I have been reading a paper where there is the following objective function to be minimised by choosing $s$:
$\displaystyle \alpha \int_{s}^{1} t(s_{2}) g(s_{1}+s_{2})ds_{2} + \alpha \int_{0}^{s} t(s)g(s_{1}+s_{2})ds_{2}+\int_{0}^{s}(s_{2}-\frac{s}{2})^{2}ds_{2}$
The first order condition that the paper arrives at is
$\displaystyle t'(s)\alpha[G(s_{1}+s)-G(s_{1})]+\frac{s^2}{4}=0$
where $s_{1}, s_{2} \in [0, 1]$ are constants and $G(x)=\int_{0}^{x} g(y)dy$. $\alpha$ is also a constant and $g(\bullet)$ is a strictly convex arbitrary function. An application of the fundamental theorem of calculus would not produce $t'(s)$ or $G(\bullet)$. I have also tried using integration by parts but I arrive at the same expression as when I used the fundamental theorem of calculus.
Any ideas on how the FOC is derived? There are a number of other similar first order conditions that are in a similar form, so I am hoping that by understanding this derivation, it will help me understand the rest of the paper. Thanks
The result is correct. Here's how I computed it: $$ \frac{d}{ds} \int_s^1 t(s_2)g(s_1+s_2)\, ds_2 =-t(s)g(s_1+s).$$ Moreover, $$ \frac{d}{ds}\int_0^s t(s)g(s_1+s_2)\,ds_2=t'(s)\int_0^s g(s_1+s_2)\, ds_2 + t(s)g(s_1+s), $$ and finally $$ \frac{d}{ds} \int_0^s(s_2-\frac{s}2)^2\, ds_2 = \frac{s^2}{4}.$$ When you multiply the first two by $\alpha$ and sum, the term $t(s)g(s_1+s)$ cancels out. And notice that $$\int_0^s g(s_1+s_2)\, ds_2= G(s_1+s)-G(s_1).$$