# Some Misconceptions on Indefinite Integrals

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:

[tex]\int \frac{1}{x} dx = \log |x|+C[/tex]

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:

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

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

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

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

[tex]\int f'(x)dx = f(x)+C[/tex]

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

[tex]\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx[/tex]

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

[tex]\int f(x)dx = \{F~\vert~F’ = f\}[/tex]

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

[tex]\int \frac{1}{x}dx = \log|x| + C = [\log|x|][/tex]

So we get that

[tex][\log|x|] = [1 + \log|x|] = [1]+[\log|x|][/tex]

So first integration by parts in our paradox is true, but only if we interpret integration by parts as

[tex]\int fg’ = [fg] + \int f’ g[/tex]

So ##fg## is an equivalence class and not a function.

Now it follows that

[tex][\log|x|] – [\log|x|]= [1]+[\log|x|] – [\log|x|][/tex]

which is also true, and from which follows

[tex][0] = [1][/tex]

which is true! But from this does not follow that ##0=1##.

Here is another example:

[tex]0 = 0\int xdx = \int 0x dx = \int 0 dx = C[/tex]

See if you can resolve this paradox using the equivalence class formalism.

Great Insight MM!

very interesting!

Great as always @micromass!

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

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?

[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.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?

I don't get why should we treat 0=C as paradox? I mean obviously

it isbut if you follow some basic rules for integration you will always get the right answer 0=00=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

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

Can you justify that step?

[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.