Really thanks Professor. I learned a lot.

I guess this is the last part of my question on this example. Please see the attachment.

After we have got function M(s), which is the moment generating function of Z(t), we need to get the probability density function. The first tool in my mind is the Laplace transform by definition of Laplace transform and moment generating function. So I just replace with -s and apply InverseLaplaceTransform, but Mathematica does give me anything:

So may I ask what shall I do to get the result, which is defective density probability function f in the paper?

I checked the book about A&S he listed in the reference, however, I was not inspired. This book could be easily achieved from the Internet. (Sorry, it is too large, I cannot upload it as the attachment.) Abramowitz, M. & Stegun, I. (1969). Handbook of mathematical functions. New York: Dover.

By the way, Mathematica does recognize the exponential integral function:

which is exactly the same as the book.

Attachments:

Hi guys, really thanks for such timely help. I learned quite a lot.

This question comes from a paper, (I put the question part in the attachment). while that paper uses Maple. When I tried myself on Mathematica, I cannot achieve. Here are the consequent questions, I put the question up at first:

Then I need to differentiate M[t] function with respect to s. So how can I define a new function whose variable is s based on the same equation of M? As you can see I use copy and paste here to define a new two variables function.

While the second question is to differentiate w.r.s s once and twice and feed s=0 in. Mathematic does not give the result here, however, Maple does give which is

The first equation is the differentiation w.r.s s once and feed s=0 in. You may have already found M is actually a moment generating function with variable s in the case. For more details, please see the attachment.

At last, please feel free to send my the note book if you think it is more easier for communication.

Attachments:

At least in Version 10 in a fresh notebook

$Assumptions = s < 1;
Simplify[DSolve[{m''[t]==1/100(2 s E^(-t/100)-3)/(1-s E^(-t/100))m'[t]+1/10000 s E^(-t/100)/(1-s E^(-t/100))m[t],
   m[0] == 1, m'[0] == 0}, m[t], t]]

seems to rapidly find a solution