What's Flawed in This 0=1 Proof Using Basic Calculus?

[protonchain]

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.

<br /> <br /> \int \frac{1}{x} dx = \int \frac{1}{x} dx<br />

<br /> u = \frac{1}{x}<br />

<br /> dv = dx<br />

<br /> du = \frac{-1}{x^2} dx<br />

<br /> v = x<br />

<br /> \int \frac{1}{x} dx = u * v - \int v du<br />

<br /> \int \frac{1}{x} dx = \frac{1}{x} * x - \int x * \frac{-1}{x^2} dx<br />

<br /> \int \frac{1}{x} dx = 1 - \int \frac{-1}{x} dx<br />

<br /> \int \frac{1}{x} dx = 1 + \int \frac{1}{x} dx<br />

<br /> 0 = 1<br />

 

[Pengwuino]

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]

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

This is what it should look like

<br /> u = \frac{1}{x}<br />

<br /> dv = dx<br />

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]

\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]

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]

I only got to the end of the second sentence

 

[D H]

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]

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

 

[gel]

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]

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