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Integrals where the integrand contains an absolute value can be very hard or impossible to express in closed form. Computing the zero’s of the integrand might help. But what if the zero’s have no closed form or there are infinitely many ? We can express any single zero by a contour integral but it is not so clear how to use that.

To understand it better I have come up with an example

$$ \int_R | \frac{ \cos(x) + \cos( \sqrt 2 x) + \cos( \sqrt 3 x)}{1 + x^2} | dx $$

Where $ R $ means taking the integral over all reals.

Notice the integrand (without the absolute value) is not complicated but lacks a closed form inverse for its zero’s. Also there are infinitely many zero’s.

We can simplify a tiny bit by noting the symmetry and reduce to $ 2 \int_{0}^{+\infty} ... $ but beyond that no simple simplifications exist as far as I see.

I think I have seen these type of integrals in a physics book mainly about quantum mechanics. And some math papers too. Unfortunately I do not recall any of them precisely.

I wonder if they occur in classical physics too.

But to stay with math , do they occur naturally from differential equations or things not obviously related to integrals ? Analytic number theory or dynamics ?

The main question is ofcourse the closed form of the integral, and how it was achieved. And in case there is no closed form , why not.

mick
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    Does this help? – NoName Feb 06 '20 at 12:03
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    @ Noname no I do not think so. Not even for the generalization to $a x$. The reason being that a faster change in $x$ ( $ ax $ ) leads to faster change in derivatives caused by the zero’s and the absolute values of the sum of cosines. Therefore the same differential equation will probably not hold. In fact I doubt any simple differential equation will hold as a function of $a$. This also hints at the difficulty of substitution methods in general for these kind of problems; the absolute value and the non-bijective nature are problematic. Hence the raison d’être of this question. But nice link. – mick Feb 19 '20 at 12:43

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