Integrals & Discontinuities

[MarkSA]

Hello,

I'm a little confused about evaluating integrals when there is a discontinuity. In my class, my teacher talked like an integral such as this would not exist:
integral -2 to 2 of: 1/x

But then she also said something about how splitting the integral up into two parts could allow it to be evaluated, since the distance below a discontinuous point is zero?

Can anyone clarify? When does an integral not exist if there is a discontinuity? Does it only not exist when there is an infinite discontinuity? But jump discontinuities and removable discontinuities can still be evaluated by splitting the integral up? Thanks

 

[daon]

The Integral is a limit of the area of the small rectangles that make up the region below a function and above the x-axis. If it so happens a function is discontinuous at a value v then you could break up the sum into the area under the first part, plus the area under/below v, plus the area under the second part.

As \(\displaystyle dx \rightarrow 0\) The area under the point f(v) at width \(\displaystyle dx\) gets arbitrarily small (as dx approaches zero so does f(v)dx). This also true for any finite number of discontinuities, and I believe is true for even a countable amount of discontinuities.

 

[MarkSA]

I think I understand what you mean. Hmm, that will be tricky for the test coming up. I wonder if I should put "does not exist" or actually evaluate it by breaking it up.

I guess when she said the integral mentioned above did not exist, she probably meant that the integral as it was written in that form does not exist, since it could still be evaluated by breaking it up?

 

[daon]

I think I understand what you mean. Hmm, that will be tricky for the test coming up. I wonder if I should put "does not exist" or actually evaluate it by breaking it up.

I guess when she said the integral mentioned above did not exist, she probably meant that the integral as it was written in that form does not exist, since it could still be evaluated by breaking it up?

The integral you posted has no problems that I can see. If you integrate you get ln|x| which is certainly defined over any interval [a,b].

Another way to look at it... Since your function, 1/x, is an odd function, the integral over [-a,a] for any a is zero. An odd function means f(-x)=-f(x). If it is not over a symmetric interval, you could break it up as follows:

\(\displaystyle \L \int_{-5}^2 \frac{1}{x} dx = -\int_{0}^{5}\frac{1}{x}dx + \int_0^{2}\frac{1}{x}dx = -\int _{2}^5 \frac{1}{x}dx{\)

 

[MarkSA]

Thanks for the reply. Maybe I picked a bad one to use as an example?

Here are the two actual integrals from class that the teachers did on the board that they came up with 'does not exist' as an answer. I also typed the explanation given for the first one:

1) Integral from -2 to 3 of: x^(-5)dx
"Does not exist since f(x)=x^(-5) has an infinite discontinuity in [-2,3]"
and
2) Integral from 0 to 4 of: [1/([x-2]^3)]dx
"..Does not exist"

Can both of these integrals be evaluated too by splitting them up? If so, maybe the teachers just expect us to put 'does not exist' for these at this point in the course...

This kind of reminds me of limits that go to infinity.. I lost a point on our first test because I put only that a particular limit went to infinity, when I was also supposed to write that it does not exist

 

[pka]

Since your function, 1/x, is an odd function, the integral over [-a,a] for any a is zero.

That is not true.
The integral of the function 1/x over [-a,a], a>0, does not exist.
The theorem is: If the function is integrable and odd then the integral over [-a,a] for any a is zero.

 

[galactus]

Even though this may appear odd, it is actually divergent and doesn't exist.

If we define:

\(\displaystyle \L\\\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx\), then if either one of those on the right diverges, then we can say

that \(\displaystyle \L\\\int_{a}^{b}f(x)dx\) diverges.

That's because:

\(\displaystyle \L\\\int_{-2}^{0}\frac{1}{x^{5}}dx\)

\(\displaystyle \L\\\lim_{L\to\0^{-}}\int_{-2}^{L}\frac{dx}{x^{5}}\)

\(\displaystyle \L\\=\lim_{L\to\0^{-}}\left[\frac{-1}{4L^{4}}+\frac{1}{64}\right]\)

\(\displaystyle \L\\={-\infty}\)


When approaching 0 from the right:

\(\displaystyle \L\\\lim_{L\to\0^{+}}\int_{L}^{3}\frac{dx}{x^{5}}=\lim_{L\to\0^{+}}\left[\frac{-1}{324}+\frac{1}{4L^{4}}\right]={\infty}\)

\(\displaystyle \L\\{\infty}-{\infty}\) is not defined as 0.

 

[pka]

This entire thread is dancing around the ape in the room.
Why do any of you think that it is call an improper integral?
I think I can tell you: The standard theorems do not apply.
There is a whole body of literature on this subject.
If it is read, then this confusion is dispelled.

 

[MarkSA]

So if I understand all of this right...

If there is an infinite discontinuity in an integral, ie:
integral from -2 to 2 of: 1/(x^5)

And the limit from the left and right approach opposite infinities (neg and pos infinity), then it does not exist?

But if it approaches the same infinity from both left and right, it does exist and can be evaluated by breaking the integral up into two pieces?

And removable and jump discontinuities can always be evaluated by breaking the integral up into two pieces? Is this correct?

 

[pka]

So if I understand all of this right...
If there is an infinite discontinuity in an integral, ie:
integral from -2 to 2 of: 1/(x^5)
And the limit from the left and right approach opposite infinities (neg and pos infinity), then it does not exist? Is this correct?

Yes that is clearly correct.

if it approaches the same infinity from both left and right, it does exist and can be evaluated by breaking the integral up into two pieces?

No, that is not correct. Just consider the following.
\(\displaystyle \int\limits_{ - 1}^2 {\frac{{dx}}{{x^4 }}}\) does not exist. Yet the limit \(\displaystyle \lim _{x \to o} \frac{1}{{x^4 }} = \infty\).

Here is a quote from Angus Taylors’ Advanced Calculus. “In Riemann Theory (the integration found in basis analysis) the functions are assumed to be bounded, and the intervals of integration are assumed to bounded. If a function or an interval in not bounded and an integral is defined by the Riemann limiting process it is called an improper integral.”

Suppose that \(\displaystyle f(b) = 0,\,a < b < c\) and \(\displaystyle f(x) \not= 0,\,a < x \not= b < c\).
The improper integral exists if and only if both
\(\displaystyle \lim _{x \to b^ - } \int\limits_a^x {\frac{{dx}}{{f(x)}}}\) and \(\displaystyle \lim _{x \to b^ + } \int\limits_x^c {\frac{{dx}}{{f(x)}}}\) are well defined.
Then \(\displaystyle \int\limits_a^c {\frac{{dx}}{{f(x)}}} = \lim _{x \to b^ - } \int\limits_a^x {\frac{{dx}}{{f(x)}}} + \lim _{x \to b^ + } \int\limits_x^c {\frac{{dx}}{{f(x)}}}\).

Your instructor in saying that there are other means to evaluate something like
\(\displaystyle \int\limits_{ - 1}^2 {\frac{{dx}}{x}}\)
may be thinking about the Cauchy principle value which is in this case ln(2).
However, this is in no way the integral in any ordinary way not even in the usual improper integral sense.

 

[MarkSA]

OK, thanks for explaining about this.