Finding zeros of function by integration: a novel relationship or not?

It seems that in certain cases one can find the zero of a function by solving an integration problem instead. This surprises me, and I am wondering to what extent this (1) has been studied, and/or (2) is interesting.

For example, consider $f(x) = x^3 +ax -1$. This has a (unique) zero on the positive real axis, $x_0 \in \mathbb R^+$ (for any value of $a \in \mathbb R$). I am surprised to find that one can express this zero as follows: $$x_0 = \frac{3}{4a} \left( 2 - \frac{1}{\mathcal I} \right) \qquad \textrm{ where }\qquad \mathcal I := \frac{1}{\pi} \int_0^\infty \frac{1}{1+(y^{2/3}-a)^2 \; y^{2/3}} \mathrm d y.$$

I was indeed able to prove this using some unconventional methods (which can be found in a physicist's toolbox). Is the above correspondence surprising to mathematicians?

(By the way, the fact that the $f(x)$ I chose above is a third-order polynomial is a red herring, as I found similar relationships where $f(x)$ contains, for example, logarithms.)

• You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– user3417
Commented Jul 20, 2018 at 4:29
• If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem). Commented Jul 20, 2018 at 17:46