Calculate this integral $\int_a^b (\int_a^b \frac{f(t) \overline{f(s)}}{1-ts} \,ds) \, dt$

Question

How to calculate this kind of integrals?

$$\int_a^b \left(\int_a^b \frac{f(t) \overline{f(s)}}{1-ts} \, ds\right) dt$$

$a=0$, $0<b<1$, $t,s \in [a,b]$ are real, and $f$ "lives" in $C([a,b], \mathbb{C})$

I have to find that it's equal to $\sum_{n=0}^{+\infty} \left|\int_a^b f(t) t^n \, dt\right|^2.$

I just know that $\sum\limits_n (st)^n = \dfrac{1}{1-ts} \dots$

Could someone help me?

Answer 1

You're almost there. You have for $|a|<1$ and $|b|<1$

$$\int_a^b\int_a^b \frac{f(t)\overline{f(s)}}{1-ts}\,ds\,dt= \int_a^b\int_a^b \sum_{n=0}^\infty f(t)t^n\overline{f(s)}s^n\,ds\,dt$$

Next, note that

$$\begin{align} \lim_{N\to\infty}\int_a^b\int_a^b f(t)\overline{f(s)}\,\left(\frac{1-(ts)^{N+1}}{1-ts}\right)\,ds\,dt&=\sum_{n=0}^\infty \int_a^b\int_a^b f(t)t^n\overline{f(s)}s^n\,ds\,dt\\\\ &=\sum_{n=0}^\infty \left(\int_a^b f(t)t^n\,dt\right)\left(\overline{\int_a^b f(t)t^n\,dt}\right)\\\\ &=\sum_{n=0}^\infty \left|\int_a^b f(t)t^n\,dt\right|^2 \end{align}$$

Since $f$ is continuous, then its magnitude is bounded and the Dominated Convergence Theorem guarantees that we can pass the limit under the integral to arrive at

$$\int_a^b\int_a^b \frac{f(t)\overline{f(s)}}{1-ts}\,ds\,dt=\sum_{n=0}^\infty \left|\int_a^b f(t)t^n\,dt\right|^2$$

as was to be shown!

Answer 2

\begin{align} \int_a^b \left(\int_a^b \frac{f(t) \overline{f(s)}}{1-ts} \, ds\right) dt = {} & \int_a^b \left( \int_a^b f(t)\overline{f(s)} \, \sum_{n=0}^\infty (st)^n \right) dt \\[10pt] = {} & \sum_{n=0}^\infty \int_a^b \left( \int_a^b f(t)\overline{f(s)}(st)^n \, ds \right) dt \\[10pt] = {} & \sum_{n=0}^\infty \int_a^b\left( t^nf(t) \int_a^b s^n\, \overline{f(s)} \,ds \right) dt \\ & \text{This can be done because $f(t)$ does not change as} \\ & \text{$s$ goes from $a$ to $b,$ i.e. for present purposes, $f(t)$ is} \\ & \text{a “constant.''} \\[10pt] = {} & \sum_{n=0}^\infty\left( \int_a^b t^nf(t)\,dt \cdot \int_a^b s^n\,\overline{f(s)} \, ds \right) \\ & \text{This can be done because the integral with respect} \\ & \text{to $s$ does not change as $t$ goes from $a$ to $b$, i.e it is a} \\ & \text{“constant'' that can be pulled out of the integral} \\ & \text{with respect to $t.$} \\[10pt] = {} & \sum_{n=0}^\infty \int_a^b t^n f(t)\,dt \int_a^b \overline{s^n f(s)} \, ds \quad \text{because $s$ is real} \\[10pt] = {} & \sum_{n=0}^\infty \int_a^b t^n f(t)\,dt \cdot \overline{\int_a^b s^n f(s) \, ds} \\[10pt] = {} & \sum_{n=0}^\infty \int_a^b t^n f(t)\,dt \cdot \overline{\int_a^b t^n f(t) \, dt} \\ & \text{because $s$ is a bound variable and can be renamed} \\ & \text{in this context} \\[10pt] = {} & \sum_{n=0}^\infty \left| \int_a^b t^n f(t)\,dt \right|^2. \end{align}