Integral Equation: $\frac{1}{\lambda(y)} = c_1 \int_0^\infty \lambda(x) \exp(-c_2 y x) \, dx$ As presented in the question's title, I wish to find a function $\lambda(\cdot): [0, \infty) \to [0, \infty)$ which satisfies the integral equation:
\begin{equation}
  \frac{1}{\lambda(y)} = c_1 \int_0^\infty \lambda(x) \exp(-c_2 y x) \, dx
\end{equation}
where $c_1$ and $c_2$ are positive constants.
Unfortunately, I have no clear idea about how to systematically tackle this question. Any help is greatly appreciated!
 A: Let me show that no such function exists. We argue this by contradiction. Assume that there is a function $\lambda : [0, \infty) \to [0, \infty)$ satisfying
$$ \frac{1}{\lambda(s)} = c_1 \int_{0}^{\infty} \lambda(x) e^{-c_2 sx} \, dx \quad \forall s \geq 0 \tag{*}$$
with the convention that $1/0 = \infty$. Then
Step 1. In this step, we normalize $\lambda$ and reveal some useful facts on it.


*

*Since the right-hand side of $\text{(*)}$ is decreasing, $\lambda$ is increasing.

*Since the left-hand side of $\text{(*)}$ is always positive (or possibly infinite), $\lambda$ cannot be identically zero.

*From these two properties, $\alpha$ defined by
$$ \alpha := \inf \{ s > 0 : \lambda(s) > 0 \} $$
Is a non-zero real number such that $\lambda(s) = 0$ for all $s \in [0, \alpha)$. Moreover, if $\alpha > 0$ then the modified version $\tilde{\lambda}(s) = e^{-c_2 \alpha s}\lambda(s+\alpha)$ satisfies
\begin{align*}
\frac{1}{\tilde{\lambda}(s)}
= \frac{e^{c_2 \alpha s}}{\lambda(s+\alpha)}
&= c_1 e^{c_2 \alpha s} \int_{\alpha}^{\infty} \lambda(x) e^{-c_2 (s+\alpha) x} \, dx \\
 &= c_1 e^{c_2 \alpha s} \int_{0}^{\infty} \lambda(x+\alpha) e^{-c_2 (s+\alpha)(x+\alpha)} \, dx \\
&= c_1 e^{-c_2 \alpha^2} \int_{0}^{\infty} \tilde{\lambda}(x) e^{-c_2 sx} \, dx.
\end{align*}
So by changing the value of $c_1$ if needed, we may assume that $\alpha = 0$. Then by @guestDiego's computation, we may further assume that $c_1 = c_2 = 1$ and we do so.

*If $\lambda(0) > 0$, then
$$ \infty > \frac{1}{\lambda(0)}
= \int_{0}^{\infty} \lambda(x) \, dx
\geq \int_{0}^{\infty} \lambda(0) \, dx = \infty $$
and we get a contradiction. Thus $\lambda(0) = 0$.

*From the standard theory of Laplace transform, if $f \geq 0$ is measurable and
$$\mathcal{L}\{f\}(s) := \int_{0}^{\infty} f(x)e^{-sx} \, dx$$
Is finite for all $s > 0$, then $\mathcal{L}\{f\}(s)$ converges for $\Re(s) > 0$ and defines an analytic function on the same region. Moreover, differentiation can be computed by using Leibniz's integral rule:
$$ \frac{d^n}{ds^n} \mathcal{L}\{f\}(s) = (-1)^n \int_{0}^{\infty} x^n f(x) e^{-sx} \, dx. $$
Step 2. Now we are ready to establish a contradiction. First, we have $\lambda'(s) \geq 0$ because $\lambda$ is increasing. Then by the Tonelli's theorem,
\begin{align*}
\frac{s}{\lambda(s)}
&= \int_{0}^{\infty} \lambda(t) s e^{-st} \, dt \\
&= \int_{0}^{\infty} \bigg( \int_{0}^{t} \lambda'(x) \, dx \bigg)  s e^{-st} \, dt \\
&= \int_{0}^{\infty} \lambda'(x) \bigg( \int_{x}^{\infty} s e^{-st} \, dt \bigg) dx \\
&= \int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx.
\end{align*}
Taking log-differentiation to both sides, we get
$$ \frac{\lambda'(s)}{\lambda(s)} = \frac{\int_{0}^{\infty} x \lambda'(x) e^{-sx} \, dx}{\int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx} + \frac{1}{s}. \tag{1}$$
This is our key ingredient toward a contradiction. Using this, we inductively prove that

Claim. For any $n = 1, 2, 3, \cdots$ and $s > 0$, we have $\frac{\lambda'(s)}{\lambda(s)} \geq \frac{n}{s}$.

The base case $n = 1$ is straightforward from $\text{(1)}$ since the ratio between two integra in the RHS of $\text{(1)}$ is non-negative. Next, assuming the claim for $n$, we have
$$
\frac{\lambda'(s)}{\lambda(s)}
\geq \frac{\int_{0}^{\infty} n \lambda(x) e^{-sx} \, dx}{\int_{0}^{\infty} \lambda'(x) e^{-sx} \, dx} + \frac{1}{s}
= \frac{n/\lambda(s)}{s/\lambda(s)} + \frac{1}{s}
= \frac{n+1}{s}.
$$
Then the claim follows from mathematical induction.
Now the contradiction is obvious: $\lambda'(s)/\lambda(s)$ is finite for $s > 0$ while $n/s$ can be arbitrary large! Therefore no such $\lambda$ can exist.
A: I make explicit the hint of iamavegan.
If $\lambda_1(y)=\sqrt{c_1}\lambda(y)$, then 
$$
 \frac{1}{\lambda_1(y)} = \int_0^\infty \lambda_1(x) \exp(-c_2 y x) \, dx= \int_0^\infty \frac{\lambda_1(s/\sqrt{c_2})}{\sqrt{c_2}} \exp(-\sqrt{c_2} y s) \, ds
$$
Define
$$
\lambda_2(z)=\frac{\lambda_1(z/\sqrt{c_2})}{c_2^{1/4}}.
$$
Then, with $z=y\sqrt{c_2} $,
$$
 \frac{1}{\lambda_2(z)} = \int_0^\infty \lambda_2(s) \exp(-sz) \, ds.
$$
and
$$
\lambda(y)=c_1^{-1/2}c_2^{1/4}\lambda_2(y c_2^{1/2})
$$
A: HINT: iterate the $\lambda$ into the integral.
\begin{align*}
\frac1{\lambda(y)}
&=c_1\int_0^{+\infty}\lambda(x)e^{-c_2xy}\,dx\\
&=c_1\int_0^{+\infty}\frac{e^{-c_2xy}}{c_1\int_0^{+\infty}\lambda(z)e^{-c_2zx}\,dz\\}\,dx\\
&=\int_0^{+\infty}\frac{e^{-c_2xy}}{\int_0^{+\infty}\lambda(z)e^{-c_2zx}\,dz\\}\,dx\\
\end{align*}
go on from here
