I am using this function (which is type of Bounded Pareto Distribution) with domain of support $x \in[1, H]$:
$$f(x) =\frac{\alpha x^{-(\alpha +1)}}{1-\left(\frac{1}{H}\right)^{\alpha }}$$
and am attempting to find $E[Y]$, where $Y=\min\{X_1,\ldots,X_n\}$ is the minimum of an i.i.d. sample $(X_1,\ldots,X_n)$ of size $n$ with distribution $f$, hence $Y$ has the following density function:
$$g(x) = \frac{\alpha n H^{\alpha } \left(H^{\alpha }-1\right)^{-n} \left(\left(\frac{H}{x}\right)^{\alpha }-1\right)^n}{x H^{\alpha }-x^{\alpha +1}}$$
Now, I am having no luck with the following integral:
$$E[Y]=\int_1^H \frac{\alpha n H^{\alpha } \left(H^{\alpha }-1\right)^{-n} \left(\left(\frac{H}{x}\right)^{\alpha }-1\right)^n}{H^{\alpha }-x^{\alpha }} \, dx$$
Typically, I run into these types of integrals often - Does anyone know how to solve this? Any help (in general) would be appreciated. Also, a way to approximate this would be helpful as well.
So far, your calculations appear to be correct. We then observe $$\frac{H^\alpha ((H/x)^\alpha-1)^n}{H^\alpha-x^\alpha} = (H/x)^\alpha \cdot \frac{((H/x)^\alpha-1)^n}{(H/x)^\alpha-1} = (H/x)^\alpha ((H/x)^\alpha - 1)^{n-1}.$$ We can now proceed to integrate term-by-term the respective binomial expansion $$\begin{align*} \operatorname{E}[Y] &= \frac{\alpha n}{(H^\alpha - 1)^n} \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \int_{x=1}^H \left(\frac{H}{x}\right)^{\alpha(n-k)} \, dx \\ &= \frac{\alpha n}{(H^\alpha-1)^n} \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \begin{cases} \frac{H^{\alpha(n-k)} - H}{\alpha(n-k) - 1}, & \alpha(n-k) \ne 1, \\ H \log H, & \alpha(n-k) = 1. \end{cases} \end{align*} $$ From this point, I relied on Mathematica to complete the summation when $\alpha$ is not the reciprocal of some integer between $1$ and $n$; and was able to express the answer in terms of the beta and incomplete beta functions: $$\operatorname{E}[Y] = \frac{Hn}{(H^\alpha-1)^n} \left((-1)^n B(n, 1 - 1/\alpha) - B(1/H; 1/\alpha-n, n) \right),$$ where $$B(z;a,b) = \int_{t=0}^z t^{a-1} (1-t)^{b-1} \, dt,$$ and $$B(a,b) = B(1;a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ When $\alpha = 1/m$ for some $m \in \{1, \ldots, n\}$, then the above formula will be undefined, but you can, I think, either attempt a limiting approach or try to achieve some kind of cancellation in the above expression.