Распределение Лапласа

${\displaystyle \operatorname {E} \xi =\int \limits _{-\infty }^{+\infty }xf(x)dx={\frac {\alpha }{2}}\int \limits _{-\infty }^{\beta }xe^{\alpha (x-\beta )}dx+{\frac {\alpha }{2}}\int \limits _{\beta }^{+\infty }xe^{-\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {\alpha }{2}}{\frac {1}{\alpha }}xe^{\alpha (x-\beta )}{\bigg |}_{-\infty }^{\beta }-{\frac {\alpha }{2}}{\frac {1}{\alpha }}\int \limits _{-\infty }^{\beta }e^{\alpha (x-\beta )}dx-{\frac {\alpha }{2}}{\frac {1}{\alpha }}xe^{-\alpha (x-\beta )}{\bigg |}_{\beta }^{+\infty }+{\frac {\alpha }{2}}{\frac {1}{\alpha }}\int \limits _{\beta }^{+\infty }e^{-\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {\beta }{2}}-{\frac {1}{2\alpha }}e^{\alpha (x-\beta )}{\bigg |}_{-\infty }^{\beta }+{\frac {\beta }{2}}-{\frac {1}{2\alpha }}e^{-\alpha (x-\beta )}{\bigg |}_{\beta }^{+\infty }=\beta -{\frac {1}{2\alpha }}+{\frac {1}{2\alpha }}=\beta }$

${\displaystyle \operatorname {E} \xi ^{2}=\int \limits _{-\infty }^{+\infty }x^{2}f(x)dx={\frac {\alpha }{2}}\int \limits _{-\infty }^{\beta }x^{2}e^{\alpha (x-\beta )}dx+{\frac {\alpha }{2}}\int \limits _{\beta }^{+\infty }x^{2}e^{-\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {\alpha }{2}}{\frac {x^{2}e^{\alpha (x-\beta )}}{\alpha }}{\bigg |}_{-\infty }^{\beta }-{\frac {\alpha }{2}}{\frac {2}{\alpha }}\int \limits _{-\infty }^{\beta }xe^{\alpha (x-\beta )}dx+{\frac {\alpha }{2}}{\frac {x^{2}e^{-\alpha (x-\beta )}}{-\alpha }}{\bigg |}_{\beta }^{+\infty }+{\frac {\alpha }{2}}{\frac {2}{\alpha }}\int \limits _{\beta }^{+\infty }xe^{-\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {\beta ^{2}}{2}}-{\frac {\beta }{\alpha }}+{\frac {1}{\alpha ^{2}}}+{\frac {\beta ^{2}}{2}}+{\frac {\beta }{\alpha }}+{\frac {1}{\alpha ^{2}}}=\beta ^{2}+{\frac {2}{\alpha ^{2}}}}$

${\displaystyle \operatorname {E} \xi ^{k}=\int \limits _{-\infty }^{+\infty }x^{k}f(x)dx={\frac {\alpha }{2}}\int \limits _{-\infty }^{\beta }x^{k}e^{\alpha (x-\beta )}dx+{\frac {\alpha }{2}}\int \limits _{\beta }^{+\infty }x^{k}e^{-\alpha (x-\beta )}dx}$

Применяя формулу интегрирования по частям несколько раз, получаем:

${\displaystyle \int x^{k}e^{\alpha (x-\beta )}dx={\frac {1}{\alpha }}x^{k}e^{\alpha (x-\beta )}-{\frac {k}{\alpha ^{2}}}x^{k-1}e^{\alpha (x-\beta )}+{\frac {k(k-1)}{\alpha ^{3}}}x^{k-2}e^{\alpha (x-\beta )}-}$${\displaystyle \ldots +(-1)^{k-1}{\frac {k(k-1)\cdots 3\cdot 2}{\alpha ^{k}}}xe^{\alpha (x-\beta )}+(-1)^{k}{\frac {k(k-1)\cdots 2\cdot 1}{\alpha ^{k+1}}}e^{\alpha (x-\beta )}}$

${\displaystyle \int x^{k}e^{-\alpha (x-\beta )}dx=-{\frac {1}{\alpha }}x^{k}e^{-\alpha (x-\beta )}-{\frac {k}{\alpha ^{2}}}x^{k-1}e^{-\alpha (x-\beta )}-{\frac {k(k-1)}{\alpha ^{3}}}x^{k-2}e^{-\alpha (x-\beta )}-}$${\displaystyle \ldots -{\frac {k(k-1)\cdots 3\cdot 2}{\alpha ^{k}}}xe^{-\alpha (x-\beta )}-{\frac {k(k-1)\cdots 2\cdot 1}{\alpha ^{k+1}}}e^{-\alpha (x-\beta )}}$

После подстановок пределов интегрирования:

${\displaystyle \int \limits _{-\infty }^{\beta }x^{k}e^{\alpha (x-\beta )}dx={\frac {1}{\alpha }}\beta ^{k}-{\frac {k}{\alpha ^{2}}}\beta ^{k-1}+{\frac {k(k-1)}{\alpha ^{3}}}\beta ^{k-2}-}$ ${\displaystyle \ldots +(-1)^{k-1}{\frac {k(k-1)\cdots 3\cdot 2}{\alpha ^{k}}}\beta +(-1)^{k}{\frac {k(k-1)\cdots 2\cdot 1}{\alpha ^{k+1}}}}$

${\displaystyle \int \limits _{\beta }^{+\infty }x^{k}e^{-\alpha (x-\beta )}dx={\frac {1}{\alpha }}\beta ^{k}+{\frac {k}{\alpha ^{2}}}\beta ^{k-1}+{\frac {k(k-1)}{\alpha ^{3}}}\beta ^{k-2}+\ldots +{\frac {k(k-1)\cdots 3\cdot 2}{\alpha ^{k}}}\beta +{\frac {k(k-1)\cdots 2\cdot 1}{\alpha ^{k+1}}}}$

Так как первый интеграл зависит от чётности k рассматриваются два случая: k — чётное и k — нечётное:

${\displaystyle \operatorname {E} \xi ^{k}={\begin{cases}\beta ^{k}+{\frac {k(k-1)}{\alpha ^{2}}}\beta ^{k-2}+\ldots +{\frac {k(k-1)\cdots 2\cdot 1}{\alpha ^{k}}},&k=2n\\\beta ^{k}+{\frac {k(k-1)}{\alpha ^{2}}}\beta ^{k-2}+\ldots +{\frac {k(k-1)\cdots 3\cdot 2}{\alpha ^{k-1}}}\beta ,&k=2n+1\end{cases}}}$

Или, в общем виде:

${\displaystyle \operatorname {E} \xi ^{k}=\sum _{i=0}^{\left\lfloor k/2\right\rfloor }{\frac {\beta ^{k-2i}}{\alpha ^{2i}}}{\frac {k!}{(k-2i)!}}}$, где ${\displaystyle \left\lfloor s\right\rfloor }$ — целая часть s.

${\displaystyle \phi (t)=\int \limits _{-\infty }^{+\infty }e^{itx}f(x)dx={\frac {\alpha }{2}}\int \limits _{-\infty }^{\beta }e^{itx}e^{\alpha (x-\beta )}dx+{\frac {\alpha }{2}}\int \limits _{\beta }^{+\infty }e^{itx}e^{-\alpha (x-\beta )}dx}$

Оба интеграла находятся, используя формулу Эйлера ${\displaystyle \textstyle e^{ix}=\cos x+i\sin x}$ и классический пример нахождения интегралов вида ${\displaystyle \int e^{\alpha x}\sin \beta x\,dx}$ и ${\displaystyle \int e^{\alpha x}\cos \beta x\,dx}$ (см. Интегрирование по частям:Примеры):

${\displaystyle \int \limits _{-\infty }^{\beta }e^{itx}e^{\alpha (x-\beta )}dx=\int \limits _{-\infty }^{\beta }(\cos tx+i\sin tx)e^{\alpha (x-\beta )}dx=\int \limits _{-\infty }^{\beta }\cos tx\,e^{\alpha (x-\beta )}dx+i\int \limits _{-\infty }^{\beta }\sin tx\,e^{\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {e^{\alpha (x-\beta )}}{\alpha ^{2}+t^{2}}}{\Big (}\alpha \cos tx+t\sin tx{\Big )}{\bigg |}_{-\infty }^{\beta }+i{\frac {e^{\alpha (x-\beta )}}{\alpha ^{2}+t^{2}}}{\Big (}a\sin tx-t\cos tx{\Big )}{\bigg |}_{-\infty }^{\beta }=}$ ${\displaystyle ={\frac {1}{\alpha ^{2}+t^{2}}}{\Big (}\alpha e^{it\beta }+t(i\cos t\beta -\sin t\beta ){\Big )}}$

${\displaystyle \int \limits _{\beta }^{+\infty }e^{itx}e^{-\alpha (x-\beta )}dx=\int \limits _{\beta }^{+\infty }(\cos tx+i\sin tx)e^{-\alpha (x-\beta )}dx=\int \limits _{\beta }^{+\infty }\cos tx\,e^{-\alpha (x-\beta )}dx+i\int \limits _{\beta }^{+\infty }\sin tx\,e^{-\alpha (x-\beta )}dx=}$ ${\displaystyle ={\frac {e^{-\alpha (x-\beta )}}{\alpha ^{2}+t^{2}}}{\Big (}-\alpha \cos tx+t\sin tx{\Big )}{\bigg |}_{\beta }^{+\infty }+i{\frac {e^{-\alpha (x-\beta )}}{\alpha ^{2}+t^{2}}}{\Big (}-a\sin tx-t\cos tx{\Big )}{\bigg |}_{\beta }^{+\infty }=}$ ${\displaystyle {\frac {1}{\alpha ^{2}+t^{2}}}{\Big (}\alpha e^{it\beta }-t(i\cos t\beta -\sin t\beta ){\Big )}}$

Окончательно характеристическая функция есть:

${\displaystyle \phi (t)={\frac {\alpha }{2}}{\frac {1}{\alpha ^{2}+t^{2}}}{\Big (}ae^{it\beta }+t(i\cos t\beta -\sin t\beta ){\Big )}+{\frac {\alpha }{2}}{\frac {1}{\alpha ^{2}+t^{2}}}{\Big (}\alpha e^{it\beta }-t(i\cos t\beta -\sin t\beta ){\Big )}={\frac {\alpha ^{2}e^{it\beta }}{\alpha ^{2}+t^{2}}}}$