functional $F(a) = \int_{0}^{\infty}e^{-kx}\operatorname{ln}(a(x))dx$ maximization

Question

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.

Answer 1

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 $$\partial_a\frac{\delta L}{\delta a}=-e^{-kx}a^{-2}<0.$$

Answer 2

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: $$a(x)=k e^{-kx}$$