# Thread: E[X] and E[X^2] for function of Weibull RV

1. ## E[X] and E[X^2] for function of Weibull RV

I'm tryng to solve the following integrals:

1) (x/t)*(e^(-x/t))*(1+e^(-x/t))^-2

2) ((x^2)/t)*(e^(-x/t))*(1+e^(-x/t))^-2

Both on negative infinity to positive infinity.

You can see that they are both expectations of functions of a weibull random variable with theta of t and beta of 1.

1) Y = X(1+e^(-X/t))^-2
2) Y = (X^2)(1+e^(-X/t))^-2

I'm not sure if you need to use that or if I just can't solve the integrals properly. Any help?

2. Originally Posted by dsmullan
I'm tryng to solve the following integrals:

1) (x/t)*(e^(-x/t))*(1+e^(-x/t))^-2

2) ((x^2)/t)*(e^(-x/t))*(1+e^(-x/t))^-2

Both on negative infinity to positive infinity.

You can see that they are both expectations of functions of a weibull random variable with theta of t and beta of 1.

1) Y = X(1+e^(-X/t))^-2
2) Y = (X^2)(1+e^(-X/t))^-2

I'm not sure if you need to use that or if I just can't solve the integrals properly. Any help?
See here:

Wolfram Mathematica Online Integrator

3. Originally Posted by Anonymous1
Yes, I've done that but when doing part (2), it integrates to a polylog function. I doubt that a prof would expect us to manually integrate to that so I'm wondering if I missed anything.

4. There may be some slick manipulation you can use.
You may not need it though.
Look at the definition of polylog.
Specifically, examine its behavior when you plug in $\pm \infty.$