基础条件
$X$，$Y$服从$U(a,b)$，$Z=X+Y$，求$E(e^{jz})$
分析两个相同的均匀分布的和
$$Z=X+Y$$
$$f_X=f_Y=\frac{1}{b-a}$$
当$z\le a+b$时，
$$\begin{array}{rl}{F}_Z(z)& ={\int }_a^{z-a}{\int }_a^{z-x}{f}_X(x)f_Y(y)dydx\\ & ={\int }_a^{z-a}\frac{1}{b-a}{\int }_a^{z-x}\frac{1}{b-a}dydx\\ & =\frac{1}{(b-a{)}^2}{\int }_a^{z-a}(z-x-a)dx\\ & =\frac{\frac{1}{2}{z}^2+2a^2-2az}{(b-a{)}^2}\end{array}$$
当$z\ge a+b$时，
$$\begin{array}{rl}{F}_Z(z)& =1-{\int }_{z-b}^b{\int }_{z-x}^b{f}_X(x)f_Y(y)dydx\\ & =1-\frac{1}{(b-a{)}^2}{\int }_{z-b}^{b}1{\int }_{z-x}^{b}1dydx\\ & =1-\frac{1}{(b-a{)}^2}{\int }_{z-b}^b(b-z+x)dx\\ & =1-\frac{\frac{1}{2}{z}^2+2b^2-2bz}{(b-a{)}^2}\end{array}$$
$$f_Z(z)=\frac{d{F}_Z(z)}{dz}=\{\begin{array}{rlr}\frac{z-2a}{(b-a{)}^2},& & z\le a+b\\ \frac{2b-z}{(b-a{)}^2},& & z\ge a+b\end{array}$$
欧拉公式
$e^{ix}=cosx+isinx$，$E[e^{jz}]=E[cosz+jsinz]=E[cosz]+jE[sinz]$
$$\begin{array}{rl}E[cosz]& ={\int }_{-\infty }^{\infty }cos(z)f_Z(z)dz\\ & =\frac{1}{(b-a{)}^2}({\int }_{2a}^{a+b}cos(z)(z-2a)dz+{\int }_{a+b}^{2b}cos(z)(2b-z)dz)\\ & =\frac{1}{(b-a{)}^2}((z-2a)sin(z){|}_{2a}^{a+b}-{\int }_{2a}^{a+b}sin(z)dz+(2b-z)sin(z){|}_{a+b}^{2b}+{\int }_{a+b}^{2b}sin(z)dz)\\ & =\frac{1}{(b-a{)}^2}(-{\int }_{2a}^{a+b}sin(z)dz+{\int }_{a+b}^{2b}sin(z)dz)\\ & =\frac{1}{(b-a{)}^2}(cos(x){|}_{2a}^{a+b}-cos(x){|}_{a+b}^{2b})\\ & =\frac{1}{(b-a{)}^2}(2cos(a+b)-cos(2a)-cos(2b))\end{array}$$
同理，
$$\begin{array}{rl}E[sinz]& ={\int }_{-\infty }^{\infty }sin(z)f_Z(z)dz\\ & =\frac{1}{(b-a{)}^2}({\int }_{2a}^{a+b}sin(z)(z-2a)dz+{\int }_{a+b}^{2b}sin(z)(2b-z)dz)\\ & =\frac{1}{(b-a{)}^2}((z-2a)cos(z){|}_{a+b}^{2a}+{\int }_{2a}^{a+b}cos(z)dz-(2b-z)cos(z){|}_{a+b}^{2b}-{\int }_{a+b}^{2b}cos(z)dz)\\ & =\frac{1}{(b-a{)}^2}({\int }_{2a}^{a+b}cos(z)dz-{\int }_{a+b}^{2b}cos(z)dz)\\ & =\frac{1}{(b-a{)}^2}(sin(x){|}_{2a}^{a+b}-sin(x){|}_{a+b}^{2b})\\ & =\frac{1}{(b-a{)}^2}(2sin(a+b)-sin(2a)-sin(2b))\end{array}$$
当$a=-\pi ,b=\pi$时，$E[e^{jz}]=E[cosz]+jE[sinz]=0+0j$