Suppose one has the following function $f: \mathbb{R}^+ \to \mathbb{R}^+$:
$$ f(t) = \prod_{k =1}^{K} \left[p_k(t) \right]^{m_{k}}\frac{1}{t}, $$ where $m_k = 0, 1, 2, \ldots, M$ such that $\sum_k m_k = M$, and $$ 0 \leq p_k(t) := \sum_{v = 1}^V w_{kv} \exp(t\lambda_v) \leq 1, $$ with $w_k \in \mathbb{R}$ and $\lambda_k \leq 0$. Also, we have $\sum_k p_k(t) = 1$ because the $p_k$s come from a stochastic matrix.
I want suspect that $$ \int_0^\infty f(t) \,dt = \infty.$$
Attempt at proof:
The idea is to find a lower bound $g(t)$ for $f(t)$ and show that $ \int_0^\infty g(t) \,dt = \infty$.
To that end, for each $k$ take
$$ h_k(t) = \alpha_k\exp(-\beta_kt) := \min_v w_{kv} \exp(t\lambda_v) \leq p_k(t) ,$$
but notice that it's possible $h_k(t) \leq 0$.
Now we can write
$$g(t) = \prod_{k =1}^{K} \left[ h_k(t) \right]^{m_{k}}\frac{1}{t} = \frac{\alpha^\prime\exp(-\beta^\prime t_i)}{t}, $$
with $\alpha^\prime = \prod_k \alpha_k^{m_k}$ and $\beta^\prime = \sum_k m_k \beta_k$.
Finally, recall that a function $g$ integrable iff $|g|$ is integrable and compute
$$ \int_0^\infty \left|\frac{\alpha^\prime \exp(-\beta^\prime t)}{t} \right| \, dt = \infty, $$
to conclude the (presumed) proof.
Question: is this correct? If not, what other approach(es) could I take? It might be that $f$ is indeed integrable, but, in that case, how do I show it?
Example
One simple special case is
$$ f(t) = \left[ \frac{1}{4} + \frac{3}{4} e^{-\alpha t} \right]^{N-x} \left[ \frac{3}{4} - \frac{3}{4} e^{-\alpha t} \right]^x \frac{1}{t} $$ with $\alpha = 4/3$ and $x , N \in \mathbb{N}$ and $x \leq N$.
A further idea
I think this can be shown some other way. Namely, by using integration by parts. Let $g(t) = \prod_{k =1}^{K} \left[p_k(t) \right]^{m_{k}}$ and then compute $$ \int_0^\infty \frac{g(t)}{t} \, dt = \left|g(t)\log(t)\right|_0^\infty - \int_0^\infty g^\prime(t) \log(t)\, dt.$$
So if one could show that $\lim_{x \to 0^+} g(x) = 0$ and $\lim_{x \to \infty} g(x) = a \leq 1$, which I believe is true, one could then show that the integral of interest diverges.
The value of the integral can differ depending on the choice of the $w_{kv}$.
If we have $w_{kv} \equiv 0$ for one $k$ we get $f\equiv 0$ what's obviously integrable.
If we have $w_{kv} > 0$ for all $k,v$ we get:
$f(t) \ge 0$ hence $$\int_0^\infty f(t) \,dt \ge \int_1^\infty f(t) \,dt$$
But for $t \in [1,\infty)$ it holds: $$p_k(t) := \sum_{v = 1}^V w_{kv} \exp(t\lambda_v) \ge \sum_{v = 1}^V w_{kv} \exp(\lambda_v) = p_k(1)$$
And so with $$c:= \prod_{k =1}^{K} \left[p_k(1) \right]^{m_{k}}$$ it holds
$$\int_0^\infty f(t) \,dt \ge \int_1^\infty f(t) \,dt \ge c\int_1^\infty\frac{1}{t} \,dt = +\infty$$