Using Integration By Parts results in 0 = 1

Question

I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation. I start with the following equation:

$$\int \frac{1}{f} \frac{df}{dx} dx$$

I let:

$$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$

Then I find:

$$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$

I can then substitute into the usual IBP formula:

$$\int udv = uv - \int v du$$

$$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$

$$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$

Then subtracting the integral from both sides, I've now shown that:

$$0 = 1$$

Obviously there must be a problem in my derivation here... What wrong assumption have I made, or what error have I made? I'm baffled.

Answer 1

Hint: Constant of integration.

Answer 2

This line $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ should be $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[\frac{1}{f} f\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ so $$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[1\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$ and $\left[1\right]_a^b=0$

Answer 3

As an exercise, look at this proof of $0=1$

Differentiate both sides $wrt $ $x$

$0=0$

Which is true, hence proved.

If you can find the error here, so you can in your above question.

Answer 4

You have correctly derived that $0 = 1$... modulo constants.

Antiderivatives are only well-defined modulo constants*; e.g. both $x$ and $x+1$ are antiderivatives (with respect to $x$) of $1$. The equation you wrote is implicitly only meant to be an equation modulo constants; that is, the two sides of the equation don't have to be equal: they're allowed to differ by a constant.

This is traditionally worked around by adding a "constant of integration" in an ad-hoc manner rather than trying to introduce modular arithmetic. This ad-hoc fix can be tricky to get right in a nontrivial algebraic calculation if you don't fully understand what's going on, as your calculation shows.

When you cancel out the two copies of $\int\frac{1}{f} \frac{df}{dx} \, dx$, that doesn't get rid of the fact that the equation is still only meant to hold modulo constants: you've merely eliminated your mental cue (the presence of an antiderivative) to remind you of that fact.

*: Technically, I should say "locally constant functions in the integration variable" rather than "constants"

Answer 5

The problem here is that when you applied the By Parts Formula $$uv-\int \frac{du}{dx}vdv $$

you took $u=\frac 1f$ and $dv=\frac {df}{dx}dx$ Now when you use by parts formula your first task is to get $v$ and for that you need to do this $\int dv$, right? You have done all righty, but the problem comes when you write $\int df =f$ Here you also need to add the constant C, Or your proof will get pretty much messed up. In the end where you got $0=1$ , you are lacking that $C$ in this equation here, when you do antiderivatives you can't just provide one answer.If you write $\int df=f$ only then you are only getting one of the infinite solutions because $\frac {d}{df}(f)=1$ but $\frac {d}{df}(f+2)$ is also 1, infact take the derivative of $f+c$ (w.r.t f) where $C$ is any constant, you will always end up with 1.So I guess that's the error.Next time be careful with constants.

Hope It answered your question.