Simplifying square of integral in general

Question

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?

Answer 1

Probably the best you can do in general is $$\left(\int_a^bf(x)dx\right)^2=\int_a^b\int_a^bf(x)f(y)dxdy$$ One place where this is useful is evaluating $\int_{-\infty}^{\infty} e^{-x^2}dx$

Answer 2

$$\begin{align}\left(\int^b_a f(x)\text{ d}x\right)^2 &=\left(\int^b_a f(x)\text{ d}x\right) \left(\int^b_a f(x)\text{ d}x\right)\\ &=\left(\int^b_a f(x)\text{ d}x\right) \left(\int^b_a f(y)\text{ d}y\right)\\ &=\int^b_a \int^b_a f(x) \cdot f(y)\text{ d}x\text{ d}y.\end{align}$$

Answer 3

Suppose you were able to find such a function g(x) to have $$\int_{a}^{b} g(x)dx =\left[ \int_{a}^{b} f(x)dx \right]^{2}$$

How can this process makes it any easier to find $$ \left[ \int_{a}^{b} f(x)dx \right]^{2}$$

You either have to integrate f(x) and square it or integrate g(x).

In either case there is only one integration involved.

The process of finding g(x) from f(x) is the extra task imposed on us if we want to integrate g(x) instead of f(x).

We may use double integrals.

With the exception of some special cases, where polar coordinates is used, the resulting double integrals are not any easier than the original integral of f(x).

Answer 4

Double integrals:

$$\left(\int_a^bf(x)\,\Bbb dx\right)^2=\int_a^b\int_a^bf(x)f(y)\,\Bbb dx\,\Bbb dy$$

Polar coordinates

$$\left(\int_a^bf(x)\,\Bbb dx\right)^2=\int_a^b\int_a^bf(x)f(y)\,\Bbb dx\,\Bbb dy=\int\int_{\text{bounds change}}rf\big(r\cos(\theta)\big)f\big(r\sin(\theta)\big)\,\Bbb dr\,\Bbb d\theta$$

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 $$\int_{\Omega} \left[\int_0^t X(s) \,\Bbb dW_s\right]^2 \,\Bbb dP = \int_{\Omega} \int_0^t X^2(s) \,\Bbb ds \,\Bbb dP$$

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:

$$\left(\sum_{i=1}^{n}a_i\right)^2 = \sum_{i=1}^{n}\sum_{j=1}^{n}a_ia_j$$

Discrete infinite analogue with changing indices:

$$\left(\sum_{i=0}^{\infty}a_i\right)^2 = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ia_j = \sum_{k=0}^{\infty}\sum_{n=0}^{k}a_na_{k-n}$$

Maybe you could do the same for $\int \int$.