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

Question

Cómo calcular este tipo de integrales?

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

$a=0$, $0<b<1$, $t,s \in [a,b]$ son reales, y $f$ "vive" en $C([a,b], \mathbb{C})$

Tengo que encontrar que es igual a $\sum_{n=0}^{+\infty} \left|\int_a^b f(t) t^n \, dt\right|^2.$

Yo sólo sé que $\sum\limits_n (st)^n = \dfrac{1}{1-ts} \dots$

Podría alguien ayudarme?

Answer 1

Usted está casi allí. Tiene por $|a|<1$ $|b|<1$

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

Siguiente, tenga en cuenta que

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

Desde $f$ es continua, entonces su magnitud es limitado y el Teorema de Convergencia Dominada garantiza que se puede pasar el límite bajo de la integral para llegar a

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

como iba a ser mostrado!

Answer 2

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