integration formulae

[MrAl]

Hi,

I believe it's a common mistake. If you think about it just briefly, how can a function that has discontinuities be continuous?
And if we do allow that, then how do we explain a function that does NOT have those 'finite' discontinuities?

The simplest definition i ever heard of was that if you could draw the function with a pencil on paper without picking up the pencil even once then it is continuous.

What definition did you want to use here exactly? Maybe it is ok in some contexts?

I would have to wonder how we would integrate some of those functions i guess...
y=1, -10<x<0
y=1, 0<x<+10
y=2, x=0

 

[steveB]

What definition did you want to use here exactly? 0

I'm using the definition that all of my books use, but I want to use the definition that PGs book uses. A statement of the type PG referenced only has meaning if the term "piecewise continuous" has a precise definition.

So I'll await PGs scan of the definition from the book that he is quoting. Otherwise we will just be arguing preferred definitions, which isn't helpful to PG.

 

[MrAl]

Hello again,

Ok sure. What books did you see it in and how exactly does it state the definition of a piecewise continuous function?

I found definitions suggesting both ways on the web!

But here's something to think about...

What is the difference between a piecewise function and a piecewise continuous function?
Also, what condition would make a piecewise function discontinuous? Or stated another way, what could we do to a piecewise continuous function to make it discontinuous?

 

[PG1995]

Hi

The scan I used was from this formulae book. That line just crossed my eyes and I wanted to know the reason why the author was specifically using that term.

The topic being discussed could be found in many of the books I have but unfortunately right now I can't really remember the exact books and page numbers. But one book I'm using for signals and systems says **broken link removed**.

Moreover, I can recall one thing that some of the books I have seen define continuous and discontinuity differently. So, I'm not surprised to see both of you arguing about this. For this current debate I would say, "each to his own". Thank you.


Best regards
PG

 

[MrAl]

Hi again,


I've also seen the phrase "piecewise smooth" function.
I can see Steve's point too, where we can probably find the area under most functions that are piecewise defined where the two endpoint limits of the 'joining' sections just have to be finite, rather than equal. This would mean that functions that are split up in time with gaps would also be called piecewise continuous. But there has to be a reason for using the qualifier "continuous" rather than just "piecewise" alone i would think.

My take was that they were trying to say that the way they were performing the task would only work with certain functions but not others. Why would they have to state piecewise continuous if anything just plain old piecewise would work. They dont have a need to clarify that it is a function so that cant be it either.

You cant find anything written in the text that talks about piecewise functions?

 

[steveB]

I've attached extracts from two different books. The first two pages are from "Complex Variables and the Laplace Transform for Engineers", by Wilbur R. LePage (I highly recommend this book BTW). And, the third page is from "Calculus of Variations", by Robert Weinstock. I have other books that conform to this definition also, but this is the definition I've always seen and used.

To me the term "continuous" automatically includes our normal well behaved functions, but also includes functions with kinks (first derivative has jump discontinuities). However, it does not include jump discontinuities. This is the reason for the term "Piecewise Continuous". It allows a class of functions that are continuous in a finite number of regions, with jumps at the boundaries. This becomes important for evaluating Laplace transforms of many discontinuous function we deal with in EE and physics.

In the absence of a direct quote from PGs book, I would recommend using this usual definition because, to the best of my knowledge, it is the definition used by mathematicians generally.

 

[MrAl]

Hello again,

Yes that makes sense, but how it is used in a given purpose i think is just adding a bias (note the first reference uses Dirichlet as well, which seems to imply that they are seeking results that are not necessarily coming from pure math).

But the second reference is interesting in that it appears to come from a math book on Calculus. They clearly state that it is ok to call a function a piecewise continuous function if it has finite jumps as long as the two limits of the joints are both finite. That does make sense too.
I think this definition might be more intuitive also.

They only question at least for me then is how do they expect to integrate a function using a single value of delta x over an interval that may not even include a given point. But perhaps i am being a little too nit picky here.


But yes now i have to agree that without any further definition from PG's actual texts that your stated definition is the better one

 

[steveB]

They only question at least for me then is how do they expect to integrate a function using a single value of delta x over an interval that may not even include a given point. But perhaps i am being a little too nit picky here.

I don't think this is too nit picky from a mathematicians point of view. I think physicists/engineers intuitively have no problem with this since, in the limit, the delta x gets infinitesimally small. Also, intuitively, we can just break down the integral into a summation of integrals over the continuous sections, which results in different delta x for each region. But, these are examples of the kinds of thinking that make the mathematicians throw spears at us.

 

[PG1995]

Hi

Re: Post #11

If you leave it inside the integral and then realize that x=sqrt(u-1), you would have a new integral to solve.

I just did it and found that the relation x=sqrt(u-1) isn't correct, at least as far as I can confirm it.

Kindly let me know if I still have it wrong or you just made a slip. And I'm sorry for asking you this now even though that problem is months old. Thanks a lot.



For my own reference:
This is what MrAl was saying in the post #6.

Regards
PG

 

[steveB]

Kindly let me know if I still have it wrong or you just made a slip.

It looks like what you wrote is correct, and it seems i made a slip of the pen. If you define u=2(x^2+1), then x=±sqrt(u/2-1). I believe that I was writing u=(x^2+1) when I sketched it out myself, and then I somehow applied my choice into your discussion.

 

[PG1995]

Hi

Could you please help me with these queries? Q1 is about the formula 16.5 and Q2 is about the 16.6. I think if could give me a simple example for each formula then I would understand it. Thank you.

Regards
PG

 

[steveB]

I think Q1 you must have used many times, but maybe you just aren't recognizing it. Try this example for that one.

f(u)=exp(u), with u=ax and du=a dx

[latex] \int e^{ax} dx= \frac{1}{a} \int a c^{ax} dx =\frac{1}{a} \int c^{u} du [/latex]

For Q2, try F(u)=exp(u), with u=sin(x), and du=cos(x) dx

[latex]\int e^{\sin{x}} dx=\int \frac{e^u}{\cos{x}} du [/latex]