Minimising $\int_{-\infty}^{x_0}\cos\left(C-f(x)\right)e^{x-x_0}\mathop{\mathrm dx}$

Question

For an analytic and bounded $f:\mathbb{R}_{\leq x_0}\rightarrow\mathbb{R}$, consider the integral \begin{equation} I=\int_{-\infty}^{x_0}\cos\left(C-f(x)\right)e^{x-x_0}\mathop{\mathrm dx}, \end{equation} for $C,x_0\in\mathbb{R}$. What choice of $f$ would minimise $I$? From the Euler-Lagrange equations, I have \begin{align} 0&=\frac{\partial L}{\partial f},\\ &=e^{x-x_0}\sin\left(C-f(x)\right), \end{align} which implies $f=C$ is a solution. However, this is contradictory when substituting into $I$ (one can easily construct a new function $f':=A\sin(x)$ if $C=x_0=1$, for example, such that $I'<I$ for some $A$). Unless $\partial L/\partial f$ is of a different form, I'm unsure of the problem here.

Answer 1

The main point is that the Euler-Lagrange equations are necessary, not sufficient. Thus if a minimizer exists it satisfies the EL equations , not that every solution of the equations is a minimizer.

The thing is that:

$$ sin(C-f(x)) = 0 $$

has plenty of solutions, it just that $f(x)=C$ is not the proper one. The apropriate choice is actually $f(x) = C - 2 \pi k - \pi $ where $k \in \mathbb{Z}$.

To see this, notice that you only have control over the part $(cos(C-f(x)))$ and the least you can hope to achieve for $I$ is whenever: $cos(C-f(x)) = -1 $ which happens precisely in the above choice of $f$ , so there is no contradiction with the Euler Lagrange equations.

Answer 2

$\int_0^{\infty}e^{-x}\cos f(x)dx$ is minimum for $f(x)\equiv\pi.$