Differentiate an integral (Lebesgue integral) Let $f:[0,1]\to\mathbb{R}$ be a bounded (Lebesgue) measurable function.
Consider the function $$w(p)=\int_0^1|f|^p\,d\mu$$.
Is $w(p)$ differentiable at any $0<p<\infty$? I.e. does $w'(p)$ exist for all  (not just almost all) $0<p<\infty$?
I hope this is not too easy a question. I have asked on Math.SE (few days ago with bounty), classmates, none of them know how to prove or disprove it. It is kind of a research question since it is not known whether this result is true or false.
Thanks.

My attempt: I can prove it with an additional assumption that $|f|\geq\epsilon$ for some $\epsilon>0$. But this technique does not work for the general case.
First let $E=\{x\in [0,1]: f(x)>0\}$. Then $w(p)=\int_E |f|^p\,d\mu$.
Basically I hope to apply "differentiation under the integral" Theorem 2 in http://planetmath.org/differentiationundertheintegralsign.
We check the conditions:


*

*$|f|^p$ is measurable since $f$ is, it is clearly integrable since it is bounded, and $|E|<\infty$.

*$\frac{\partial}{\partial p}|f|^p=|f|^p\ln|f|$ exists since on $E$, $|f|>0$.

*$|f|^p\ln|f|$ is also bounded (since $f$ is bounded, and $|f|\geq\epsilon$) so it is dominated by its upper bound, which is integrable over $|E|<\infty$. 
So we apply "Differentiation under the integral", $w'(p)=\int_E |f|^p\ln|f|\,d\mu$, which exists again since $|f|^p\ln|f|$ is measurable (composition of $f$ with continuous $\phi=x^p\ln x$), and also bounded.
 A: In fact, if $f$ is any measurable function on a measure space $(X,\mathcal{S},\mu)$, and wlog  $f(x)>0$ on $X$, if for real numbers $a<b$  one has $\int_X f^a d\mu<+\infty$ and $\int_X f^b d\mu<+\infty$ then $$w(p):=  \int_X f^p d\mu$$ is finite and continuous on $[a,b]$ (by Dominated Convergence, since $f^p\le f^a+f^b$),   and analytic on $]a,b[$: indeed for any $a<p<b$  and $|s|<\min\{p-a,b-p\}$ we have
$$w(p+s)=\sum_{k=0}^\infty \Big(\int_X f^p\log^k f\,  d\mu\Big)\frac{s^k}{k!}$$
as it follows immediately by standard integration by series, since one has 
$$ \sum_{k=0}^\infty \int_X\Big|\, f^p\log^k f\,   \frac{s^k}{k!} \bigg|\, d\mu\le \int_X f^a d\mu+\int_X f^b d\mu<+\infty\, .$$
(the latter follows plainly splitting the integration domain as $X=\{f\le1\}\cup\{f>1\}$).
A: Yes, we need the boundedness of $|f|^p \ln|f|$ when both $x$ and $p$ vary, but important note is that $p$ may vary on a given segment $[p_1,p_2]$, $1<p_1<p_2$ (this would imply that the derivative $d/dp(\int)$ exists at all points $p\in (p_1,p_2)$, and since the segment is arbitrary, it exists for all $p>1$.) Now this boundedness is pretty clear. 
