Find function $f(x)$ satisfying $\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$ I am looking for a non-trivial function $f(x)\in L_2(0,\infty)$ independent of the parameter $n$ (a natural number) satisfying the following integral equation: $$\displaystyle\int_{0}^{\infty} \frac{f(x)}{1+e^{nx}}dx=0$$
or prove that $f(x)=0$ is the only solution.
The similar question is here but there are no parameters in the integral and the answer is based upon hit and trial method. 
I want to know what would be the nice approach to tackle this problem.
EDIT: Such a function may exist, here is an example due to Stieltjes. A function $f(x) = \exp(-x^{1/4}) \sin x^{1/4}$ satisfies $\int_0^{\infty} f(x) x^n dx = 0$ for all integers $n \ge 0$.
We use the substitution $x=u^4$ to write
$I_n = \int_0^{\infty} f(x) x^n dx = 4 \int_0^{\infty} e^{-u} \sin(u) u^{4n+3} du$;
then integrate by parts four times (differentiating the power of $u$, and integrating the rest) to show that $I_n$ is proportional to $I_{n-1}$,
and finally check that $I_0=0$.(the Edit copied from here)

The issue condition can be written in the form of
$$\int_0^\infty f\left(\dfrac zn\right)\dfrac{dz}{1+e^z} = 0.\tag1$$
Let the Maclaurin series of $f(x)$ exists, then 
$$f(z) = \sum\limits_{k=0}^\infty a_kz^k.$$
Taking in account the integral representation of the Riemann zeta function in the form of

$$\int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} =
 (1-2^{-k})k!\zeta(k+1)\tag3$$

(see also Wolfram Alpha),
one can get
$$I_n = \sum\limits_{k=0}^\infty \dfrac{a_k}{n^k} \int\limits_0^\infty\dfrac {z^k\,dz}{1+e^z} = \sum\limits_{k=0}^\infty \dfrac{1-2^{-k}}{n^k}k!\zeta(k+1) = 0,$$
$$I_n = \sum\limits_{k=0}^\infty \dfrac{b_k}{n^k} = 0\quad \forall (n\in \mathbb N),\tag4$$
where 
$$b_k = (1-2^{-k})k!\zeta(k+1).\tag5$$
The system $(4)$ is linear relatively to the coefficients $b_k.$ On the other hand,
determinant of this system is
$$ \Delta = 
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 
1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dfrac1{2^4} & \dots & \dfrac1{2^k} &\dots\\
1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dfrac1{3^4} & \dots & \dfrac1{3^k} &\dots\\
1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dfrac1{4^4} & \dots & \dfrac1{4^k} &\dots\\
&&&&\dots\\
1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dfrac1{n^4} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix}\\[6pt]
=
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 
0 & -\dfrac12 & -\dfrac1{2^2} & -\dfrac1{2^3} & -\dfrac1{2^4} & \dots & -\dfrac1{2^k} &\dots\\
1 & -\dfrac23 & -\dfrac2{3^2} & \dfrac2{3^3} & \dfrac2{3^4} & \dots & -\dfrac2{3^k} &\dots\\
1 & -\dfrac34 & \dfrac3{4^2} & -\dfrac3{4^3} & -\dfrac3{4^4} & \dots & -\dfrac3{4^k} &\dots\\
&&&&\dots\\
1 & -\dfrac{n-1}n & -\dfrac{n-1}{n^2} & -\dfrac{n-1}{n^3} & -\dfrac{n-1}{n^4} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix}\\[6pt]
\sim
\begin{vmatrix}
1 & 1 & 1 & 1 & 1 & \dots & 1 & \dots \\ 
0 & 1 & \dfrac12 & \dfrac1{2^2} & \dfrac1{2^3} & \dots & \dfrac1{2^k} &\dots\\
0 & 1 & \dfrac13 & \dfrac1{3^2} & \dfrac1{3^3} & \dots & \dfrac1{3^k} &\dots\\
0 & 1 & \dfrac14 & \dfrac1{4^2} & \dfrac1{4^3} & \dots & \dfrac1{4^k} &\dots\\
&&&&\dots\\
0 & 1 & \dfrac1n & \dfrac1{n^2} & \dfrac1{n^3} & \dots & \dfrac1{n^k} &\dots\\
\end{vmatrix} \not = 0.$$
If the homogenius linear system has unzero determinant, then zero solution is single.
Looks that consideration of the determinant sequences allows to assume that $\textbf{the issue task has only the  trivial solution}$.
