As an excercise of Measure Theory, I have to calculate an expression for $\varphi(t)=\int_0^1 \frac{x^t-1}{\ln(x)} dx, t > -1$ using the Differentiation under the integral sign Theorem, which says:
Let $(X, \mathcal{A}, \mu)$ be a measure space, $g: X \rightarrow [0, +\infty)$ an integrable function, $I \subseteq \Bbb{R}$ an open interval and $f: X \times I \rightarrow \Bbb{R}$ a function such that $\forall t \in I, f_t(x):=f(x,t)$ is integrable. If $F(t):= \int_X f(x,t) d\mu$ then:
- If $\forall x \in X, f_x:=f(x,t)$ is continuous and $|f(x,t)| \leq g(x)$ on $X \times I$, $F$ is continuous.
- If $\forall x \in X, f_x$ is diffentiable on $I$ and $|f_x'(t)| \leq g(x) $ on $X \times I$, $F$ is differentiable and $F'(t) = \int_X f'_x(t) d\mu$
In order to apply it to my function $\varphi$, I am trying to search the proper $g$, but I am having many issues. I can see $g$'s that satisfies $|f(x,t)| \leq g(x)$ but that are not integrable. Is there any inequality or way to proceed in this cases which helps? I am not very good on delimit functions, so any advice in particular or in general would be appreciated.
Thanks in advance :)
The tightest bounding function $g$ (the most likely to be integrable) is $g(x)=\sup_t|f(x,t)|$, but the (double) problem is that $\lim_{t\to-1}|f(x,t)|=\frac{1-\frac1x}{\ln x}\sim_{x\to0}\frac1{x|\ln x|}$ is not integrable at $x=0^+$, and $\lim_{t\to+\infty}|f(x,t)|=\frac1{|\ln x|}$ is not integrable at $x=1^-$.
However, in order to prove the continuity of $\varphi$ on $(-1,+\infty)$, it suffices to prove it on $(a,b)$ for every $-1<a<0<b<+\infty$. You can then choose $$\forall x\in(0,1)\quad g_{a,b}(x):=\sup_{a<t<b}|f(x,t)|=\max\left(\frac{1-x^a}{\ln x},\frac{x^b-1}{\ln x}\right)$$ which is continuous and integrable, since $g_{a,b}(x)\sim_{x\to0^+}\frac{x^a}{|\ln x|}$ is integrable at $0^+$, and $\lim_{x\to1^-}g_{a,b}(x)=\max(-a,b)<\infty$.