1
$\begingroup$

In finding this anti-derivative: $$\int\frac{1}{f(x)}\,dx,$$ with $f(x) \not \neq0$ not everywhere, i.e., there is $x_0$ such that $f(x_0)=0$;

would it be possible to prove that the anti-derivative must contain a logarithmic term, i.e., either: $$\ln(f(x)) \text{ or } \log(f(x)) \text{?}$$

I don't know if this proposition is even true, but the result of a problem that I am working on implies that this must be the case.

Thank you.

$\endgroup$
  • $\begingroup$ Do you mean f'/f instead of 1/f? $\endgroup$ – randomgirl Apr 10 '17 at 4:00
  • $\begingroup$ What if $f(x) = {1 \over x}$? $\endgroup$ – copper.hat Apr 10 '17 at 4:00
  • 1
    $\begingroup$ Let $f(x)= \sqrt x$ for $x\ge 0.$ Any antiderivative for $1/f$ on $(0,\infty)$ has the form $2\sqrt x + C.$ $\endgroup$ – zhw. Apr 10 '17 at 5:49
1
$\begingroup$

Suppose $f(x) = 1+x^2.$ Then $$ \int \frac 1 {f(x)} \, dx = \int \frac 1 {1+x^2} \, dx = \arctan x + C. $$ That is not $\log f(x).$

Note that in this context $\text{“}{\ln}\text{''}$ and $\text{“}{\log}\text{''}$ typically both mean the same thing, i.e. the logarithmic function whose base is $e\approx2.71828\ldots\,.$

$\endgroup$
  • $\begingroup$ Thank you a lot. Could you please look at my edited question in which I imposed a constraint on $f(x)$? $\endgroup$ – A Slow Learner Apr 10 '17 at 4:37
  • $\begingroup$ @Marmousi It is still false. Consider $f(x)=x^2$. $\endgroup$ – Olivier Apr 10 '17 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.