2
$\begingroup$

I recently saw in another post that $(\ln{f})'=\frac{f'}{f}$

where $f=f(x)$

From which it follows that $\int\frac{f'}{f}\mathrm dx=\int(\ln{f})'\mathrm dx=\ln{f}+C$

What about integrating the inverse of this?

I.e. what about $\int\frac{f}{f'}\mathrm dx=\int\frac{1}{(\ln{f})'}\mathrm dx=?$

3 Answers 3

5

\int\frac{f'}{f}dx=\int(\ln{f})'dx=\ln{f}+C is actually just a substitution :$u=f(x) \,;\, du =f'(x) dx$.

About the second question:

\int\frac{f}{f'}dx=\int\frac{1}{(\ln{f})'}dx

Suppose that such formula exists. Let $g(x)$ be any function and let $f(x)=e^{g(x)}$.

Then the above formula would yield a general formula for

\int \frac{1}{g'(x)} dx \,.

Conversely if a formula for \int \frac{1}{g'(x)} dx \,. exists then you can get your formula by defining $g(x):= \ln |f(x) | \,.$

The question you ask is equivalent to the existence of a formula for

\int \frac{1}{g'(x)} dx \,.

I highly doubt that this is true, but couldn't find the right transcendent function, I am sure someone smarter will ;)

3

We give an explicit example that completes the argument of @user9176. Note first that $\dfrac{e^x}{x}$ does not have an elementary antiderivative. For a proof, see this. But in the notation of @user9176, \frac{e^x}{x}=\frac{1}{g'(x)}, where g'(x)=xe^{-x}. Since $\int xe^{-x}\,dx=-(xe^{-x}+e^{-x})+C$, the function $g(x)$ is an elementary function, and therefore so is $f(x)$, where $f(x)=e^{g(x)}$.

  • 0
    @lhf: There was a link there, but I couldn't figure out how to make it work. Just now realized the http was missing.2011-10-30
1

I wonder if you know about change of variables for an indefinite integral. You can read about it on the linked website, but the main idea is that \int G'(f(x))f'(x)dx = G(f(x))+c.

In your case, $G(f) = \log f$ so \displaystyle{G'(f) = \frac1f} and hence \displaystyle{\frac{f'}f = G'(f)f'}.

On the other hand, there is no such a function $G$ that \displaystyle{G(f)f' = \frac{f}{f'}} for any function $f$ which is, say differentiable. The naive explanation is that if such function would exist then \displaystyle{G(f) = \frac f{f'^2}} which is impossible since the LHS depends only on $f$ while RHS depends on both $f$ and f'.