Joel Kammet

Posted on Tuesday, 04 February, 2003 - 12:10 am:

In my textbook the derivation of the solution to one form of linear equation with variable coefficients goes like this:

$dy / dt + a y = g (t)$

$e^{a t} dy / dt + a e^{a t} y = e^{a t} g (t)$

$d / dt (e^{a t} y) = e^{a t} g (t)$ so far, so good

$e^{a t} y = \int e^{a s} g (s) ds + c$

Whoa, where did $s$ come from? The text says ''Note that we have used $s$ to denote the integration variable to distinguish it from the independent variable $t$ .''

Integration variable? Can someone please explain that better?

David Loeffler

Posted on Tuesday, 04 February, 2003 - 12:14 am:

Perhaps writing the integral as

$\int_{t_{0}}^{t} e^{a s} g (s) ds$

will make it clearer.

David

Joel Kammet

Posted on Tuesday, 04 February, 2003 - 04:52 am:

Does that substitution relate specifically to using a computer math program?

Stephen Burgess

Posted on Tuesday, 04 February, 2003 - 12:51 pm:

It's not a substitution.

Basically, $\int_{t_{0}}^{t} f (s) ds$

is the same as $\int_{t_{0}}^{t} f (u) du$

is the same as $\int_{t_{0}}^{t} f (v) dv$

and so on.

(if you can't read the limits, increase your font size)

Once you've put in the limits of the integral, it's not important what letter you originally used inside the integral. The reason they have changed the letter here is because it is considered poor style and poor mathematics to use the same letter for both.

It's like the dummy $i$ in $\sum_{i = 1}^{10} a_{i}$ - you wouldn't use the same $i$ for the end of your sum as the one which you are varying.

The dummy letter, in your case $s$ , is known as the integration variable.

Joel Kammet

Posted on Wednesday, 05 February, 2003 - 08:51 pm:

Sorry, I'm still not getting the point. Taking a specific example:

$dy / dt + (1 / 2) y = 2 + t$

$e^{t / 2} dy / dt + (1 / 2) e^{t / 2} y = 2 e^{t / 2} + t e^{t / 2}$

$d / dt (e^{t / 2} y) = 2 e^{t / 2} + t e^{t / 2}$

$e^{t / 2} y = \int (2 e^{t / 2} + t e^{t / 2})$

$e^{t / 2} y = 4 e^{t / 2} + 2 t e^{t / 2} - 4 e^{t / 2} + c$

$y = 2 t + c e^{- t / 2}$

This seems pretty straightforward and I don't seem to have any problem with the mechanics of it. What I don't get is why isn't the 4th line (of the general case, in the first post) just

$e^{a t} y = \int e^{a t} g (t)$
and what is this business about the "integration variable", and distinguishing it from the independent variable. This is the first time I've come across the term "integration variable", and I can't find it mentioned in any textbook other than this particular Elementary Differential Equations textbook (Boyce/DiPrima) (and even this one doesn't explain it other than the 1-sentence comment I typed in the first post.

If anyone could tell me what is the general topic that I have to look for, or better yet point me to an online source that elaborates on this topic to some length, that would be very helpful.

Thanks.

Joel

David Loeffler

Posted on Wednesday, 05 February, 2003 - 10:05 pm:

Well, technically the right-hand side of that line as you wrote it doesn't make sense, as when you do an integral, you need to specify its limits. The upper limit of this integral is

t

, while the lower limit is some arbitrary fixed value; so we can't really have

t

appearing in the interval as well, so something else has to be used, and Boyce + DiPrima just happen to have called it

s

Most A-level textbooks will smudge this distinction by specifying an integral without any limits at all and with $t$ in the integral itself; this is technically entirely meaningless. Boyce and DiPrime are doing this properly.

Do you see what Stephen was saying about $\int_{t_{0}}^{t} f (s) ds = \int_{t_{0}}^{t} f (u) du =$ etc? It doesn't matter what you call it, it's a dummy. To return to his example of sums, both $\sum_{i = 1}^{5} f (i)$ and $\sum_{n = 1}^{5} f (n)$ just mean $f (1) + f (2) + f (3) + f (4) + f (5)$ , and $n$ (or $i$ 0 has disappeared when I write it out in full. It's just the same for integrals (not surprisingly, as integrals are defined in terms of limits of sums).

David

Joel Kammet

Posted on Thursday, 06 February, 2003 - 04:06 am:

I'm not worried about whether you call it s or u or i or whatever, I just don't understand the problem with calling it t.

Why do you say, "you need to specify its limits"? Isn't this an indefinite integral (an antiderivative) of g(t) with respect to t? Why is it necessary to specify limits? Or is it a definite integral? Is that the point I'm missing?

(Maybe now we're getting somewhere.)

Joel

David Loeffler

Posted on Thursday, 06 February, 2003 - 10:09 am:

In higher mathematics all integrals are definite. The notion of an indefinite integral is an ill-defined invention of A-level textbook writers. That is the point I am trying to make.

David

Joel Kammet

Posted on Thursday, 06 February, 2003 - 08:45 pm:

I'm still puzzling over that statement, David.

As you know, we have been taught that given a function $s (x)$ , its indefinite integral $\int s (x)$ is that family of functions $t (x)$ such that $t < quote / > (x) = s (x)$ . Why is that ill-defined? Are you just saying to call it an antiderivative, not an integral?

Or is there more to your point than that?

Do you agree that whatever you call it, it is a useful tool for solving certain kinds of problems?

Joel

David Loeffler

Posted on Friday, 07 February, 2003 - 09:39 am:

There is an ambiguity of definition here. If you define the indefinite integral that way, how do you know that, given a function

f

, there exists any function

F

such that

F < quote / > = f

That is why in higher mathematics one takes a slightly different approach. One defines the derivative, and defines the antiderivative as the family of functions whose derivatives are the given function, and shows that any two such must differ by a constant. However, at this stage one cannot prove that such antiderivatives exist at all, except where they may be found explicity.

Then one defines, entirely separately, the notion of an integral, as a limit of Riemann sums, which make sense only when the limits of integration are given.

Then, having done this, one establishes that for functions satisfying more restrictive conditions (i.e. continuity), if one defines $F (t) = \int_{t_{0}}^{t} f (s) ds$ for some fixed $t_{0}$ , then $F < quote / > (t) = f (t)$ . So this allows one to construct antiderivatives of functions by means of an integral, thus proving existence.

So the 'indefinite integral' that you refer to is what I am calling an antiderivative; but one must be careful to note that the definition of an antiderivative has nothing to do with integrals, and to define an integral one must have upper and lower limits.

I don't deny that this discussion has been entirely about notation. The integral is a stupendously powerful tool and only a fool would deny this. However, one must bear in mind how it is defined, and avoid writing down things that do not make sense given the definition.

David

Joel Kammet

Posted on Friday, 07 February, 2003 - 09:42 pm:

There is an ambiguity of definition here. If you define the indefinite integral that way, how do you know that, given a function f, there exists any function F such that F' = f?

I don't, unless I know how to find it. (Or one of you guys shows me how to find it.)

That is why in higher mathematics one takes a slightly different approach. One defines the derivative, and defines the antiderivative as the family of functions whose derivatives are the given function, and shows that any two such must differ by a constant. However, at this stage one cannot prove that such antiderivatives exist at all, except where they may be found explicitly.

Then one defines, entirely separately, the notion of an integral, as a limit of Riemann sums, which make sense only when the limits of integration are given.

That's essentially what we did in first year calculus -- not as rigorously as you would, I'm sure, but the underlying concepts are the same.

Then, having done this, one establishes that for functions satisfying certain more restrictive conditions (i.e continuity),

anything else besides continuity?

if one defines F(t) for some fixed t0, then F'(t) = f(t). So this allows one to construct antiderivatives of functions by means of an integral, thus proving existence.

I recognize this as the fundamental theorem of calculus. We were taught only that it shows that integration and differentiation are inverse processes. If I am reading you correctly, a much more significant implication is that it enables us to prove that a particular integral exists even if we can't find it explicitly. Correct?

So the 'indefinite integral' that you refer to is what I am calling an antiderivative; but one must be careful to note that the definition of an antiderivative has nothing to do with integrals, and to define an integral one must have upper and lower limits.

Well, to be fair, I think that wherever I have read about the 'indefinite integral', the author has carefully defined that term as being synonymous with 'antiderivative', but I guess that we (the 1st and 2nd year math student readers) do tend to blur the distinction between definite & indefinite integrals, and maybe lose sight of the precise definition of integral. So thanks for taking the trouble to point this out.

David Loeffler

Posted on Saturday, 08 February, 2003 - 04:09 pm:

I would only take issue with one statement in this message. 'If I am reading you correctly, a much more significant implication is that it enables us to prove that a particular integral exists'.

Substitute the word 'antiderivative' for 'integral' here. The FTC allows us to prove that antiderivatives exist, by showing that integrals constitute antiderivatives.

As for the conditions under which a function is the derivative of its integral, continuity on an open interval enclosing the point of interest should be sufficient - I confess I can't exactly remember.

David

Joel Kammet

Posted on Saturday, 08 February, 2003 - 04:43 pm:

Picky, picky.

Thanks again, David.

Joel