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?

  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ Commented Mar 12, 2018 at 17:39
  • $\begingroup$ No problem :) I'll be careful next time. $\endgroup$ Commented Mar 12, 2018 at 17:40
  • $\begingroup$ $\overline{f(s)}$ is the complex conjugate? $\endgroup$ Commented Mar 12, 2018 at 17:43
  • $\begingroup$ Formally, $\int_a^b \left(\int_a^b \frac{f(t) \overline{f(s)}}{1-ts} \, ds\right) dt = \int_a^b \left(\int_a^b \sum_n s^nt^n f(t) \overline{f(s)} \, ds\right) dt = ...$. Expand, and it all comes out. Then you still have to worry about convergence. $\endgroup$ Commented Mar 12, 2018 at 17:44

2 Answers 2


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!


\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}

  • $\begingroup$ There was no justification for the first step in the developemnt in which the integral and summation are interchanged. This is perhaps the most challenging part to show. $\endgroup$
    – Mark Viola
    Commented Mar 12, 2018 at 21:34

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