So the Logistic Distribution pdf (w/ mean = 0 and shape parameter = 1) looks like this:
$$f_X(x)=\frac{e^{x}}{({1+e^{x}})^2}\;\;, \;\;-\infty<x<\infty$$ Now, I am interested in getting its moment-generating function (mgf). So here is what I have:
$$M_X(t)=\int_{-\infty}^\infty e^{tx}\frac{e^{x}}{({1+e^{x}})^2}dx$$
So I let$$u=\frac{1}{1+e^x}\;\;,\;\;du=-\frac{e^x}{({1+e^{x}})^2}dx$$
Thus, I get
$$\int_0^1\left(\frac{1}{u}-1\right)^tdu$$
Now, I do not know what to do next. The answer was given to be $B(t+1,t-1)\;,\;-1<t<1$ where $B(\dot\;,\dot\;)$ is the Beta function.
Can anyone help me? I will really appreciate it a lot. Thanks.
Best Answer
By definition,
$$B(x,y)=\int_0^1 u^{x-1}(1-u)^{y-1} \mathrm{d}u$$
Rewrite your last step as $$\int_0^1 u^{-t}(1-u)^t \mathrm{d}u = B(1-t,t+1)=B(1+t,1-t) $$ where the last step follows from the symmetry of the Beta function.