Computing eigenfunction of expectational operator Let $A$ be an operator from $\textbf{L}^2$ to itself defined by $$Af(x) = \mathbb{E}[e^{-W \gamma}\cdot f(x\cdot e^W)]$$ where $W$ is a $\mathbb{N}(0,1)$ distributed random variable and $\gamma \in [3,10]$ is a parameter. 
I believe there is no analytical method to compute the principal eigenfunction and eigenvalue of this operator. 
By principal eigenfunction, I mean the positive function $f(x)$ such that $$Af(x) = \kappa f(x)$$ $\forall x \in \mathbb{R}$.
What kind of numerical methods are used for problems of this sort?
Also, if anyone can recommend some textbook covering this kind of problems that would be amazing.
Thank you!
Edit: the source for this kind of problem is an Economics publication on long run risk exposure (Hansen and Scheinkman (2009))
Edit2: Uranix gave great tips to approach this specific problem but also made me realize that there is not much hope of solving this analytically in general (when the factor multiplying $f(x\cdot e^W)$ is a general function of $W$). Does anyone know which numerical techniques can be used in this context?  
 A: This is not a answer, but just some ideas.
So basically the operator in question acts as
$$
(Af)(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-W\gamma} f(x e^W) e^{-W^2/2} dW
$$
Let $g(x) = (Af)(x)$ and let's split the $f$'s domain into three pieces:
$$
f(y) = \begin{cases}
f_{-}(y), &y < 0\\
f_0, & y = 0\\
f_{+}(y), &y > 0
\end{cases}
$$
One can easily see that $g_(x)$ is completely determined by $f_{-}(y)$, similarly $g_{+}(x)$ is determined by $f_{+}(y)$ and
$$
g_0 = f_0 e^{\gamma^2/2}.
$$
In other words, unless $f_0$ is zero, the eigenvalue should be equal to $e^{\gamma^2/2}$.
Assume that $f(y), \kappa$ is the principal eigenpair. It is necessary that
$$
(Af)(x) = \kappa f(x) 
$$ 
for $x > 0$. Also it is sufficient, since we can easily extend $f(x)$ is the following way (not the only possible):
$$
f_{-}(y) = f_{+}(y), \qquad f_0 = 0.
$$
So, assume now that $x > 0$, $f: \mathbb R_{+} \to \mathbb R$ and 
$$
\int_0^{\infty} f^2(y) dy < \infty.
$$
Introduce $s = x e^W > 0$. By changing variables,
$$
(Af)(x) = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} \left(\frac{x}{s}\right)^\gamma f(s) \exp\left(-\frac{(\log s - \log x)^2}{2}\right) \frac{ds}{s}
$$
which can be treated as an integral transform with kernel
$$
K(s,x) = \frac{x^\gamma}{\sqrt{2\pi}s^{\gamma+1}} \exp\left(-\frac{(\log s - \log x)^2}{2}\right)\\
(Af)(x) = \int_0^\infty K(s, x) f(s) ds.
$$
This kernel is a homogenous function:
$$
K(\lambda s, \lambda x) = \frac1\lambda K(s, x),\qquad \lambda > 0.
$$
Hence,
$$
(Af)(\lambda x) = \int_0^\infty K(s, \lambda x) f(s) ds = 
\int_0^{\infty} K(\lambda u, \lambda x) f(\lambda u) 
\lambda du = \int_0^\infty K(u, x) f(\lambda u) du = (Ah)(x)\\
h(u) \equiv f(\lambda u).
$$
If $f(x)$ is an eigenfunction, so is $h(x) = f(\lambda x)$, and they both have the same eigenvalue. Due to linearity of the transform, any linear combination also will be an eigenfunction. So should be
$$
k(x) = \int_{0}^{\infty} p(\lambda) f(\lambda x) d\lambda
$$
for any integrable function $p(\lambda)$. 
IDEA. Treat
$$
\int_{0}^{\infty} f(\lambda x) p(\lambda) d\lambda = k(x) \equiv 1
$$
as an integral equation for $p(\lambda)$. If it has a solution, then $k(x) \equiv 1$ is an eigenfunction of the $A$ with same as $f(x)$ eigenvalue.
From now I strongly suspect that the only possible eigenfunction is $f(y) = \operatorname{const}$ with the eigenvalue equal to $e^{\gamma^2/2}$.
