I'm trying to derive the characteristic function for the Laplace distribution with density $$\frac{1}{2}\exp\{-|x|\}$$
My attempt: $$\frac{1}{2}\int_{\Omega}e^{itx-|x|}\mathrm{d}x$$ $$=\frac{1}{2}\int_0^\infty e^{(-it+1)-x}\mathrm{d}x+\frac{1}{2}\int_{-\infty}^0e^{(it+1)x}\mathrm{d}x$$ $$=\frac{1}{2}e^{(-it+1)}\int_0^\infty e^{-x}\mathrm{d}x+\frac{1}{2}e^{(it+1)}\int_{-\infty}^0e^{x}\mathrm{d}x$$ $$\frac{1}{2}e^{-it+1}+\frac{1}{2}e^{(it+1)}$$
$$\frac{1}{2}e(e^{-it}+e^{it})$$ But this doesn't look like the characteristic function on Wikipedia.
$\endgroup$2 Answers
$\begingroup$In the third line you wrote substantially that $$ e^{(it+1)x}=e^{(it+1)x}e^{x} $$ which is extremely wrong. In the second line you also wrote $e^{(-i t +1)-x} $ which is inconsistent with the first line.
$\endgroup$ 3 $\begingroup$First I will derive what we will need later using Euler's formula.
\begin{align*} \int \cos(ux) e^{-x} dx + i \int \sin(ux) e^{-x} dx&= \int (\cos(ux) + i \cdot \sin(ux)) e^{-x} dx \\ &= \int e^{iux} \cdot e^{-x} dx \\ &= \int e^{(iu-1)x} dx \\ &= \frac{1}{iu-1} e^{(iu-1)x} \\ &= \frac{1}{iu-1} \frac{(iu+1)}{(iu+1)} e^{(iu-1)x} \\ &= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot e^{iux} \\ &= -\frac{e^{-x}}{u^2+1} (iu+1) \cdot (\cos(ux) + i \sin(ux)) \\ &= -\frac{e^{-x}}{u^2+1} (iu\cos(ux) - u\sin(ux) + \cos(ux) + i \sin(ux)) \\ &= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) + i (-\frac{e^{-x}}{u^2+1} (u\cos(ux) + \sin(ux))). \end{align*}
Taking the real term,$\int \cos(ux) e^{-x} dx = -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)). \qquad (1)$
Now, the characteristic function is\begin{align*} \varphi_{X} (u) &= E[e^{iuX}] \\ &= \int_{-\infty}^{\infty} e^{iux} \cdot \frac{1}{2} e^{-|x|} dx \\ &= \frac{1}{2} \int_{-\infty}^{0} e^{iux} \cdot e^{-(-x)} dx + \frac{1}{2} \int_{0}^{\infty} e^{iux} \cdot e^{-x} dx \\ &= \frac{1}{2} \int_{-\infty}^{0} (\cos(ux) + i \sin(ux)) \cdot e^{x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\ &= \frac{1}{2} \int_{0}^{\infty} (\cos(ux) - i \sin(ux)) \cdot e^{-x} dx + \frac{1}{2} \int_{0}^{\infty} (\cos(ux) + i \sin(ux)) \cdot e^{-x} dx \\ &= \int_{0}^{\infty} \cos(ux) \cdot e^{-x} dx \\ &= -\frac{e^{-x}}{u^2+1} (\cos(ux) - u\sin(ux)) \Big|_{0}^{\infty} \qquad (1) \\ &= 0 - (- \frac{1}{u^2+1} (1 - 0)) \\ &= \frac{1}{u^2+1}. \end{align*}
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