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)
I'll prove a partial result that hopefully someone can complete to showing that $f(x) \equiv 0$ is the unique continuous and $L^2(\mathbb{R}_{&gt;0})$ solution. Beyond that, perhaps one can use that continuous functions are dense in $L^2$ to get the requested result.  The main tool I want to contribute is the following lemma:

$\mathbf{Lemma}:$ Consider $f \in C(\mathbb{R}_{&gt;0}) \cap L^2(\mathbb{R}_{&gt;0})$. Then for any choice of $0&lt;x_0&lt;y_0 \leq \infty$, we have 

$$f(x_0) = \lim_{n \to \infty} ne^{nx_0}\int_{x_0}^{y_0}\frac{f(x)}{1+e^{nx}}dx $$

$\mathbf{Proof}:$ Fix $\epsilon&gt;0$ and, by continuity, choose $\delta&gt;0$ such that $|x-x_0| &lt; \delta \implies|f(x)-f(x_0)| &lt; \epsilon$. Note that

$$\lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{1}{1+e^{nx}}dx = 1$$

and

\begin{align}
\lim_{n \to \infty} \left | n e^{nx_0} \int_{x_0+\delta}^{y_0}\frac{f(x)}{1+e^{nx}} dx \right | 
&amp; \leq  \lim_{n \to \infty} n e^{nx_0} \int_{x_0+\delta}^{y_0} e^{-nx} |f(x)| dx \\
&amp; \leq  \lim_{n \to \infty} n e^{nx_0} e^{-(n-1)(x_0+\delta)} \int_{x_0+\delta}^{y_0} e^{-x} |f(x)| dx \\
&amp; \leq  \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}f(x)\|_1 \\
&amp; \leq \lim_{n \to \infty} n e^{x_0+\delta} e^{-n\delta} \|e^{-x}\|_2 \cdot \|f\|_2  \\
&amp; = 0
\end{align}

by Holder's inequality (neither of these depended on the choice of $\delta$-- they're just straightforward computation/comparison). Given these results, we see that 

\begin{align}
\lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{y_0} \frac{f(x)}{1+e^{nx}}dx - f(x_0) \right |
&amp; = \lim_{n \to \infty} \left |ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{f(x)-f(x_0)}{1+e^{nx}}dx \right | \\ 
&amp; \leq \lim_{n \to \infty} ne^{nx_0} \int_{x_0}^{x_0+\delta} \frac{|f(x)-f(x_0)|}{1+e^{nx}}dx \\
&amp; \leq \epsilon 
\end{align}

Since $\epsilon&gt;0$ was arbitrary, this proves the lemma.
$$\tag*{$\blacksquare$}$$

This lemma gives us some insight into the behavior of functions satisfying your condition. In particular, if $f$ satisfies these hypotheses and the condition under consideration, we must have for any $x_0 &gt; 0$ that

$$0 = \lim_{n \to \infty} ne^{nx_0} \int_{0}^\infty \frac{f(x)}{1+e^{nx}}dx = f(x_0) + \lim_{n \to \infty} ne^{nx_0} \int_{0}^{x_0} \frac{f(x)}{1+e^{nx}}dx$$
Or

$$f(x_0) = -\lim_{n \to \infty} ne^{nx_0} \int_0^{x_0} \frac{f(x)}{1+e^{nx}}dx $$

From this result, we have $\forall$ $0&lt;x_0&lt;y_0&lt;\infty$ that

$$\lim_{n \to \infty} ne^{nx_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx = \lim_{n \to \infty} e^{-n(y_0-x_0)} \left [ ne^{ny_0} \int_0^{y_0} \frac{f(x)}{1+e^{nx}}dx \right ] = 0 \cdot (-f(y_0)) = 0 $$

and similarly $\forall$ $0 &lt; w_0 &lt; x_0$ with $f(w_0) \neq 0$, the expression

$$ne^{nx_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx = e^{n(x_0-w_0)} \left [ ne^{nw_0} \int_0^{w_0} \frac{f(x)}{1+e^{nx}}dx \right ]  $$
diverges to $-\text{sgn}(f(w_0)) \cdot \infty$ as $n \to \infty$, since the term in brackets approaches $-f(w_0)$.

Now, suppose $f$ is not identically $0$, so the set

$$\{t &gt;0 | f \equiv 0 \text{ on } (0,t) \}$$

is bounded above; define $T_1$ as the supremum of this set if it is nonempty and $0$ if it is empty. Suppose $\exists T_2&gt;T_1$ such that 
$$ T_1 &lt; t &lt; T_2 \implies f(t) \neq 0$$ 
Then $f$ is nonzero on $(T_1,T_2)$ by definition, so by the Intermediate Value Theorem $f$ cannot change sign in this interval. Thus either $f(x) \geq 0$ or $-f(x) \geq 0$ on $(0,T_2)$-- WLOG assume the former, so that for any $x_0 \in (T_1,T_2)$ the first identity after the lemma gives that $f(x_0) \leq 0$, showing $f(x_0)=0$, a contradiction. We therefore conclude that no such $T_2$ exists, i.e. $\forall t&gt;T_1$ $\exists s \in (T_1,t)$ such that $f(s)=0$. That is to say, on any interval containing $0$ on which $f$ is not identically zero, $f$ must have infinitely many roots greater than $T_1$. In particular, the suggestion $f(x)=e^{-x^{1/4}} \sin(x^{1/4})$ cannot be a solution.

Note that if one can use the above results to show that $T_1 = 0$, it would establish that no continuous solution with compact support on $\mathbb{R}_{&gt;0}$ exists (relevant because such functions are dense in $L^2(\mathbb{R}_{&gt;0})$).
There is no one continuous function that will satisfy that for all n. 

By trivial solution I assume you mean $f(x)=0$. One potential (although also fairly trivial, or at least boring) answer is 

$$f(x)=(1+e^{nx})g(x)$$

where

$$\int_0^{\infty} g(x) \, dx=0$$

say maybe $\sin(x)$ or indeed any periodic function that acts in this way. However, this of course is not strictly defined as our $\epsilon-\delta$ definition will not hold for convergence, i.e. 

$$\lim_{x\to\infty} \int_0^n \sin(x) \, dx$$

is not defined, so that is not an option either.

Let us now  consider to look at this graphically. Imagine, if you can, the graph of such a function, say defined from $0$ to $\infty$ such that our integral is zero. Without a loss of generality, let us say that this function is positive to start with, so at some point, say $x_0$ this function will have to cross the $x$ axis and become negative. 

This means that our function has to be periodic and increasing at exactly the same rate and time as $1+e^{nx}$, thus resulting in only the trivial solution as above, or that the finitely bounded area before this critical value $x_0$ has identical area to our unbounded area beyond $x_0$.

This means that our function $f(x)$ satisfies $$\int_0^{x_0}f(x) \, dx=-\int_{x_0}^\infty f(x) \, dx$$

As the function must satisfy this for all n this answer should suffice to show that the function $f(x)$ must be trivial, as if $f(x)$ satisfies this for some n, the the denominator will grow at a different rate for a different n, thus showing that there is no such $f(x)$.