# Lelong numbers and integrability of psh functions

Let $$\varphi$$ be a plurisubharmonic function in the unit ball $$B_1\subset \mathbb{C}^n$$ with $$\varphi\le 0$$. Suppose that the Lelong number $$\nu(\varphi,0) for some $$k>0$$. Does it follow that there exists $$\alpha>0$$, possibly depending on $$k$$, such that $$\int_{B_{\frac{1}{2}}}e^{-\alpha\varphi}dvol(z)\le C$$?

Note that this does not immediately follow from the definition of Lelong number. Since $$\nu(\varphi,0)=\lim\inf_{z\rightarrow0}\frac{\varphi(z)}{\log|z|}$$, so $$\nu(\varphi,0) apriori only implies $$\varphi(z)-k\log|z|$$ bounded from below along some sequence $$z_i\rightarrow0$$.

On the other hand, if we know $$\nu(\varphi,0)=0$$, then above is indeed true, since under this assumption, we would have $$\varphi(0)$$ bounded from below, and the result above follows from a Lemma of Hormander(Lemma 4.4 in his book on several complex variables).

If the Lelong number of a plurisubharmonic function $$\varphi$$ at a point $$a$$ satisfies the condition $$\nu(\varphi,a) < 2$$, then the function $$e^{-\varphi}$$ is locally integrable with respect to the Lebesgue measure in a neighbourhood of $$a$$.
H. Skoda, Sous-ensembles analytiques d’ordre fini ou infini dans $$\mathbb{C}^n$$, Bull. Soc. Math. de France 100 (1972), 353-408.
If you want $$\alpha$$ to not depend on more than $$\nu(\varphi,0)$$ and you want it to hold in $$B_{1/2}(0)$$, as I have interpreted your question now, then the answer is no:
In $$\mathbb{C}$$, you can take $$\varphi_m = \log |z-1/4|^m-\log (5/4)^m$$, which is $$\leq 0$$ on $$B_1(0)$$, and such that $$\nu(\varphi_m,0) = 0$$, while $$e^{-\alpha\varphi_m}$$ is integrable on $$B_{1/2}(0)$$ only for $$\alpha < 2/m$$.