Antiderivative of 1/x

1. Nov 4, 2015

Cosmophile

We are going over antiderivatives in my calculus course and reached a question regarding $f(x) = \frac {1}{x}$.

My instructor went on to say that $\int \frac {1}{x}dx = \ln |x| + C$. This makes sense to me, but only to a certain point. For $f(x) = \frac {1}{x}$, $f$ is defined $\forall x \neq 0$. So we should have two interavls which we are looking at: $x < 0$ and $x > 0$. Because of this, we would then have:

$$\int \frac {1}{x}dx = \ln |x| + C_1 \qquad x > 0$$
$$\text {and}$$
$$\int \frac {1}{x}dx = \ln (-x) + C_2 = \ln |x| + C_2 \qquad x < 0$$

The second integral comes into being because $-x > 0$ when $x<0$. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am.

Thanks!

2. Nov 4, 2015

pwsnafu

You are correct. The "constant of integration" is constant over connected components of the domain. See also Wikipedia's article on antiderivative.

3. Nov 4, 2015

fzero

It's an interesting idea, but the problem is that
$$\frac{d}{dx} \ln (-x) = - \frac{1}{x},$$
instead of $1/x$, so this isn't an antiderivative.

4. Nov 4, 2015

pwsnafu

No $\frac{d}{dx} \ln(-x) = \frac{1}{x}$

5. Nov 4, 2015

Cosmophile

If we are considering $x < 0$, $-x > 0$ so
$$\frac {d}{dx} \ln (-x) = \frac {1}{x}$$

Also, by the chain rule where $u = -x$, $\frac {du}{dx} = -1$, and

$$\frac {d}{dx} \ln(-x) = \frac {1}{-x}(-1) = \frac {1}{x}$$

6. Nov 4, 2015

micromass

You are completely correct. There are two different constants of integration. Your teacher must not be very good if he doesn't know this.

7. Nov 4, 2015

pwsnafu

Seconded. This
worries me. Mathematics is not concerned with with whether something "serves a purpose". There is plenty of mathematical research that serves no purpose other than itself.

8. Nov 4, 2015

Cosmophile

I appreciate the replies thus far. I suppose I'm having a hard time developing an argument for my case to propose to him.

9. Nov 4, 2015

PeroK

Essentially, you are correct. You can demonstrate this by taking:

$f(x) = ln|x| + 1 \ (x < 0)$ and $f(x) = ln|x| + 2 \ (x > 0)$ and checking that $f'(x) = \frac{1}{x} \ (x \ne 0)$

Usually, however, you are only dealing with one half of the function : $x < 0$ or $x > 0$. This is because an integral is defined for a function defined on an interval. For the function $\frac{1}{x}$, you can't integrate it from on, say, $[-1, 1]$ because it's not defined at $x = 0$.

So, strictly speaking, what the integration tables are saying is:

a) For the function $\frac{1}{x}$ defined on the interval $(0, +\infty)$, the antiderivative is $ln|x| + C$

b) For the function $\frac{1}{x}$ defined on the interval $(-\infty, 0)$, the antiderivative is $ln|x| + C$

In that sense, you don't need different constants of integration.

10. Nov 4, 2015

micromass

I see no reason not to define the undefined integral on more general sets.

11. Nov 4, 2015

Cosmophile

Why did you arbitrarily chose $C_1 = 1$ and $C_2 = 2$ in the first part of your response?

I understand that the antiderivative is defined only on intervals where the base function is defined, which means our integral is only defined on $(-\infty, 0) \cup (0, \infty)$. However, when I think of a constant added to a function, I simply think of a vertical shift. I understand that $f(x) = \frac {1}{x}$ has to have two independent antiderivatives as a consequence of the discontinuity, but I cannot see why it is necessary that the constants have to be different. Of course, a difference in constants is the only way that the two sides can be different, because we've already shown that, ignoring the added constant, the two sides have identical antiderivatives.

Am I making any sense? Sorry, and thanks again.

12. Nov 4, 2015

PeroK

Why not?

A couple of technical points:

1)"The" antiderivative is actually an equivalence class of functions. "An" antiderivative is one of the functions from that class. For example:

$sin(x) + C$ (where $C$ is an arbitrary constant) is the antiderivative of $cos(x)$; and $sin(x) + 6$ is an antiderivative (one particular function from the antiderivative class).

2) There is no such thing as "the" function $1/x$, as a function depends on its domain. $1/x$ defined on $(0, \infty)$, $1/x$ defined on $(-\infty, 0)$ and $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ are three different functions.

The antiderivative of $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ is $ln|x| + C_1 \ (x < 0)$; $ln|x| + C_2 \ (x > 0)$. Where $C_1$ and $C_2$ are arbitrary constants. This is the full class of functions which, when differentiated, give $1/x$.

The two separate functions $1/x$ defined on $(0, \infty)$, and $(-\infty, 0)$ both have the antiderivate $ln|x| + C$, where $C$ is an arbitrary constant, defined on the appropriate interval.

There is, therefore, a subtle difference.

13. Nov 4, 2015

HallsofIvy

Look at f(x)= ln(|x|)+ 9 for x> 0 and ln(|x|)+ 4 for x< 0. That function is differentiable for all non-zero x and its derivative is 1/x.

14. Nov 4, 2015

Cosmophile

Your second point was very well said, and is certainly something I'll carry with me. So unless a particular domain is specified, such as $(0, \infty)$, I have to include the $\ln|x| +C_1 \quad (x < 0); \quad \ln |x| + C_2 \quad (x > 0)$.

I suppose my issue is coming from the fact that, when I see $f(x) = \frac {1}{x}$, I imagine the standard hyperbola $f(x) = \frac {1}{x}$ defined on $(- \infty, 0) \cup (0, \infty)$. Now, when I imagine the function $f(x) = \ln x$, I think automatically of this graph:

http://www.wolframalpha.com/share/img?i=d41d8cd98f00b204e9800998ecf8427eq1tuvuvmvh&f=HBQTQYZYGY4TMNJQMQ2WEZTBGUYDCNJQMQ3DAODGGAZTAMBQGI4Qaaaa But I've also seen this graph:

Which should I be thinking of for this problem?

Last edited: Nov 5, 2015
15. Nov 4, 2015

Staff: Mentor

The second graph looks like it is probably f(x) = ln|x|.

16. Nov 5, 2015

Cosmophile

You're right. I wonder why Wolfram interprets it that way, but only uses the right-hand side of the graph when I set it to show real values.

17. Nov 5, 2015

Staff: Mentor

What was the equation you graphed? If you entered the integral in post #1 of this thread, the antiderivative is ln|x| (plus the constant).

18. Nov 5, 2015

Cosmophile

I graphed $f(x) = \frac {1}{x}$ and got the funky results. Graphing ln|x| gave me what I needed. But if I'm given $f(x) = \frac {1}{x}$, with no specification of the domain, I end up with $F(x) = \ln |x| + C_1, \quad (x < 0); \ln|x| + C_2, \quad (x > 0).$

What's weird is that when I look at the graph of ln|x|, the two halfs seem connected (obviously, they aren't - there's a discontinuity at $x=0$), so it's hard for me to imagine why the two sides need different constants of integration. That would mean the two bits move around independently, and I can't at all think of an example of that if $f(x) = \frac {1}{x}$ with nothing else added.

I hope that makes some sense. I'd love to get some graphical insights - I think that would help me out a bit. Sorry for asking so much on what seems to be a fairly simple topic!

(Also, thanks for chiming in, Mark44. I can always count on you and Micromass to respond).

19. Nov 5, 2015

PeroK

I suggest that you are getting yourself a bit confused over this. And, I guess this is why your maths instructor is saying it's not useful. There is nothing special about $1/x$. Any function $1/x^n$ has the same issue. Wolfram Alpha, for example, doesn't get into this subtlety of having different constants of integration. The same with tan(x) - you could have a different constant of integration on each interval upon which it's defined.

The key point is this. Suppose I have a differential equation:

$f'(x) = x^2$ and $f(1) = 0$ $(-\infty < x < \infty)$

This specifies a unique function: $f(x) = \frac{1}{3}(x^3 - 1)$

But, if we have:

$f'(x) = 1/x$ and $f(1) = 0$ $(x \ne 0)$

This does not specify a unique function. We also need a value for some $x < 0$. For example:

$f'(x) = 1/x$, $f(1) = 0$ $f(-1) = 1$ $(x \ne 0)$

The differential equation, therefore, splits into two:

For $x > 0$ we have $f(x) = ln|x| + C$, $f(1) = C = 0$, hence $f(x) = ln|x| (x > 0)$

For $x < 0$ we have $f(x) = ln|x| + C$, $f(-1) = C = 1$, hence $f(x) = ln|x| + 1 (x < 0)$

Now, of course, you could use $C_1$ and $C_2$ instead of reusing the symbol $C$ as I have done. But, it hardly matters. The important point is that the integration and differentiation for a function like $1/x$ must be done separately on each contiguous interval on which it's defined.

One final point. You can have the same situation with any function. Suppose we have:

$f'(x) = x \ (-2 < x < -1$ and $1 < x < 2)$

Then we are specifying a differential equation across two discontiguous intervals, hence we need two initial values, as there is a different constant of integration on each interval.

20. Nov 5, 2015

micromass

About the wolfram issue. Wolfram alpha is a very good software but not infallible. In particular, here it is (strictly speaking) wrong.

When asked how to integrate $1/x$, wofram alpha somehow involves complex numbers. Furthermore, it will only show one particular primitive.
So when you ask for the primitive of $1/x$, wolfram accurately calculates the following primitive: $f(x) = \text{ln}(x)$ for $x>0$ and $f(x) = \text{ln}(x) + \pi i$ for $x<0$. (The explanation for why $\pi i$ is involved requires complex analysis). Anyway, this is a correct primitive as you can differentiate it to get $1/x$. Wolfram alpha is wrong however to say that any other primitive is of the form $f(x) + C$. Rather, it is of the form $f(x) + C_1$ for $x>0$ and $f(x) + C_2$ for $x<0$. In particular, we can choose $C_1 = 0$ and $C_2 = -\pi i$, to get $\text{ln}|x|$.

Then this:

So if you want only the real values, then wolframalpha works with the same primitive $f$ that has already been computed. But this primitive is complex for $x<0$ and thus wolfram ignores it. So that is why it only shows $f(x)$ for $x>0$. There is a primitive that takes only real values (namely $\text{ln}|x|$), but wolfram does not compute it and erroneously states that the real primitive only exists for $x>0$.

21. Nov 24, 2015

Cosmophile

Hey, all. Quck update on everything:

My teacher still doesn't agree with me. Essentially, he says that $\ln x$ isn't defined for $x < 0$ so it's senseless to worry about it. I then tried contradicting him by asking why then we include the absolute value in $\ln |x| + C$ if, as he says, we don't look at $x < 0$ at all.

Anyway, I did some additional research and it all makes sense to me now. Instead of viewing $C$ as a constant value, we can instead view it to be a locally constant function. An example I found that works perfectly for showing my case is the heaviside step function. It's locally constant for $x < 0$ and also $x > 0$, but has different values for each interval.

So, when I plot $f(x) = \ln |x| + H(x)$ where $H(x)$ is the heaviside step function, I get this:

http://www5b.wolframalpha.com/Calculate/MSP/MSP4681b2fb3c00ga6b660000068g8caeiag20dba6?MSPStoreType=image/gif&s=63&w=300.&h=185.&cdf=RangeControl [Broken]

So here we have a function which has two different constants for $x < 0$ and $x > 0$ respectively. Now all I have to show is that the derivative of this function is $\frac {1}{x}$. Doing so proves that, by definition, my function is an antiderivative of $\frac {1}{x}$.

Of course, $\frac {d}{dx} H(x) = 0$ for all $x$ in the domain of $\frac {1}{x}$. Really, the derivative of the heaviside step function is a Dirac Delta function. Still, my understanding is that, because the intervals of continuity are the same, this works fine. So,

$$\frac {d}{dx} \ln |x| + H(x) = \frac {1}{x}$$

What do you all think? Is there anything I need to make more clear or be more careful about?

Last edited by a moderator: May 7, 2017
22. Nov 25, 2015

Staff: Mentor

But that isn't the point. At the start of this thread, you asked about the antiderivative(s) of 1/x; IOW, $\int \frac {dx} x$. The answer is $\ln|x| + C$, where, of course, C is an arbitrary constant.

The check is simple enough -- differentiate $\ln|x| + C$, which is $\frac{1}{|x|} \frac{d}{dx}|x| = \frac{1}{|x|} \frac{|x|} x = \frac 1 x$.
I don't think you need any of this "locally constant" machinery or Heaviside functions.

Last edited by a moderator: May 7, 2017
23. Nov 25, 2015

Cosmophile

$\ln |x| + C$ isn't the full answer when trying to find $\int \frac {dx}{x}$, though. That's the answer if the constants of integration for $x<0$ and $x>0$ are the same, but that needn't be the case, which is what I showed using the heaviside step function.

24. Nov 25, 2015

Staff: Mentor

$\ln|x| + C$ is the full answer to the question, "What is $\int \frac {dx}{x}$?"
The constant is arbitrary. If you pick any real number for C, $\frac{d}{dx}(\ln |x| + C) = \frac 1 x$. This shows that the family of functions $\ln |x| + C$ represents all of the antiderivatives of 1/x.

25. Nov 25, 2015

Cosmophile

Was there more to what you were going to say? The quote at the end threw me off.