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?

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    $\begingroup$ The answer to your question is "probably not". If you have some particular application in mind where you know more about $f$ then maybe. $\endgroup$ – Ethan Bolker Mar 16 '18 at 0:04
  • $\begingroup$ In stochastic calculus, we can do this $$\int [\int_0^t f(s) dW_s]^2 dP = \int \int_0^t f^2(s) ds dP$$ $\endgroup$ – BCLC Mar 16 '18 at 6:47

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 $\int_{-\infty}^{\infty} e^{-x^2}dx$

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    $\begingroup$ Good example to show how polar coordinates come to rescue. $\endgroup$ – Mohammad Riazi-Kermani Mar 16 '18 at 1:16
  • $\begingroup$ @MohammadRiazi-Kermani: what polar coordinates? $\endgroup$ – tomasz Mar 16 '18 at 10:53
  • $\begingroup$ @tomasz Using a trick attributed to poisson, converting the double integral to polar coordinates are how that conversion helps you evaluate that improper integral. Unfortunately, that trick essentially only helps for this integral. For more on both of these things, see Robert Dawson's article "On a 'Singular' Integration Technique of Poisson" at cs.smu.ca/~dawson/Poisson.pdf $\endgroup$ – Mark S. Mar 16 '18 at 11:01
  • $\begingroup$ @MattSamuel Many thanks. (1) What if we had another case, in which we started with $\int_{a}^{b}(f)^{2}dx$ and wish to simplify to some form without the squaring over the integrand? Any ideas? And (2) what if we wanted to put it equal to zero, as in the condition $\int_{a}^{b}(f)^{2}dx=0$, but the squaring was making calculation too difficult; could there be a method to reduce this to a form (to work with $f$ or related functions without the squaring) that would satisfy this condition? (Sorry I didn't include this in original post, but can update it, if we could find some answers). $\endgroup$ – user135626 Mar 19 '18 at 12:06
  • $\begingroup$ @used Without knowing the function, I think there's nothing you can do if the square is inside the integral. $\endgroup$ – Matt Samuel Mar 19 '18 at 20:39

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


Double integrals:


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 $\int \int$.

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    $\begingroup$ Thanks for this. The Jensen inequality you cited: isn't it rather the Cauchy-Schwarz inequality on two real functions with the second function being constant? $\endgroup$ – user135626 Mar 16 '18 at 15:06
  • $\begingroup$ @user135626 perhaps. I was sleepy awhile ago and am sleepy now but I recall I triple checked C-S ineq. I don't think it's C-S. I think it's Jensen's, which happens to look like C-S. I remember because the first time I learned Jensen's, my classmates observed it looked liked C-S. $\endgroup$ – BCLC Mar 16 '18 at 16:52
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    $\begingroup$ I don't know why this answer got downvoted. $\endgroup$ – Matt Samuel Apr 2 '18 at 3:15
  • $\begingroup$ @user135626 no, you're right. Edited. $\endgroup$ – BCLC Apr 2 '18 at 3:48
  • $\begingroup$ @user135626 edited again. I was thinking of Hölder not Jensen. Hölder is related to CBS though. $\endgroup$ – BCLC Apr 2 '18 at 16:15

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).

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    $\begingroup$ Actually sometimes they are easier as double integrals. $\endgroup$ – Matt Samuel Mar 16 '18 at 0:50
  • $\begingroup$ Only if we are lucky to have such functions to integrate. In this particular case we have to actually integrate f(x) twice instead of integrating it once and square it. $\endgroup$ – Mohammad Riazi-Kermani Mar 16 '18 at 0:55
  • $\begingroup$ This isn't a particular case, it's the general case. And there are particular cases where it's easier. Sometimes we can do the double integral but can't do the original integral, so integrating it once and squaring it is impossible. $\endgroup$ – Matt Samuel Mar 16 '18 at 1:00
  • $\begingroup$ Correct, for example when we use polar coordinate to find double integrals. $\endgroup$ – Mohammad Riazi-Kermani Mar 16 '18 at 1:13

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