Let $X$ follow Pareto distribution with parameters $\alpha, a, h$. That is, $X\sim Pa(\alpha,a,h)$, where $\alpha>0$ is the shape parameter, $-\infty < a < \infty$ is the location parameter, and $h>0$ is the scale parameter.
Then the pdf of $X$ is $p(x) = \left(\alpha\cdot h^{\alpha}\right)/(x-a)^{\alpha+1}$; $x\geq a+h$.
Entropy, $H(X) = -\int_{a + h}^{\infty}p(x)\log p(x)\,dx$
$$H(X) = -\int_{a + h}^{\infty}\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \log\left(\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \right)\,dx$$
$$= -\int_{a +h}^{\infty} \frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \left[\log(\alpha h^{\alpha}) -(\alpha+1)\log(x-a) \right]\,dx$$
$$= -\alpha h^{\alpha}\int_{a + h}^{\infty}\frac{1}{(x-a)^{\alpha+1}} \left[\log(\alpha) + \alpha \log(h) -(\alpha+1) \log(x-a) \right]\,dx$$
$$= -\alpha h^{\alpha}\int_{a + h}^{\infty} \frac{ \log(\alpha) + \alpha \log(h)}{(x-a)^{\alpha+1}} dx + \alpha h^{\alpha}\int_{a + h}^{\infty} \frac{(\alpha+1) \log(x-a)}{(x-a)^{\alpha+1}}dx$$
$$= -\alpha h^{\alpha}(\log(\alpha) + \alpha \log(h))\int_{a + h}^{\infty} \frac{ 1}{(x-a)^{\alpha+1}} dx + \alpha h^{\alpha}(\alpha+1)\int_{a + h}^{\infty} \frac{ \log(x-a)}{(x-a)^{\alpha+1}}dx $$
Then how can I end up with the result
$$H(X) = -(1+2\alpha)\log(h) – \log(\alpha) + 1 + (1/\alpha)$$
By using the following (which can be computed using standard integrals, integration by parts and this limit)
$$\int^\infty_{a+h}\frac{1}{(x-a)^{\alpha+1}}dx=\left.-\frac{1}{\alpha(x-a)^\alpha}\right|^\infty_{a+h}=\frac{1}{\alpha h^\alpha},$$
$$\int^\infty_{a+h}\frac{\log(x-a)}{(x-a)^{\alpha+1}}dx=\left.-\frac{\alpha\log(x-a)+1}{\alpha^2(x-a)^\alpha}\right|^\infty_{a+h}=\frac{\alpha\log(h)+1}{\alpha^2h^\alpha},$$
I get that $H(x)=-\log(\alpha)+\log(h)+1+\frac{1}{\alpha}$ which seems to match up with the entropy listed on wikipedia - are you sure that the end result you list above is correct? I could have mucked a minus sign somewhere - but I can't seem to find it.