decimal approximate of this fraction, we would get the following Maple output:

> 1/(9.*(2^9));

                        0.0002170138889

Since this value and the decimal approximation agree to 13 decimal places, we might be content to consider this a valid conjecture.  While we may not be able to confirm this conjecture absolutely, we might further illustrate that our conjecture work for other situations.  If we move the lower limit of integration to the value c, where c > 1, we have made the following conjecture concerning our improper integral: .

 

            Another example would be to evaluate the improper integral . 

Based on our conjecture, we would guess that the value of the improper integral would be equal to .  According to the Maple output, the decimal approximation for this fraction would be

> 1./16384;

                        0.00006103515625

If we used Maple to evaluate the improper integral, we would get the following decimal approximation:

> evalf(Int(1/(x^5),x=8..infinity));

                        0.00006103515625

Once again we have approximations that are equivalent to 13 decimal places.  We could repeat the process as often as we needed to convince ourselves that we might have a valid conjecture for natural numbers.

            Let us now try to test our conjecture on a rational number that is not a natural number for the lower limit of integration.  We will evaluate the improper integral . 

From our conjecture, we will estimate the value of this expression to be .  If we calculate the value of the improper integral using the Maple software program, we would have the following output:

> evalf(Int(1/(x^2),x=2.5..infinity));

                          0.4000000000

Since our values once again match to 13 decimal places, we are once again satisfied by our results. 

 

            For one last attempt, we will investigate a real number that is not a rational number for our lower limit of integration.  We will evaluate the improper integral . 

From our conjecture, we will have the following our conjectured solution: .  Using the Maple output, the decimal approximation for our conjectured solution is

> 1./48;

                         0.02083333333

Now we will compare this approximation to the evaluation of the Maple output for the improper integral.

> evalf(Int(1/(x^7),x=sqrt(2)..infinity));

                         0.02083333333

Once again, we have our decimal approximations being equivalent for the first 13 decimal places.  While it does not constitute proof, we can be encouraged by the thought that our conjecture is yielding desired results. 

 

            Now we want to see what would happen if we were to move the lower limit of integration to the left of 1, in other words, decrease the value of the lower limit of integration.  The first place to test our conjecture would be to let the lower limit of integration equal 0.  Let us evaluate the improper integral .  Let us first look at the Graphing Calculator output for this program to determine the region for which we are trying to find the area. 

In the Graphing Calculator output, it is hard to see the graph of the line that represents the lower limit of integration, x = 0.  However, we do know that this equation gives us the y-axis.  What is interesting to note about this graph is that it does look like that the y-axis serves as a vertical asymptote.  If indeed the y-axis is an asymptote, then we would not be able to determine the value of the improper integral.  According to our Maple output, we would have the following value of the improper integral

> evalf(Int(1/(x^3),x=0..infinity));

                        Float(infinity)

We are given an answer that does not have much meaning associated with it.  No longer are we getting decimal approximations, but rather a somewhat cryptic statement regarding infinity.  Let us now look at our conjecture and see what possible solution we might receive:

we might end up with a statement that does not make much sense either: dividing by 0.  From this, we have learned that we must clarify our conjecture.  We know that negative numbers and zero would not work with this conjecture.  Moving forward, the only numbers we want to check are positive numbers only. 

            Let us now check a positive number in the interval (0, 1) for our lower limit of integration.  We will investigate a rational number and an irrational number in this interval to get an understanding if our conjecture might still hold.  Let us first look at the expression .

Based on our conjecture, we will have the conjectured solution of .  If we evaluate the improper integral using Maple, we will get the following result:

> evalf(Int(1/(x^6),x=(1/2)..infinity));

6.400000000

Once again, we have an instance for which the decimal approximation of the improper conjecture satisfies the conjectured solution to 13 places.  While it does not confirm results, this method does indeed add evidence to our claim.

            We could also try an irrational number between 0 and 1 to see if our conjecture stills holds. Let us look at the improper integral .

 According to our conjecture, we would expect the following value: , which would give us the following decimal approximation in Maple:

> 1./(7*(sqrt(3./4))^7);

                          0.3910097061

Now let us compare that approximation to the decimal approximation that Maple gives for the improper integral:

> evalf(Int(1/(x^8),x=sqrt(3/4)..infinity));

                          0.3910097061

Once again, we have our conjecture matching the Maple output to 13 decimal places, meaning we have yet to dispel our conjecture. 

 

             So far we have only investigated natural number values of the exponents in the denominator.  For our next set of investigations, we are going to break free from this restriction and see what results may come from there. 

 

Last Steps: Investigate Non-Natural Number Values for the Exponent in the Denominator

 

            Every investigation we have examined to this point has always involved a natural number exponent in the denominator of our original expression. We will investigate the extension of our conjecture by looking at other values for our exponent.  We will continue the investigation in the same manner we extended our investigation for the lower limit of integration: investigate a value of the exponent less than 1, rational values of the exponent greater than 1, and irrational values of the exponent greater than 1.  

            First let us look into the value of the exponents less than 1.  We will do that by evaluating the improper integral . 

Determining our conjectured solution, we would get , which provides an interesting situation.  From our graph, we see that the region lies above the x-axis, implying that the value of any integral, improper or not, should be positive.  However, the conjectured value is negative.  We have reached another situation for which we must refine our conjecture.  Our conjecture will not work for positive values of the exponent less than 1.  It might be worth our while to confirm that our conjecture would not work for negative exponent values.  Let us look at the improper integral .  If we examine the graph of y = x bounded by the x-axis and the line x = 1, we can get the following Graphing Calculator output:

We notice that the functional values for y = x are increasing without bound as the values of x increase without bound.  This would suggest that we are finding an area of a region that seemingly increases without bound.  If so, we would find it difficult—if not impossible—to associate the area of this region with some finite value.  As such, we will not use our conjecture on expressions in the form , where p < 1. 

            Now let us focus our attention back to the values of the exponent that are greater than 1.  This time we will focus on rational numbers.  While it is not possible to prove our conjecture for all rational numbers, we will try to assure ourselves that the conjecture will not fail for randomly selected rational numbers.  Let us look at the improper integral .

Since the lower limit of integration is 1, we can use our original version to state that our conjectured solution will be .  If we use Maple to evaluate the improper integral , we get the following output:

> evalf(Int(1/(x^2.5),x=1..infinity));

                          0.6666666667

Since the Maple output has rounded the repeating decimal to .6666666666667, we will say that our conjecture has not failed for this particular instance.  We can move our lower limit of integration to a value greater than 1 and investigate what occurs.  Let us look at the improper integral

Our conjecture solution for this particular expression would be .  We would need Maple to compute the decimal approximation for this particular value. 

> 1/((4./3)*3^(4./3));

                          0.1733403187

When we use Maple to evaluate the improper integral , we would get the following output:

> evalf(Int(1/(x^(7/3)),x=3..infinity));

                          0.1733403186

This is the first instance in our work where we have not received an exact match for all 13 decimal places in each solution.  It is my belief that since we are working with the constant multiple of the reciprocal of an irrational number, it is possible that there are some differences in rounding.  As a result, to have the two results differ by one hundred-trillionth will be acceptable in this instance. 

 

            Now let us look at an irrational number for the exponent.  Let us look at the integral .  The region were are investigating can be seen in the following Graphing Calculator picture:

We will conjecture that our solution will be .  We will use Maple to find the decimal approximation of this particular expression. 

> 1/((sqrt(5.)-1)*(1^(sqrt(5.)-1)));

 

                          0.8090169947

If we use Maple to find the value of the decimal approximation of the improper integral , we will get the following:

 

> evalf(Int(1/x^(sqrt(5)),x=1..infinity));

                          0.8090169944

Once again, because we are working with reciprocals of irrational numbers, we will be more forgiving in comparing our decimal approximations.  We will say that since the difference of the two approximations is only three hundred-trillionths, that we have not failed in our conjecture. 

            The last situation we will investigate is the irrational exponent and a lower limit of integration greater than 1.  For this, we will evaluate the improper integral , which has a region that can be found in the following Graphing Calculator output

 

Our conjectured solution will be , which has the following decimal approximation according to Maple:

> 1/(((10.)^(1/3)-1)*(2^((10.)^(1/3)-1)));

                          0.3891447375

If we evaluate the improper integral , we get the following decimal approximation:

> evalf(Int(1/(x^(10^(1/3))),x=2..infinity));

                          0.3891447375

Since we were able to get our approximations to 13 decimal places, we will be satisfied that our conjecture provides a reasonable determination of our improper integral evaluation. 

 

Future Steps: Can a Computer Perform the Necessary Symbolic Manipulations?

 

            All of the work we have done so far has not made a proof.  We have shown in a few isolated instances for a handful of cases, we were able to get our conjecture to work.  However, the question remains can we use software like Maple to show us that our conjecture holds for all values (which have been modified throughout our investigations) in all cases (which, likewise, have been modified throughout our investigations).  In this fateful class session two years ago, we believe we had worked out proof using our knowledge of calculus.  We were hoping to find verification elsewhere.  It was my intention to see if technology could provide that verification I was looking for.  While it did confirm a select number of situations, we are missing the proof that guarantees our conjecture is true.  For the moment, I will say that the technology has somewhat cemented my beliefs, but that as technology improvements, it might be possible that the little conjecture might take hold.