2
$\begingroup$

If I have a monotonic function, say ln, can I bring it inside an integral?

in other words, is $$\ln\left[ \int f(x)\, dx\right] = \int \ln(f(x))\, dx.$$

My limits of integration don't depend on $x,$ so I think I can.

Could I move expectation inside the integral?

(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)

$\endgroup$
  • 1
    $\begingroup$ You can say, I think, that $\ln\int f\ge \int\ln f,$ since $\ln$ is a concave function. See Jensen's Inequality. $\endgroup$ – Adrian Keister Jun 13 at 21:27
  • $\begingroup$ To see this intuitively, consider the integrals as a limit of a specific sum: $\int f(x) = \lim \sum \cdots$. Your claim is then equivalent to $\ln(\lim \sum \cdots) = \lim \sum \ln(\cdots)$. Clearly, this should not hold for operations, which do not commute with summations, such as $\ln$. In contrast, if you take $g(x) := 2\cdot x$ as your function applied on the integral, you clearly have $g(\int f(x)) = \int g(f(x))$. Indeed, $g$ commutes with summation. $\endgroup$ – ComFreek Jun 14 at 7:31
6
$\begingroup$

$\ln\left(\int_0^1 x\,dx\right)=\ln\left(\frac{1}{2}\right),$ but $\int_0^1 \ln x\,dx=-1$. So no, it's not that simple.

$\endgroup$
2
$\begingroup$

No. As a counterexample, take $f(x)=e^x$.

$\endgroup$

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.