# Some Misconceptions on Indefinite Integrals

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Integration is an incredibly useful technique taught in all calculus classes. Nevertheless, there are certain paradoxes involved with integration that are not easily solved. At least, I asked in my topology class whether anybody could resolve the paradox, and nobody found the correct answer.

The first paradox arises in the following integral:
$$\int \frac{1}{x} dx = \log |x|+C$$
This is taught in any calculus class, but it is actually wrong. Indeed, how is this usually interpreted? It is interpreted as all the functions ##f(x)## whose derivative is ##1/x## is of the form ##\log|x|## plus some constant. That this interpretation is false can be seen by simply noting that the following function is not of that form but the derivative is ##1/x## nevertheless:

$$f(x) = \left\{\begin{array}{ll} \log x & \text{if}~x>0\\ 1+\log(-x) & \text{if}~x<0\end{array}\right.$$

In fact, all the functions whose derivative is ##1/x## have the form

$$f(x) = \left\{\begin{array}{ll} \log(x) + C & \text{if}~x>0\\ \log(-x) +C’& \text{if}~x<0\end{array}\right.$$

So there are two arbitrary constants involved, not one!

What goes wrong here? Well, everything relies on the theorem ##f’=g’## implies ##f=g+C##. This follows easily from the mean-value theorem. But the mean-value theorem is proven for functions ##f:[a,b]\rightarrow \mathbb{R}## which are continuous on ##[a,b]## and differentiable on ##(a,b)##. In our example with ##1/x##, the domain is ##\mathbb{R}\setminus\{0\}##, which is not an interval and cannot be written as an interval. So the traditional formula

$$\int f'(x)dx = f(x)+C$$

is only valid if the domain is an interval. Otherwise, the structure of the integral is harder.

The second paradox involves an integration by parts. We have

$$\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx$$

Cancelling both sides yields ##1=0##. What is going on?

Now we need to go back to the very definition of an indefinite integral. An indefinite integral is not just one function, it is a set of functions. Thus

$$\int f(x)dx = \{F~\vert~F’ = f\}$$

What about our ##f(x)+C## notation then? That must be seen as an equivalence class. So we define on the set of functions the following: ##f\sim g## iff ##f’ = g’##. Then we can write the equivalence classes ##[f] = \{g~\vert~f\sim g\}##. It is now exactly the point that an indefinite integral is such an equivalence class, but we write this equivalence class (wrongly) as ##[f] = f+C##.

The operations on the equivalence classes are easy and natural:

• ##[f]+[g] = [f+g]##
• ##\alpha [f] = [\alpha f]##
• ##[\alpha] = [0]##
• ##0[f] = [0]##

with ##\alpha\in \mathbb{R}##

In our (wrong notation), things are a bit more awkward:

• ##(f+C) + (g+C) = (f+g)+C##
• ##\alpha (f+C) = (\alpha f) + C##
• ##\alpha + C = C##
• ##0(f+C) = C##

These look very weird, but solves our problem. Indeed, we know that
$$\int \frac{1}{x}dx = \log|x| + C = [\log|x|]$$

So we get that
$$[\log|x|] = [1 + \log|x|] = [1]+[\log|x|]$$

So first integration by parts in our paradox is true, but only if we interpret integration by parts as
$$\int fg’ = [fg] + \int f’ g$$
So ##fg## is an equivalence class and not a function.

Now it follows that
$$[\log|x|] – [\log|x|]= [1]+[\log|x|] – [\log|x|]$$
which is also true, and from which follows
$$[0] = [1]$$
which is true! But from this does not follow that ##0=1##.

Here is another example:
$$0 = 0\int xdx = \int 0x dx = \int 0 dx = C$$
See if you can resolve this paradox using the equivalence class formalism.

40 replies
1. I
ibkev says:

The first paradox was very cool but where did the equation come from that the second paradox starts with?

2. S
sholton says:

When we did integrals, we didn't look "backward" and consider the integral to be "interpreted as all the functions…".  Instead, it was interpreted as :the area under the curve of the specified function; in this case, the function is 1/x – in which case, the RHS is correct.

Going back the other way doesn't work, b/c the integral doesn't actually represent "all the functions" – just the specified one. And you can't take the derivative of a discontinuous function (or a "not smooth" function) as represented by your RHS?

3. N
NathanaelNolk says:

[QUOTE="sholton, post: 5550888, member: 602119"]When we did integrals, we didn't look "backward" and consider the integral to be "interpreted as all the functions…".  Instead, it was interpreted as :the area under the curve of the specified function;[/QUOTE]

That is indeed true for a definite integral, but this insight is considering indefinite integrals, which deals with the problem of finding antiderivatives.

4. I
ibkev says:

Can someone explain where the equation comes from that the second paradox starts with? The one where the integrals on both sides cancel leaving 0 = 1?

5. L
LLT71 says:

I don't get why should we treat 0=C as paradox? I mean obviously it is but if you follow some basic rules for integration you will always get the right answer 0=0

0=0*∫xdx=∫0xdx)=∫0dx=0∫dx=0(x+C)=0*x+0*C=0+0=0

0=0*∫f(x)dx=0*(F(x)+C)=0*F(x)+0*C=0+0=0

EDIT: I get it now…

let a=const. => a∫f(x)dx=a*(F(x)+C)= a*F(x) + a*C = a*F(x) + C

6. J
jbriggs444 says:

[QUOTE="LLT71, post: 5564976, member: 587786"]0*F(x)+0*C=0+C[/QUOTE]

Can you justify that step?

7. L
LLT71 says:

[QUOTE="jbriggs444, post: 5565424, member: 422467"]Can you justify that step?[/QUOTE]

i am not mathemathician but let me try my highscool version:

0*F(x)+0*C=0+C which says that in order to this equation to be true C must be zero, which is obvious. but at the same time if you let 0*C=0 equation is still valid 0=0+0.

for example

∫2*xdx=2*∫xdx=2*(x^2/2+C)=x^2+2*C

(constant times constant = constant, which I think is the paradox part)

if you differentiate x^2+2*C you will get function 2*x which is function inside the integral. same thing will happen if you let 2*C=C. no matter if you see it as n*C (n=const.) or just C it can still be treat as "any constant" cause if you differentiate any constant function C or 2*C  you will get zero.