I want to follow the following integral: $$\frac{1}{C}\int_0^\infty \log(z)\,z^{p-1}\exp\left(-\frac{az+b/z}{2}\right)\,dz$$ where C is the normalising constant.
The following might be useful (expectation of logarithm under gamma distribution): http://en.wikipedia.org/wiki/Gamma_distribution#Logarithmic_expectation
Thanks in advance, Sachin
This is what I managed so far, and the basic idea is to use sufficient statistics and also take note of the constant term as well.
Let the integral of GIG, $$I=\int_0^\infty \frac{(a/b)^{p/2}}{2K_p(\sqrt{ab})} z^{p-1}\exp\left(-\frac{1}{2}(az+b/z) \right)dz=1$$ (http://en.wikipedia.org/wiki/Generalized_inverse_Gaussian_distribution) where, $K_p$ is the modified bessel function of the second kind.
Let us differentiate the above w.r.t. "p". Note that $z^{p-1}=\exp((p-1)\ln z)$ and $(a/b)^{p/2}=\exp(\frac{p}{2}\ln(a/b))$. $$\frac{1}{2}\ln(a/b)\,I-\frac{\frac{dK_p(\sqrt{ab})}{dp}}{(K_p(\sqrt{ab}))^2}\int_0^\infty \frac{(a/b)^{p/2}}{2} \dots dz+\int_0^\infty \ln(z)\frac{(a/b)^{p/2}}{2K_p(\sqrt{ab})} \dots dz =0 \\ \therefore \frac{1}{2}\ln(a/b)-\frac{\frac{dK_p(\sqrt{ab})}{dp}}{K_p(\sqrt{ab})} +E(\ln(z))=0 \\ E(\ln(z))=\frac{\frac{dK_p(\sqrt{ab})}{dp}}{K_p(\sqrt{ab})}-\frac{1}{2}\ln(a/b)$$
Note that there is no analytic solution or function, that could be found at the time of writing with regards to the differential of the Modified Bessel function (w.r.t. the order p). Numeric differentiation would be suggested. This is the closest solution I could find to this differentiation: http://dlmf.nist.gov/10.38