# Valid and Invalid Substitutions in Indefinite Integration

Discussion in 'Physics & Math' started by Brandon9000, May 22, 2007.

1. ### Brandon9000Registered Senior Member

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Suppose that I am attempting to perform an indefinite integration, that all functions involved are well behaved, and that the integration involves only one variable, x, which is real. Further suppose that I attempt to make a variable substitution, as is usually done when performing integration, and that the substitution has the form:

x = f(u)

If the range of f(u) doesn't match the range of values for which the integrand is defined, will I get the wrong anti-derivative, even if I make no errors in my algebra?

3. ### ZephyrHumans are ONERegistered Senior Member

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3,371
Probably.

Simple test: differentiate your integral and see if it gives the original function.

5. ### Brandon9000Registered Senior Member

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172
I don't have a particular integral in mind. This is a theoretical question.

For example, suppose one tries to substitute x = a sin (theta) in an integrand in which x can vary from minus infinity to plus infinity.

I want to know if a substitution in which the ranges don't match, followed by perfectly correct algebra, can be expected to result in the wrong anti-derivative in indefinite integration.

7. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Hmmm. This is a good question. I spent about five minutes trying to think of an example but couldn't.

There could be some mathematical argument about functions having to have the same range, or something, or functions having the same asymptotic behavior.

I was trying to think of examples but couldn't. The best I could do is finding ever more complicated examples of the same thing, i.e.

$\int_0^{\infty} dx\left[\frac{\cos x}{x}-\frac{\sin x}{x^2}\right]e^{\frac{\sin x}{x}}$

8. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Actually I think there is something one can say about the ranges of two functions. Here's what I have so far (if I'm wrong feel free to point and laugh)...

$\int_a^b dx f(x) = F(b)-F(a)$ ---(1)

by fund thm of calculus. (Assume this holds.) Now we want to find g such that

$\int_c^d du g(u) = G(d)-G(c) = F(b)-F(a)$ ---(2).

Now the rhs of both of these is a constant, so differentiate the first one wrt x and the second wrt u.

$\int_c^d du \frac{dg}{du} = 0 = \int_a^b dx \frac{df}{dx}$ ---(3).

This is assuming that f and g are 'nice', i.e. that you can exchange the order of differentiation an integration. From the above, you can show that

$\int_a^b dx \int_c^d du \frac{dg}{du} = 0 =\int_c^d du \int_a^b dx \frac{df}{dx}$ ---(4).

Now assume you can exchange order of integration---i.e. the functions are 'nice'. Then

$\frac{dg}{du}\|_c^d = \frac{df}{dx}\|_a^b$ ---(5).

Now taking (5) and (2), you have to ask, if some (but not all) of the numbers a,b,c,d are infinite, do the relationships still hold. My hunch is no, because one of the functions, say f, has an infinite range, whereas g doesn't (as per original post).

There's probably exceptions that should be checked by hand---i.e. fundamental theorem doesn't hold, or the function has discontinuities or something...all the cases that the mathematicians point at and that the physicists sweep under the rug

Any suggestions?

Last edited: May 22, 2007
9. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Ok, I'm just not creative enough. A buddy of mine pointed out that an integral like

$I=\int_1^{\infty} \frac{1}{x^3}dx$

can be solved with

$u = \frac{1}{x}$,

and

$du = \frac{-1}{x^2}dx$.

Then the integral is
$I=\int_0^1 du u = \frac{1}{2}.$

Now check by direct integration. It gives the same answer.

(I was wrong.)

Last edited: May 23, 2007
10. ### D HSome other guyValued Senior Member

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2,257
Check again, Ben. The answer should be -1/2 (both ways yield this result).

11. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Ahh yes you're right. I flipped the limits and forgot about it.

12. ### iceauraValued Senior Member

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30,133
An indefinite integral is defined as a limit: the limit as a -> inf of F(a) - F(b), say.

The limit as a -> inf of sin(a) is not defined, in this context.

In the example of 1/x, the limits involved are defined.

So the question, in general, stands, unless I'm missing something altogether.

We fear ending up with something like lim a -> inf of cos(a) - cos(b), for an evaluation of a perfectly well defined integral, after using a substitution, yes?

13. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Hmm. Yes. Then look at equation ---(5) above---it says that the f and g have to have the same behavior at the endpoints. This isn't the case in the example you gave.

14. ### Brandon9000Registered Senior Member

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172
The standard way of doing integrals involving sqrt(a^2 - x^2) is to substitute x = a sin u. So, if I were a teacher and taught such trigonometric substitutions to my class, and then a student asked me, "can I also use x = a sin u to integrate 1/(a^2 - x^2)?" where the range of x in the integrand is clearly all reals, may I tell him, "No, that's an invalid substitution, because the range of sin u is only -1 to +1?" If I said that and nothing more, would I be giving him correct information?