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