Given constants $r, \alpha$, I would like to find the function $p: \mathbb R \to \mathbb R$ subject to the constraint $$\int_0^1p(t)r^tdt = 1$$ which minimizes $$\int_0^1e^{-\alpha p(t)}dt$$.
How can I do this?
My work so far:
- If $r = 1$, then $p$ is uniform, by symmetry. Can I somehow incorporate $r$ into the integrand, and define a new function $P(t) = p(t)/r^t$, so that $P(t)$ must be uniform?
- Can I use Lagrange multipliers? Their standard usage is finding a point where extrema occur, not a function.
- Can I use Calculus of Variations (and least action)? I have encountered these but am not proficient in them.
Working with $J(p) := \int_0^1\, e^{-ap(t)}\, dt + \lambda\int_0^1\, p(t)e^{bt}\, dt,$ the calculus of variations is the proper tool. Informally, if $p$ is a stationary point, then with $p + \varepsilon w$, for small $\varepsilon$ and pretty arbitrary function $w$, the first variation \begin{align} (J(p + \varepsilon w) - J(p))/\varepsilon &= \int_0^1\, e^{-ap(t)}\, \frac{e^{-a\varepsilon w(t)} - 1}{\varepsilon}\, dt + \lambda\int_0^1\, w(t)e^{bt}\, dt \\ &= \int_0^1\, (-a e^{-ap(t)} + \lambda e^{bt})w(t)\, dt + O(\varepsilon) \end{align} is zero. This leads to the stationary point satisfying $-a e^{-ap(t)} + \lambda e^{bt} = 0$, that is $$p(t) = -a^{-1}\log(\lambda/a) - (b/a)t,$$ with $\lambda$ being chosen so that $\int_0^1\,p(t)e^{bt}\, dt = 1$.