What Are the Pitfalls of Multiplying and Dividing in Integration?

[flyingpig]

Homework Statement

I thought I understood what it means to multipyl and divide something, but lately I think it's wrong.

Mostly in Calculus, there are functions that can't be integrated unless we multiply and divide by something. Usually trig integrals need that, but even a simple integral like

\int_{0}^{1} dx

Can go wrong if we multiply and divide by the same thing. Also most of the time we aren't just multiplying and dividing by a number, we do a variable.

For instance, to the above integral, if I go ahead and multiply and divide x (x/x) then that integral becomes undoable.

Often we take granted to use this trick, but sometimes it doesn't work.

How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?

 

[Mark44]

Homework Statement

I thought I understood what it means to multipyl and divide something, but lately I think it's wrong.

Mostly in Calculus, there are functions that can't be integrated unless we multiply and divide by something. Usually trig integrals need that, but even a simple integral like

\int_{0}^{1} dx

Can go wrong if we multiply and divide by the same thing. Also most of the time we aren't just multiplying and dividing by a number, we do a variable.

For instance, to the above integral, if I go ahead and multiply and divide x (x/x) then that integral becomes undoable.

How so?
If the integrand is now x/x, that function is continuous everywhere except at 0, and is equal to 1 for all x other than zero. Both integrals (the original one and the new one) have identical values, although you need to use a limit to evaluate the new one.

Often we take granted to use this trick, but sometimes it doesn't work.

How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?

I don't see what the problem is. The only thing to be concerned about is if you multiply by 1 in the form of some expression over itself, and the denominator expression is zero for some value in the interval of integration.

 

[flyingpig]

No but that's the point, x/x changes the function completely because we can't even integrate it

 

[HallsofIvy]

No, "multiplying by x/x" doesn't change it at all:
\int\frac{x}{x}dx= \int dx= x+ C

 

[flyingpig]

No, "multiplying by x/x" doesn't change it at all:
\int\frac{x}{x}dx= \int dx= x+ C

But I referring to the definite integral

 

[Mark44]

But I referring to the definite integral

I covered this in post #2. The two integrals below have exactly the same value, 1.
\int_{0}^{1} dx
\int_{0}^{1} \frac{x}{x}dx

 

[vela]

In the strict sense, the Riemann integral doesn't exist because the integrand isn't defined at x=0, but that's part of the reason why other types of integrals were developed. So in your example, the answer would be that if you're evaluating a Riemann integral, you don't multiply by x/x because you're not allowed to divide by 0.

How does one recognize it if the function we are "integrating" creates a problem(especially if it is very very complicated)? How do we catch the mistake?

To put it simply, you have to think carefully. You have to check for and be aware of these possibilities. That's why math teachers like to show you fake proofs of non-sensical results, like 0=1. They're typically based on dividing or multiplying by zero in a non-obvious way. Your teachers want to point out that you always need to be careful to avoid that mistake.

As far as complicated integrands go, you need to learn about how functions behave to avoid running into mistakes. There's a reason why you study curve drawing and other topics which depend on analyzing a function. It's often not good enough to just learn how to turn the crank to obtain an answer.

If you do make a mistake, what happens is usually you get a result that doesn't make sense (e.g., 0=1). It won't usually be as obvious as that example, which is why you should always check your work and check your results to see if they make sense.

 

[Antiphon]

By L'Hospital's rule the function = 1 when x = 0.

 

[flyingpig]

I covered this in post #2. The two integrals below have exactly the same value, 1.
\int_{0}^{1} dx
\int_{0}^{1} \frac{x}{x}dx

How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.

 

[Char. Limit]

How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.

I believe a definite integral needs to be defined on the open interval between the two endpoints, and x/x is defined on the interval (0,1). What you're saying, which is true, is that it's not defined on the interval [0,1]. But I believe that definite integrals only require the open interval.

 

[Antiphon]

That integral is perfectly well defined even when taken from -1 to 1.

At zero, you evaluate the integrand as [d(x)/dx]/[d(x)/dx] = 1/1 = 1.

 

[JG89]

The function x/x is Riemann integrable over [0,1]. The point x = 0 doesn't create any problems.

A set E has measure if for every \epsilon > 0 there is a countable covering of E by closed intervals [a_i, b_i] such that \sum_{i = 0}^\infty (b_i - a_i) < \epsilon as b_i - a_i is the length of the interval [a_i, b_i].

A function is Riemann integrable if and only if it is bounded and its set E of points at which it is discontinuous has measure zero.

Since the only point at which the function x/x is discontinuous is at 0, then E = {0}. For any \epsilon > 0 the closed interval [-\epsilon / 4, \epsilon / 4] covers E = {0}, and the interval has length \epsilon / 2 < \epsilon. Thus E has measure zero and x/x is integrable over [0,1]

 

[Dickfore]

Using integration by parts on:
<br /> \int_{0}^{1}{\frac{x}{x} \, dx}<br />
with u = 1/x \Rightarrow du = - dx/x^{2} and dv = x \, dx \Rightarrow v = x^{2}/2, we get:
<br /> \int_{0}^{1}{\frac{x}{x} \, dx} = \left.\frac{1}{x} \, \frac{x^{2}}{2}\right|^{1}_{0} + \frac{1}{2} \, \int_{0}^{1}{\frac{x^{2}}{x^{2}} \, dx} = \frac{1}{2} + \frac{1}{2} \, \int_{0}^{1}{\frac{x^{2}}{x^{2}} \, dx}<br />
Continuing inductively:
<br /> \int_{0}^{1}{\frac{x^{n}}{x^{n}} \, dx} = \left.\frac{1}{x^{n}} \, \frac{x^{n + 1}}{n + 1}\right|^{1}_{0} + \frac{n}{n + 1} \, \int_{0}^{1}{\frac{x^{n+1}}{x^{n + 1}} \, dx} = \frac{1}{n + 1} + \frac{n}{n + 1} \, \int_{0}^{1}{\frac{x^{n+1}}{x^{n + 1}} \, dx}<br />
we get:
<br /> \int^{1}_{0}{\frac{x}{x} \, dx} = \frac{1}{2} + \frac{1}{2} \left[\frac{1}{3} + \frac{2}{3} \, \int_{0}^{1}{\frac{x^{3}}{x^{3}} \, dx}\right] = \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{3} \, \int_{0}^{1}{\frac{x^{3}}{x^{3}} \, dx}<br />
<br /> =\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{n \cdot (n + 1)} + \frac{1}{n + 1} \, \int_{0}^{1}{\frac{x^{n + 1}}{x^{n + 1}} \, dx}<br />
The integral is actually always the same. Assuming it is finite, when we take the limit n \rightarrow \infty, the integral on the r.h.s. is multiplied by a factor 1/(n + 1), so that term tends to zero and we have the following identity:
<br /> \int_{0}^{1}{dx} = \sum_{n = 1}^{\infty}{\frac{1}{n \, (n + 1)}}<br />
If you use partial fraction decomposition:
<br /> \frac{1}{n \, (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}<br />
you will see that the sum is actually:
<br /> \sum_{n = 1}^{\infty}{\frac{1}{n \, (n + 1)}} = \sum_{n = 1}^{\infty}{\left[\frac{1}{n} - \frac{1}{n + 1}\right]} = 1<br />
as is the value of the integral.

So, nothing is wrong. You just found a method for evaluating some sums.

 

[Mark44]

I covered this in post #2. The two integrals below have exactly the same value, 1.
\int_{0}^{1} dx
\int_{0}^{1} \frac{x}{x}dx

How can they both be 1 when x = 0 isn't even defined. The area of the rectangles aren't the same.

x = 0 is defined, but x/x is not defined at 0. I'm sure that's what you meant, but it wasn't what you said.

The areas of the two rectangles are exactly the same.

As I mentioned before, the second integral above is improper, so it needs to be evaluated using a limit.
\int_{0}^{1} \frac{x}{x}dx = \lim_{a \to 0^+}\int_a^1 \frac{x}{x}dx
= \lim_{a \to 0^+}\int_a^1 dx = 1

 

[flyingpig]

x = 0 is defined, but x/x is not defined at 0. I'm sure that's what you meant, but it wasn't what you said.

The areas of the two rectangles are exactly the same.

As I mentioned before, the second integral above is improper, so it needs to be evaluated using a limit.
\int_{0}^{1} \frac{x}{x}dx = \lim_{a \to 0^+}\int_a^1 \frac{x}{x}dx
= \lim_{a \to 0^+}\int_a^1 dx = 1

Okay this makes more sense to me.

But that's only when we take the limit. The two rectangles still aren't the same, I mean one of them have a hole in the corner. I know I am being really persistent and annoying with this even though the numbers are telling me I am wrong, but I just find it very difficult to accept this

 

[HallsofIvy]

The "hole in the corner" doesn't affect the area of the rectangle. Nor does it affect the limit.

 

[flyingpig]

The "hole in the corner" doesn't affect the area of the rectangle. Nor does it affect the limit.

So you are telling me that if I decide to not just poke a hole and just poke a line of holes, essentially wiping out the line,the rectangle's area is still the same? I guess what I am trying to ask is, how many holes can I poke to change the area?

 

[HallsofIvy]

How many holes? Changing the value of a function at a countable number of points does not change the value of the integral of the function.

 

[flyingpig]

So what if I decide to create another discontinutiy by multiplying it by say (x-2)/(x-2) and x/x together?

Still no change?

What if I do

\frac{x}{x} \frac{x-1}{x-1} \frac{x-2}{x-2} \frac{x-3}{x-3}...

 

[HallsofIvy]

I answered that in post 18:

Changing the value of a function at a countable number of points does not change the value of the integral of the function.

 

[flyingpig]

I answered that in post 18:

Is this count as countable?

\frac{x}{x} \frac{x-1}{x-1} \frac{x-2}{x-2} \frac{x-3}{x-3}...

What do you mean countable anyways? Do you mean alternating holes?

 

[Mentallic]

I don't find this idea to be mathematically sound, but it's just a little something you might find useful. If you think of the integral as being the area between the x-axis and the function, if the function isn't defined at a point then the "area" will be a vertical line shooting off to infinite. Lines don't have any area so you can neglect it. So really, the integral will be no different if we have the line there or not - thus, the hole in the function doesn't matter.

 

[Mark44]

A set is countable if there is a one-to-one mapping between the members of the set and the positive integers. In your example, you are adding a countable number of discontinuities to the integrand, so the value of the integral does not change.