# An indefinite integral with no constant of integration

1. Dec 4, 2013

### yuiop

I this old thread it mentions that the indefinite integral of f'(x)/f(x) is log(|f(x)|)+C which means that there is some ambiguity about the sign of f(x). There does however, seem to be no ambiguity about the value of C as it always appears to be zero, but I have never seen this mentioned anywhere. I have only tested it using random real values, so I am not sure if it is generally applicable with complex numbers, etc.

I noticed this while carrying out the indefinite integral of an expression of the form $\frac{1}{J(r)}\frac{J(r)}{dr}$. For example consider:

$J(r) = K(r) + X$ where X is a constant. Take the derivative:

$\frac{dJ(r)}{dr} = \frac{d(K(r)+X)}{dr} = K'(r)$. Divide both sides by J(r):

$\frac{1}{J(r)}\frac{dJ(r)}{dr} = \frac{K'(r)}{K(r)+X}$. Try to recover the original expression by taking the indefinite integral:

$\int{\left(\frac{1}{J(r)}\frac{dJ(r)}{dr}\right)} dr =\int{\left(\frac{K'(r)}{K(r)+X}\right)} dr= \log(|K(r)+X|)= \log({|J(r)|})$

In other words, when the integral is taken the original constant X is recovered and the constant of integration C is not required.

I am not a mathematician so this is not intended to be a formal proof that $\int \left(f'(x)/f(x)\right) = \log {(|f(x)|)}$ rather than $\log{(|f(x)|)}+C$. I thought maybe some of you would like to check it out.

Last edited: Dec 4, 2013
2. Dec 4, 2013

### ShayanJ

A constant added to a function has no effect on the behaviour of the function in a differential equation and in that sense,it is not needed.But it does change the value of the function everywhere so in that sense it has some effect and can be useful.

3. Dec 4, 2013

### yuiop

Here is another interesting example. If we have this expression:

$f(x) = \log(-A*x)$ where A is a constant, then

$f'(x) = 1/x$.

$\int \frac{1}{x} dx = log(x)$.

It is clear that the multiplicative constant A has not been recovered so it is perhaps more accurate to say that the result should have a multiplicative constant of integration so that:

$\int \frac{1}{x} dx = log(C*|x|)$, where in this case C = -A.

If instead we find the indefinite integral of f'(x)/f(x) then the result is $\log(\log(-A*x))$ so that $\int(f'(x)/f(x)) = log(|f(x)|)$ holds true even in this case without any arbitrary multiplicative or additive constants of integration.

4. Dec 4, 2013

### JJacquelin

it doesn't mean that there is some ambiguity about the sign of f(x). It meams that it is true for f(x) and for -f(x) as well. That is : the indefinite integral of (-f(x))'/(-f(x)) is log(|-f(x)|)+C
because (-f(x))'/(-f(x)) = f'(x)/f(x) and log(|-f(x)|)+C =log(|f(x)|)+C
No, generaly C is not =0. There is a mistake in you example (C was forgotten) the correct writting is :
$\int{\left(\frac{1}{J(r)}\frac{dJ(r)}{dr}\right)} dr =\int{\left(\frac{K'(r)}{K(r)+X}\right)} dr= \log(|K(r)+X|)+C= \log({|J(r)|})+C$
May be you make a confusion between the constant C and your constant "X". Obviously :
The indefinite integral of f'(x)/f(x) is log(|f(x)|)+C
The indefinite integral of (f(x)+X)'/(f(x)+X) is log(|f(x)+X|)+C
and, because (f(x)+X)'=f'(x), it is also true that:
the indefinite integral of f'(x)/(f(x)+X) also is log(|f(x)+X|)+C
Nothing is ambiguous.

5. Dec 4, 2013

### pwsnafu

The derivative of $\log {(|f(x)|)}$ is $\frac{f'(x)}{f(x)}$.
The derivative of $\log{(|f(x)|)}+1$ is also $\frac{f'(x)}{f(x)}$.
But $\log {(|f(x)|)} \neq \log{(|f(x)|)}+1$.
Therefore you are wrong.

6. Dec 4, 2013

### yuiop

@ pwsnafu: Here is a counter example to your 'counter proof'.

Assume f(x) = x2 and x=3.

The derivative of $\log {(|x^2|)}$ is $2/x = 2/3$

and $$\frac{f'(x)}{f(x)} = \frac{2/3}{(x^2)} = \frac{2}{27}$$.

Your assertion that the derivative of $\log (|f(x)|)$ is $f'(x)/f(x)$ is shown to be false,
so the basic premise of your 'counter proof' is flawed.

7. Dec 4, 2013

### Office_Shredder

Staff Emeritus
Ummm, yuiop, isn't pwsnafu's claim exactly what you claimed you were proving in your first post though?

At any rate,
$$\frac{f'(x)}{f(x)} = \frac{2x}{x^2} = \frac{2}{x}$$
you screwed up the calculation, it really is the same as the derivative of log(x2)

8. Dec 5, 2013

### pwsnafu

:uhh:

yuiop, the first post of this thread...I'll give you the benefit of the doubt that you made a mistake and move on.

Your claim is that there is an indefinite integral such that the constant of integration is always zero. But if $F(x)$ is an antiderivative of $f(x)$ then $F(x)+1$ is also an antiderivative of $f(x)$. It doesn't matter what $f$ is, this is always true.

9. Dec 5, 2013

### yuiop

Yes, I did mess that up. That is what happens if you post while watching TV and drinking beer :P. It appears that pwsnafu's claim is correct, but there is still something still bothering me here, that I will come back to later ;)

10. Dec 5, 2013

### yuiop

I think I have got to the root of my confusion. I have been failing to discriminate between the meanings of g and g(x) and have been using them interchangeably. For example if g= (x^2 + 5), then g(x) = x^2 and they are not the same thing. Having cleared that up, what I meant to say in the OP of this thread is that the while the integral of (g'(x)/g(x)) = log(|g(x)|) + C, the integral of (g'/g) = log(|g|), without the constant of integration.

Using g = x^2+5, g(x) = x^2 and g'= g'(x) = 2x, then:

$\int \left(\frac{g'(x)}{g(x)} \right) dx =\int \left(\frac{2x}{x^2} \right) dx = \log(|x^2|) + C = log(|g(x)|) + C$

which solves to $g(x) = x^2$. In this case C cancels out. We need to introduce a second constant C2 to find g where $g = x^2 + C2$. Alternatively

$\int \left(\frac{g'}{g} \right) dx = \int \left(\frac{2x}{x^2 + 5} \right) dx = \log(|x^2+5|) = log(|g|)$

which solves to $g = x^2 + 5$ and in this case the constant of integration is not required.

Does that make more sense now?

Last edited: Dec 5, 2013
11. Dec 5, 2013

### Office_Shredder

Staff Emeritus
No, you're confused now. If g = x2+5 (which is bad notation in the first place, you should never really write it down like that), then g(x) = x2+5

All you did in your original post was integrate a function and find an antiderivative; it's just like saying the integral of x2 is x3/3.... you are still missing the constant of integration.

12. Dec 5, 2013

### Staff: Mentor

Right, they are different. g represents the function, and g(x) is the value of the function for an input x.
??
What you wrote makes no sense, so I don't believe you have overcome your confusion. Assuming from your first notation, which is not at all standard, that g maps a number x to the number x2 + 5, then g(x) = x2 + 5. You can't come back and then say that g(x) = x2.
Nope.

13. Dec 5, 2013

### JJacquelin

Why do you continue to use the same letter for two different functions ?
(x^2) is a first function. (x^2+5) is a second function. They are two different functions.
If you want to avoid a confusion, then use different symbols :
g(x)=x^2
f(x)=x^2+5
The derivative of g(x) is g'(x)=2x
The derivative of f(x) is f '(x)=2x
g(x) and f(x) have the same derivative. This is normal. You have to understand that if the dérivatives of two functions are equal, this doesn't mean that the functions are equal.
That is why the antiderivative of 2x can be x^2, or (x^2+5), or (x^2+any number). That is why the set of antiderivative of 2x is expressed on the form x^2+C, where C is any constant number.

14. Dec 5, 2013

### JJacquelin

Not at all !
You continue to make a confusion between a set of functions and a function alone.
The indefinite integral is a symbol for an infinite number of functions, which are related by an infinite number of constants. You must not forget the "C" which indicate that any fonction taken among them is likely to differ from the others because the constant C can be any number. Do not forget the "C". The correct writting is :
$\int \left(\frac{g'}{g} \right) dx = \int \left(\frac{2x}{x^2 + 5} \right) dx = \log(|x^2+5|)+C1 = log(|g|)+C2$
where C1 and C2 are any constants.

Last edited: Dec 5, 2013
15. Dec 5, 2013

### yuiop

The trouble with $\log(|x^2+5|)+C1 = log(|g|)+C2$ with $C1 \ne C2$ is that it follows that:

$\log(|x^2+5|) = log(|g|)+(C2-C1)$

$(|x^2+5|) = exp[log(|g|)+(C2-C1)]$

$(|x^2+5|) = e^{(C2-C1)} (|x^2+5|)$

The above expression is only true if $C1=C2$ so that $e^{(C2-C1)}=1$

16. Dec 5, 2013

### yuiop

OK, I accept your arguments and will abandon the notation I introduced in post #10 that distinguishes between g and g(x) and revert to g(x) = x^2+5.

I agree with what you are saying here, but it is not what I am trying to get at in this thread.

What I am really trying to establish in this thread is, if the only information I have is the expression for g'(x)/g(x), can I unambiguously determine the function g(x)?

For example if the only information I have is that $g'(x) = 2x$, then the indefinite integral is $x^2+C$ and I cannot determine the true value of $g(x)$ without some other information to establish the value of $C$.

Now if instead I am only given that $g'(x)/g(x) = 2x/(x^2+5)$ or equivalently $2/(x+5/x)$, then taking the indefinite integral:

$\int \left(\frac{g'(x)}{g(x)} \right) dx = \int \left(\frac{2x}{x^2 + 5} \right) dx$

$log(|g|) +C = \log(|x^2+5|) +C$

$g(x) = (|x^2+5|)$

Does that work? Can $g(x)$ be unambiguously determined?

Last edited: Dec 5, 2013
17. Dec 5, 2013

### economicsnerd

That is a totally precise question, and the answer is "No."

Let $g$ be some differentiable function, and fix a nonzero constant $c$. Define the function $h:=cg$, i.e. let $h$ be the function (with the same domain as $g$) that takes any $x$ in its domain to the number $h(x):=c\cdot g(x)$. This is well-defined specific function.

Now, given any $x$ with $g(x)\neq 0$ (so that the ratio makes sense to write down), one can verify that $\dfrac{h'(x)}{h(x)}=\dfrac{g'(x)}{g(x)}$.

This is true even if $c\neq 1$ (in which case $f$ and $g$ are different functions).

So the given ratio doesn't uniquely identify $g$.

18. Dec 6, 2013

### JJacquelin

The anwer is NO : $g(x)$ cannot be unambiguously determined if there is no additional condition or information.
When you write that the two integrals are equal, this means that the two infinite sets of antiderivatives are the same. But it doesn't mean that a particular function in the first set is equal to a particular function in the second set. The two "C" in your equation are not the same (but in some particular cases, a particular value C of the first integral can be equal to a particular value C of the second integral. Knowing if it is the case or not requires additional condition or information).
That is why it is better to use the notation C1 for the first integral and C2 for the second ( particular C1 can be equal to particular C2 in some cases).
Of course, people well aware of all this do not need to make the distinction and currently use a common symbol "C", implicitly meaning that all the C's might be not equal, even written in the same equation.

Last edited: Dec 6, 2013
19. Dec 6, 2013

### yuiop

.. and that is a totally precise answer. Thanks! I think I have now worked out that if F(x) is the antiderivative of f(x) and if:

$\frac{g'(x)}{g(x)} = f(x)$,

then in general:

$g(x) = K* e^{F(x)}$,

where K is an arbitrary multiplicative constant.

Thanks for you patience JJacqueline. I see that now.Thanks! People rarely explicitly show what is going on the LHS when taking the integral, but in this case I needed to know, because things are bit more involved because of the logarithimic function on the LHS. Here is what I think is going on in this case.

$\frac{g'(x)}{g(x)} = f(x)$,

$\int \left( \frac{g'(x)}{g(x)} \right) dx =\int f(x) dx$,

$\log (g(x)) + C1 = F(x) + C2$,

$\log (g(x)) = F(x) + C2 - C1 = F(x) +C$,

$g(x) = \exp( F(x) + C)$,

$g(x) = exp(C)*exp(F(x))$,

$g(x) = K* e^{F(x)}$.

Last edited: Dec 6, 2013
20. Dec 6, 2013

### JJacquelin

Fine !
All's well that's ends well.

Just a remark to end : Several times it was said that "additional conditions or information" are necessary in order to fully determine an unique antiderivative. What might be thoses conditions ? If the lower and upper limits of integration are correctly defined, then the undefinite integral becomes a definite integral and there is no longer an arbitrary constant "C".