Why is this proof wrong

  • #1

Main Question or Discussion Point

Another one of those 0 = 1 proofs.

If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned.

Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over).

Note, you need to know basic calculus.

[tex]

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

[tex]
u = \frac{1}{x}
[/tex]

[tex]
dv = dx
[/tex]

[tex]
du = \frac{-1}{x^2} dx
[/tex]

[tex]
v = x
[/tex]

[tex]
\int \frac{1}{x} dx = u * v - \int v du
[/tex]

[tex]
\int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx
[/tex]

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

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

[tex]
0 = 1
[/tex]
 
Last edited:

Answers and Replies

  • #2
Pengwuino
Gold Member
4,989
15
Right off the bat, [tex]u = \frac{{dv}}{x} = dx[/tex] makes no sense. The second line makes no sense either.... am i completely missing something?
 
  • #3
Sorry those are supposed to be separate. I will edit that in.

This is what it should look like

[tex]
u = \frac{1}{x}
[/tex]

[tex]
dv = dx
[/tex]

I have also added an extra step just to show that I am going to be using integration by parts to do the "proof"
 
  • #4
Pengwuino
Gold Member
4,989
15
[tex]\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx[/tex] isn't valid. Integration by part goes like:

[tex]\int\limits_a^b {udv} = [uv]_a^b - \int\limits_a^b {vdu} [/tex]

The term you think is 1 is actually 0.
 
  • #5
We are dealing with antiderivatives. So, given F(x) and G(x) that are antiderivatives of 1/x, it is true that F(x) = G(x) + C, for some constant C. For example, the second to last line you have can be written as

ln(x) + C = 1 + ln(x) + D for some constants C, and D. We do not then conclude that 0 = 1.

The integral of 1/x dx is a FAMILY of functions that all differ by a constant.
 
  • #6
neu
220
1
I only got to the end of the second sentence
 
  • #7
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as

[tex]\int u dv = u*v - \int v du + C[/tex]

We drop the arbitrary constant because it is implied by the very use of indefinite integrals, or antiderivatives. The inverse of the derivative is not unique.
 
  • #8
Russell and D_H are correct, the constant term is missing. Gj guys :)
 
  • #9
gel
533
5
Russell and D_H are correct, the constant term is missing. Gj guys :)
and Pengwuino too. You either put in constants of integration, or use definite integrals. Either way resolves the error.
 
  • #10
Right, but since I defined the problem in the start as indefinite integrals, I was looking for the answer that was related to indefinite integrals. Anyways. It's just a simple problem
 

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