# Minimizing a functional definite integral

I have a definite integral defined by

$$T\left(G\left(g\right)\right)=\int_{g_{1}}^{g_{2}}G(g)\mathrm{d}g$$

where $G$ is a continuous function of a variable $g$, and $g_{1}$ and $g_{2}$ are known numbers. I want to minimize $T\left(G\left(g\right)\right)$, that is I want to find a continuous function $G=f\left(g\right)$ that makes $T\left(G\left(g\right)\right)$ minimum. Ideally I would differentiate it and equate to zero, but because $T\left(G\left(g\right)\right)$ is too complicated to be obtained and then differentiated analytically, I would like to know if there is a numeric technique or any other technique by which this problem can be solved.

• I don't completely understand your problem. Your functional doesn't look "complicated" at all - it maps a function $G$ to it's integral $\int_1^2 G$. However, the 'minimisation' isn't well-defined. If you take $G(g) \equiv c \equiv \text{constant}$, you can make $T(G)$ any value you want, so what would be the minimum? Maybe I'm completely misreading the question, but in that case many others are probably having the same problem. – Vibert Dec 19 '12 at 23:30
• Look up calculus of variations (en.wikipedia.org/wiki/Calculus_of_variations), to see how these type of problems are normally dealt with. In your case, as Vibert points out, either $f(g)=0$ or $f(g) = -\inf$, depending on how you define minimisation, is the solution to your problem. – Jaime Dec 19 '12 at 23:51
• Also note that $g$ of the right hand side is swallowed, and this $T$ doesn't depend on $g$ (but on $G$ and $g_1,g_2$). – Berci Dec 20 '12 at 1:18