The Inverse Gaussian Distribution density is : $$\frac{\phi^{\frac{1}{2}}}{\sqrt{2\pi y^3}} exp[\frac{-\phi(y - \mu)^2}{2\mu^2y}]$$
Got to this integral: $$\int_0^\infty \frac{1}{y} \frac{\phi^{\frac{1}{2}}}{\sqrt{2\pi y^3}} exp[\frac{-\phi(y - \mu)^2}{2\mu^2y}]dy$$
What kind of trick do I use to get the value here? Since the score expected value must be 0 the result should be $\frac{1}{\phi}$, but the point is to show that.
