An asymptotic estimate for density of eigenvalues this is the screenshot of the useful part of the cited book
Let $\{\lambda_n\}$ be constants such that
${\lambda_n}=n^2\pi^2+\int_{0}^{1} q(t)\,dt +c_n \qquad \text{for} \quad n\rightarrow \infty$
where $(c_n)_n \in l^2$ and $q\in L^2(0,1)$.
I should prove that, if $\Lambda(u)$ denote the number of terms of the sequence $\sqrt{\lambda_n}$ which are $\leq u$,
$ \int_{1}^{v} \frac{\Lambda(u)}{u}\, du > v-\frac{1}{8}\log v + c $
for some constant c.
The problem comes from "an introduction to inverse scattering and inverse spectral problems" by Colton, Chadan, page 86 and I'm trying to apply Levinson's result.
 A: If you can't solve the general problem, try a special case and hope that solving it will give insight into the general problem.
In this situation you can first start by assuming that the potential $q=0$ and the sequence $(c_n)_n \equiv 0$ - these seem the simplest assumptions that one can make. Then your assumption is that $\lambda_n = n^2 \pi^2$ and the claim is that there is a constant $c$ such that
$$\int_1^v \frac{\Lambda(u)}{u}\text{d}u > v -\frac{1}{8} \log v + c.$$
Now $\sqrt{\lambda_n} = \pi n$. Hence, the number of eigenvalues which are less than or equal to $u$ seems to be roughly $u / \pi$. Therefore the claim is
$$\frac{(v-1)}{\pi} > v -\frac{1}{8} \log v + c,$$
which is false when $v \to \infty$.
Either I have done a mistake or the claim as presented is wrong.
(This is also what uniquesolution suggested in a comment.)
You should assume that this claim is true (for now) and continue forward. If you can establish the next result, then try to modify this claim so that it is true and suffices to prove the next result.
Maybe something like
$$\int_1^v \frac{\Lambda(u)}{u}\text{d}u > c_1 v -\frac{1}{8} \log v + c_2$$
with $c_1 > 0$, $c_2$ arbitrary, is sufficient for your purposes?
