# Can the minimum be given by an integral?

for $a,b > 0$, \begin{align} &\int_{0}^{\infty} \frac{\sin (ax) \sin (bx)}{x^{2}} \ dx \\ &= \int_{0}^{\infty} \frac{a \cos (ax) \sin (bx) + b \sin(ax) \cos(bx)}{x} \ dx \\ &= \frac{a}{2} \int_{0}^{\infty} \frac{\sin[(a+b)x] - \sin [(a-b)x]}{x} \ dx + \frac{b}{2} \int_{0}^{\infty} \frac{\sin [(a+b)x] + \sin[(a-b)x]}{x} \ dx \\ &= \frac{a}{2} \left( \frac{\pi}{2} - \frac{\pi}{2} \text{sgn} (a-b) \right) + \frac{b}{2} \left(\frac{\pi}{2} + \frac{\pi}{2} \text{sgn} (a-b) \right) \\ &= \frac{\pi}{2} \min \{a,b \} . \end{align}

So we can express the minimum of 2 positive real values by an integral.

We can for example also express the Lambert-W function by an integral where the integrand contains only elementary functions. So we are also capable of expressing the functional inverse by an integral. We can also express the $n$ th derivative of a function by an integral.

This inspired me to wonder about a combination of the above. for $x>0$ and $f$ a real-analytic function : $min(f(x))$.

Can this be given by an integral ?

Notice $min(f(x)) = f(y)$ where $y$ is one of the solutions to $f ' (y) = 0$ so the derivative and functional inverse are naturally related to the minimum.

I have never seen such an integral for a general $f(x)$ so I wonder if it exists.