# Evaluate the integral: $I(a,b,s,m)=\int_0^{\infty } \lambda ^a e^{\frac{(m-\log (\lambda ))^2}{2 s^2}-\frac{\lambda }{b}} \, d\lambda$.

$$I(a,b,s,m)=\int_0^{\infty } \lambda ^a e^{\frac{(m-\log (\lambda ))^2}{2 s^2}-\frac{\lambda }{b}} \, d\lambda$$ with $a,b,s,m>0$.

This doesn't converge. Let $u=log(\lambda)$, $d\lambda = e^udu$, and the integral becomes $I(a,b,s,m)=\int_{-\infty}^{\infty}e^{au}e^{\frac{(m-u)^2}{2s^2} - \frac{e^u}{b}}e^udu$. As $u\rightarrow -\infty$, the integrand will go to $\infty$ as the $e^{u^2}$ term will dominate all other terms. Note that as $u \rightarrow \infty$, the $e^{-e^u}$ dominates and the integrand will go to zero. But as $u\rightarrow -\infty$, $e^{-e^u}\rightarrow 1$