functional $F(a) = \int_{0}^{\infty}e^{-kx}\operatorname{ln}(a(x))dx$ maximization Good afternoon. I try to find  a function $a(x)>0$ subject to $\int_0^\infty a(x) dx=1$ maximzing the following functional
$$
F(a) = \int\limits_{0}^{\infty}e^{-kx}\operatorname{ln}(a(x))dx.
$$
But I don't know how to do this problem. Because formally we should find a derivate with respect to function. I will be gratefull for hints ideas and literature recomendation.
 A: Due to the probability requirement, we can use Lagrange multipliers. Our Lagrangian is $L:=\int_0^\infty e^{-kx}\ln a dx+\lambda (1-\int_0^\infty a dx)$, so $$0=\frac{\delta L}{\delta a}=e^{-kx}a^{-1}-\lambda.$$Thus $a\propto e^{-kx}$. By unitarity, $a=ke^{-kx}$ with $k>0$. Our stationary point is a maximum because $$\partial_a\frac{\delta L}{\delta a}=-e^{-kx}a^{-2}<0.$$
A: With Calculus of variations we have that:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial L}{\partial a'}\right)=\frac{\partial L}{\partial a}$$
Where $L$ is the Lagrange function:
$$L(a', a, x)=e^{-kx} \log(a(x))$$
It does not depend on $a'$, so the lhs is zero. So your differential equation for $a$ is:
$$0=e^{-kx} \frac{1}{a}$$
So we can't find an appropriate $a$.
After the edit, the new Lagrange function:
$$L'=L+\lambda(1-a(x))$$
$$L'=e^{-kx} \log(a(x))+\lambda(1-a(x))$$
So the new differential equation is:
$$0=e^{-kx}\frac{1}{a}-\lambda$$
$$\lambda=e^{-kx}\frac{1}{a}$$
$$a(x)=\frac{1}{\lambda} e^{-kx}$$
Substituting it back to the condition:
$$\lambda=\int_{0}^{\infty} e^{-kx} \mathrm{d} x$$
$$\lambda=\frac{1}{k}$$
So the function is:
$$a(x)=k e^{-kx}$$
