連続型確率変数(continuous random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)

$$f_X(x)=\{\begin{array}{ll}{\displaystyle \frac{1}{b-a}}& \{a\le x\le b\}\\ {\displaystyle 0}& \{x<a,b<x\}\end{array}$$ $$\begin{array}{rcl}{\displaystyle M_X(t)}& \equiv & {\displaystyle E[{\mathrm{e}}^{tX}]}\\ & =& {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{tx}{f}_X(x)\mathrm{d}x}\\ & =& {\displaystyle {\int }_a^b{\mathrm{e}}^{tx}\left(\frac{1}{b-a}\right)\mathrm{d}x}\\ & =& {\displaystyle \frac{1}{b-a}{\int }_a^b{\mathrm{e}}^{tx}\mathrm{d}x}\\ & =& {\displaystyle \frac{1}{b-a}{\int }_{ta}^{tb}{\mathrm{e}}^s\frac{1}{t}\mathrm{d}s\dots s=tx,\frac{\mathrm{d}s}{\mathrm{d}x}=t,\mathrm{d}x=\frac{1}{t}\mathrm{d}s,a\to ta,b\to tb}\\ & =& {\displaystyle \frac{1}{b-a}\frac{1}{t}{\int }_{ta}^{tb}{\mathrm{e}}^s\mathrm{d}s}\\ & =& {\displaystyle \frac{1}{t(b-a)}{\left[{\mathrm{e}}^s\right]}_{ta}^{tb}}\\ & =& {\displaystyle \frac{1}{t(b-a)}[{\mathrm{e}}^{tb}-{\mathrm{e}}^{ta}]}\\ & =& {\displaystyle \frac{{\mathrm{e}}^{tb}-{\mathrm{e}}^{ta}}{t(b-a)}}\end{array}$$