The closed form solution to the equation with sum of exponential functions

Question

I have a function

$\begin{equation}f(x)=\sum_{k=1}^{n}\left(\frac{1}{n}+\frac{ax}{k}\right)e^{-a(\frac{nx}{k}-b)}\end{equation}$,

where $a,b$ are both positive constants, $n$ is a positive integer.

$x^*$ is the positive solution to the following equation

$\begin{equation} f(x^*)=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\end{equation}$

My questions are as follows:

1. Is it possible to get a closed form $x^*$? If not, what will be best approximate solution?

2. Given another equation, $h(n)= \frac{1}{n}\sum_{k=1}^{n}e^{-a(\frac{nx^*}{k}-b)}$, then with the increase of $n$, is $h(n)$ increasing or decreasing (theoretical result)? (Note that the change of $n$ also changes $x^*$)

Some preliminary results I obtained:

1. $x^*$ exists, because when $x=0$, $f(0)=e^{ab}>1$, when $x= \infty$, $f(\infty)=0<1$. The numerical solution can be obtained using Matlab $fsolve$.

2. With the increase of $n$, $x^*$ is decreasing. To prove it, we first prove that with the increase of $n$, $f(x)$ decreases. In particular, $f(x)=\frac{1}{n}\sum_{k=1}^{n}\left(1+\frac{axn}{k}\right)e^{-a(\frac{nx}{k}-b)}$, let $g(y)=(1+\frac{ax}{y})e^{-a(\frac{x}{y}-b)}$, which is a concave and increasing function. Therefore, according to Theorem presented in related theorem, we can get $\frac{1}{n}\sum_{k=1}^{n}g(\frac{k}{n})>\frac{1}{n+1}\sum_{k=1}^{n+1}g(\frac{k}{n+1})$. Moreover, with the increase of $x$, $f(x)$ decreases, which can be obtained by taking derivative of $f(x)$ with respect to $x$, i.e., $\frac{\partial f}{\partial x}=\sum_{k=1}^{n}-\frac{a^2nx}{k^2}e^{-\frac{anx}{k}}<0 \ (x>0)$. Therefore, with the increase of $n$, $x^*$ is decreasing.

3. When $n\rightarrow \infty$, $x^*$ converges, because when $n\rightarrow \infty$, $f(x)$ becomes $\int_{0}^1g(y)dy$, $x^*$ is then the solution to $\int_{0}^1(1+\frac{ax}{y})e^{-a(\frac{x}{y}-b)}dy=1$. Similarly, $h(n)$ also converges.

4. Numerical results: I got some numerical results about the relations between $n$ and $x^*$, and between $n$ and $h(n)$, which are contained in the following link,

relation between $n$ and $x^*$

relation between $n$ and $h(n)$

Any suggestions or comments will be appreciated! Thank you in advance.

Answer 1

For $n=1$, the solution is $x = (-1 - W_{-1}(-e^{-1-ab}))/a$ where $W_{-1}$ is the "$-1$" branch of the Lambert W function. For $n > 1$, I doubt very much that there is a closed-form solution in general.

Answer 2

$$\int(1+\frac{ax}{y})\,e^{-a(\frac{x}{y}-b)}dy=y\,e^{-a(\frac{x}{y}-b)}$$ Assuming $x >0$ $$\int_{0}^1(1+\frac{ax}{y})e^{-a(\frac{x}{y}-b)}dy=e^{a (b-x)}$$ $$\int_{0}^1(1+\frac{ax}{y})e^{-a(\frac{x}{y}-b)}dy=1 \implies x^*=b$$