# How to derive the expectation of Gompertz distribution?

We are given $$h(t)=\lambda exp(\theta t)$$, h(t) is the harzard function defined in fields of biostatistics. It has relationship with f(t) in the following way:

$$h(t)=\frac{f(t)}{1-F(t)}$$

I have reached the step of finding $$f(t)=\lambda \exp(t/2-2\lambda \exp(t/2)+2\lambda)$$

But I am unable to derive the expectation by using $$E(x)=\int tf(t)dt$$

Could someone please show me how to find the expectation of the Gompertz distribution?

Substituting $$t = \ln u$$ and then integrating by parts, $$\int_0^\infty \lambda t \exp(t - \lambda e^t) \,dt = -\int_1^\infty d(e^{-\lambda u}) \ln u = \int_1^\infty \frac {e^{-\lambda u}} u du = \Gamma(0, \lambda), \\ \int_0^\infty t f(t) dt = 2 e^{2 \lambda} \Gamma(0, 2 \lambda).$$