View Full Version : 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.
\int \frac{1}{x} dx = \int \frac{1}{x} dx
u = \frac{1}{x}
dv = dx
du = \frac{-1}{x^2} dx
v = x
\int \frac{1}{x} dx = u * v - \int v du
\int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx
\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx
\int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx
0 = 1
Pengwuino
May13-09, 11:28 PM
Right off the bat, 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
u = \frac{1}{x}
dv = dx
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
\int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx isn't valid. Integration by part goes like:
\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.
I only got to the end of the second sentence
Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as
\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 :)
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
vBulletin® v3.7.6, Copyright ©2000-2009, Jelsoft Enterprises Ltd.