I would be very grateful to get some help with the following problem.
Let $X_1, ..., X_n$ be independent and uniformly distributed on the interval $(0,\theta)$ with $\theta>0$. Let the prior density of $\theta$ be the log-normal distribution with parameters $(\mu,\sigma^2)$ where $\mu \in \mathbb{R}$ and $\sigma>0$ are known constants. I'm looking for
a) the posterior density of $\ln{\theta}$,
b) the k-th posterior moment of $\theta$,
c) the maximum of the posterior density of $\theta$.
I tried to solve a) as follows.
Let $p(\theta)$ be the prior density of $\theta$ and $L(x_1, ..., x_n|\theta)$ the likelihood function of $\theta$ given the outcome $X_1=x_1, ..., X_n=x_n$. Then the posterior density $f(\theta|X_1=x_1, ..., X_n=x_n)$ is given by $$\frac{L(x_1, ..., x_n|\theta)p(\theta)}{N}$$ where $$N=\int_0^\infty L(x_1, ..., x_n|\theta)p(\theta) d\theta$$ is a normalizing constant.
I have $$p(\theta)=\frac{1}{\sqrt{2\pi}\sigma\theta}\exp \left(-\frac{(\ln{\theta}-\mu)^2}{2\sigma^2}\right)$$ and $$L(x_1, ..., x_n|\theta)=1/\theta^n$$ where $x_1, ..., x_n \in (0,\theta)$ and $0$ otherwise.
Now I need to solve that integral, but I can't figure it out. Since we're asked to give the posterior density of $\ln{\theta}$, I tried to substitute $y=\ln{\theta}$ by using $$\theta^{-(n+1)}=\exp \left(\ln{\theta^{-(n+1)}}\right)=\exp \left(-(n+1)\ln{\theta}\right)$$ and simplifying the integral to $$N=\int_0^\infty \frac{1}{\sqrt{2\pi}\sigma}\exp \left(-(n+1)y+\frac{(y-\mu)^2}{2\sigma^2}\right)dy.$$
However, I still can't solve that integral and feeding it to Wolframalpha didn't yield a result as well.
Have I done something wrong or do I need to choose a completely different approach? So far I didn't care much about b) and c) since I need to solve a) in order to do them, I think.