Prove that a limit of an integral function is finite

Question

I have to prove that the following limit is finite \begin{equation} \lim_{x \to b}\int_a^x \left(\int_\xi ^x (b-y)^{-\alpha}e^{-2y} dy \right)(b-\xi)^{-(1-\alpha)}e^{2 \xi} d \xi \end{equation} I'm in the case $0<\alpha < 1$, then I know that for every $\xi \in [a,b] \subset \mathbb{R}_+, a \ne 0$ \begin{equation} \lim_{x \to b} \int_\xi ^x (b-y)^{-\alpha}e^{-2y} dy< \infty \end{equation} and \begin{equation} \lim_{x \to b} \int_a^x (b-\xi)^{-(1-\alpha)}e^{2 \xi} d \xi< \infty \end{equation} From this two fact can I conclude that \begin{equation} \lim_{x \to b}\int_a^x \left(\int_\xi ^x (b-y)^{-\alpha}e^{-2y} dy \right)(b-\xi)^{-(1-\alpha)}e^{2 \xi} d \xi < \infty \end{equation} Or what I need to conclude?

Answer 1

The result you want in the newly stated problem is true, and it's simple. Every thing in sight is positive, so your iterated integral is an increasing function of $x.$ And for any $x\in [a,b]$ this integral is bounded above by

$$\int_a^b \int_a ^b (b-y)^{-\alpha}e^{-2y} (b-\xi)^{-(1-\alpha)}e^{2 \xi} \, dy \, d \xi .$$

Because $0< \alpha < 1,$ the last integral converges; this is easy to check, because it is the product of two convergent integrals. The result follows.

Answer 2

Take for example $[0,1]$ and $$F(x)=\begin{cases} \frac1{1-x}, & x<1\\ 0, &x=1 \end{cases}$$ and $f(\xi,x)=F(\xi)$. Clearly $f$ is real and bounded for every $\xi\in[0,1]$

Take $g(\xi)=1$ the constant function, and clearly $\int_0^xg(\xi)d\xi$ is bounded for every $x\in[0,1]$.

But $$\lim_{x\to1}\int_0^xf(\xi,x)g(\xi)d\xi=\lim_{x\to1}\int_0^xF(\xi)d\xi$$ is clearly not integrable.

My guess would be to ask for continuity of $f$.