As part of a bigger exercise I need to try and find the integral of $$1-\exp\{-(x/a)^b\} dx.$$
Note: I can't quite get LaTex to format the equation properly but the exponential should be raised to the whole bracketed term!
I am reasonably sure there is an analytical solution as the above equation is the cdf of a Weibull distribution and it is derived by integrating the pdf of the Weibull, which has a very similar term in it $$\frac{b}{a}(x/a)^{b-1}\exp\{-(x/a)^b\}$$ and provides the neat analytical answer above. The problem being that I can't find anywhere which explains the actual steps from pdf to cdf so I can apply them to this problem, it seems to just be taken as given.
Could anyone point me in the right direction of how I would do this?
Alternatively, my ultimate goal is an analytical expression to work out the expected value of a Weibull distribution specifically between two points of the distribution, say y and z, not over the whole distribution. I have used integration by parts to get the formula for the integral of x.f(x) but this leaves me needing to find the integral above. If there is another, easier way someone knows of to find this value, or could suggest that would also be extremely helpful.
Thanks!
Best Answer
According to WolframAlpha the integral can be expressed with the incomplete gamma function: $$\int\left( 1 - \exp{\left(-\left(\frac{x}{a} \right)^b\right)}\right) dx = \left(\frac{x}{b}\right)\left(\left(\frac{x}{a} \right)^b \right)^{-\frac{1}{b}} \Gamma\left(\frac{1}{b}, \left(\frac{x}{a} \right)^b\right) + x + C$$ The incomplete gamma function can be assumed to be known in the context of statistical distributions, e.g. as the CDF of the Chi-square or Gamma distribution.