# Integral of Weibull distribution

How would you find the distribution function for the following density functions (Weibull function):

$$f_{X}(x) = c\tau x^{\tau−1}e^{− cx^{\tau}}$$

for $0< x < \infty$, $\tau > 0$ and $c>0$.

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Is the edit by johnny OK? I'm guessing it should actually be $$f_{X}(x) = c\tau x^{\tau−1}e^{− cx^{\tau}}$$ Am I right? – Pedro Tamaroff Dec 8 '12 at 15:44
Certainly looks more Weibullish your way, @PeterTamaroff. :) – johnny Dec 8 '12 at 15:50
Are you looking to evaluate $$\int_0^\infty f_X(x)\text{ ? }$$ – Pedro Tamaroff Dec 8 '12 at 16:08

So $$F_X(x)=\int_0^x c\tau s^{\tau−1}e^{− cs^{\tau}} ds\\ = c\tau \int_0^x s^{\tau−1}e^{− cs^{\tau}} ds$$

If we use the substitution $s^{\tau}=u$, and $\frac{du}{ds}=\tau s^{\tau-1}$ this simplifies to

$$c\int_0^x e^{− cu} du\\ =\left[-e^{-cu}\right]_0^x\\ =1-e^{-cx}.$$

I hope that I've not given this to you too easily and that this is useful to you.

$\textbf{EDIT}$: I have assumed you were asking for the c.d.f. but the other commenters are correct your question is not entirely clear on its terminology. Also fixed my $\LaTeX$.

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thank you for editing and answering the question simon. I am new to this forum and didn't know how to write the question in the correct form – Sharingan Dec 8 '12 at 16:57
That's ok. Is this answer what you wanted? Don't forget you can accept the answer as correct by clicking the tick icon on the left side of it. – Simon Hayward Dec 8 '12 at 16:59
Ah that special feeling of rejection when you put up the right answer and you don't get a tick or even an upvote. – Simon Hayward Dec 8 '12 at 18:52
Your welcome. Thanks again for your help :-) – Sharingan Dec 11 '12 at 2:50
@Simon Hayward If you take $u=s^{\tau}$ then you change also the limits, and the result then becomes, $1-e^{-cx^{\tau}}$ or not ? – OBDA Jun 23 '14 at 13:43