1. ## integration

$\displaystyle \int_0^\infty (\frac{k}{\lambda})(\frac{x}{\lambda})^{k-1} e^{-(\frac{x}{\lambda})^{k}}dx$; $\displaystyle k$and$\displaystyle \lambda$ are positive

My first instinct was to use integration by parts but this ends up looping back on itself infinitely. I'm struggling to see some kind of pattern I can play with in the repetition.

2. Originally Posted by Rubberduckzilla
$\displaystyle \int_0^\infty (\frac{k}{\lambda})(\frac{x}{\lambda})^{k-1} e^{-(\frac{x}{\lambda})^{k}}dx$; $\displaystyle k$and$\displaystyle \lambda$ are positive

My first instinct was to use integration by parts but this ends up looping back on itself infinitely. I'm struggling to see some kind of pattern I can play with in the repetition.
Try reading this: Gamma function - Wikipedia, the free encyclopedia

3. Hello,

Substitute $\displaystyle t=x/\lambda$ and then see this : Gamma function - Wikipedia, the free encyclopedia

4. The question is related to a probability distribution function i believe. So this integral should = 1

5. Originally Posted by Rubberduckzilla
The question is related to a probability distribution function i believe. So this integral should = 1
Yes, we all realise that. But your question was how to do the integration.