TITLE: Simplifying square of integral in general
QUESTION [13 upvotes]: For a real-valued function $f=f(x)$, over the real variable $x$, with the following integral
$$ \left[ \int_{a}^{b} f(x)dx \right]^{2}, $$
is there a known general method/approach to handle this as to remove the squaring from over the integral, say by making changes to the integrand and/or interval, and then proceed with a form like $\int g(x)dx$ afterwards, were $g(x)$ is some other function?

REPLY [4 votes]: Double integrals:
$$\left(\int_a^bf(x)dx\right)^2=\int_a^b\int_a^bf(x)f(y)dxdy$$
Polar coordinates
$$\left(\int_a^bf(x)dx\right)^2=\int_a^b\int_a^bf(x)f(y)dxdy=\int\int_{\text{bounds change}}rf(r\cos(\theta))f(r\sin(\theta))drd\theta$$
In stochastic calculus, there's this How to compute expectation of square of Riemann integral of a random variable?, and there's also Ito isometry $$\int_{\Omega} [\int_0^t X(s) dW_s]^2 dP = \int_{\Omega} \int_0^t X^2(s) ds dP$$
By Cauchy–Bunyakovsky–Schwarz inequality inequality or Hölder's inequality,
$$\left(\int_a^b f\right)^2 \leqslant (b-a)\int_a^b (f)^2$$
Discrete finite analogue:
$$(\sum_{i=1}^{n}a_i)^2 = \sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j$$
Discrete infinite analogue with changing indices:
$$(\sum_{i=0}^{\infty}a_i)^2 = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ia_j = \sum_{k=0}^{\infty}\sum_{n=0}^{k}a_na_{k-n}$$
Maybe you could do the same for $\int \int$.