Why is this proof wrong

**protonchain** · May13-09, 11:19 PM

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.

$LaTeX Code: \\int \\frac{1}{x} dx = \\int \\frac{1}{x} dx $

$LaTeX Code: u = \\frac{1}{x} $

$LaTeX Code: du = \\frac{-1}{x^2} dx $

$LaTeX Code: \\int \\frac{1}{x} dx = u * v - \\int v du $

$LaTeX Code: \\int \\frac{1}{x} dx = \\frac{1}{x} * x - \\int x * \\frac{-1}{x^2} dx $

$LaTeX Code: \\int \\frac{1}{x} dx = 1 - \\int \\frac{-1}{x} dx $

$LaTeX Code: \\int \\frac{1}{x} dx = 1 + \\int \\frac{1}{x} dx $

**Pengwuino** · May13-09, 11:28 PM

Right off the bat, $LaTeX Code: u = \\frac{{dv}}{x} = dx$ makes no sense. The second line makes no sense either.... am i completely missing something?

**protonchain** · May13-09, 11:32 PM

Sorry those are supposed to be separate. I will edit that in.

This is what it should look like

$LaTeX Code: u = \\frac{1}{x} $

I have also added an extra step just to show that I am going to be using integration by parts to do the "proof"

**Pengwuino** · May14-09, 12:25 AM

$LaTeX Code: \\int \\frac{1}{x} dx = 1 - \\int \\frac{-1}{x} dx$ isn't valid. Integration by part goes like:

$LaTeX Code: \\int\\limits_a^b {udv} = [uv]_a^b - \\int\\limits_a^b {vdu}$

The term you think is 1 is actually 0.

**Russell Berty** · May14-09, 03:51 AM

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.

**neu** · May14-09, 05:04 AM

I only got to the end of the second sentence

**D H** · May14-09, 05:44 AM

Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as

$LaTeX Code: \\int u dv = u*v - \\int v du + C$

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.

**protonchain** · May14-09, 05:46 AM

Russell and D_H are correct, the constant term is missing. Gj guys :)

**gel** · May16-09, 04:16 PM

Originally Posted by protonchain

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.

**protonchain** · May16-09, 04:38 PM

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