When one wants to calculate the characteristic function of a random variable which is of normal distribution, things boil down to calculate: $\int_{-\infty}^{+\infty}e^{-\frac{(x-it)^2}{2}}dx$ There are several ways to calculate this integral. I tried to calculate this integral using contour integration: $ \oint_C f(z)dz=\int_{-a}^af(z)dz+\int_{Arc(a)}f(z)dz $ where $ f(z)=e^{-\frac{(z-z_0)^2}{2}}, z_0=it $ and $C$ is the union of a semicircle and $[-a,a]$. How can I calculate $ \lim_{a\to+\infty}\int_{Arc(a)}f(z)dz? $ Alternatively, from the very beginning, I get $ \lim_{a\to+\infty}\int_{-a-z_0}^{a-z_0}e^{-\frac{z^2}{2}}dz. $ But I have no idea how to choose contour.
Calculate $\int_{-\infty}^{+\infty}e^{-\frac{(x-it)^2}{2}}dx$ using contour integration
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0Using an arc is overkill. What integral that is "like" this integral but has a solution that you know? – 2012-10-26
3 Answers
This assumes you already know that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$. If that is not already known, this proof will not work.
For $N>0$, let $C_N$ be the rectangle curve that goes from $-N+0i$ to $N+0i$, then $N+0i$ to $N+ti$, then from $N+ti$ to $-N+ti$ and finally from $-N+ti$ to $-N+0i$.
Then $\int_{C_N} e^{-z^2} dz=0$. Note that the size of the contribution of the sides of the rectangle approach zero as $N\to\infty$, so that means that $\lim_{N\to\infty} \left(\int _{-N}^N e^{-x^2}dx - \int_{-N}^N e^{-(x+ti)^2} dx\right) = 0$
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0Fair enough, I see. Thanks. – 2012-10-26
Read the 8-th proof in this delightful paper by K. Conrad. Since he defines a complex function that seems taken out of the blue, you can read there also an explanation for it and references for past papers and books that have that idea.
Why not just do this?: $ \begin{eqnarray} I\left(t\right) &=& \int_{-\infty}^{\infty} dx \exp\left[-\frac{1}{2}\left(x-i t\right)^2\right] \\ &=& \int_{-\infty-i t}^{\infty - i t} du \exp\left(-\frac{1}{2} u^2\right) \end{eqnarray} $ Now deform the contour so that part of it goes along the real axis: $ \begin{eqnarray} I\left(t\right)&=& \lim_{r \rightarrow \infty} \left\{\int_{-t}^{0} dy \exp\left[-\frac{1}{2} \left(-r+iy\right)^2\right] + \int_{0}^{-t} dy \exp\left[-\frac{1}{2} \left(r+iy\right)^2\right]\right\} \\ &+& \int_{-\infty}^{\infty} dx \exp\left(-\frac{1}{2} x^2\right) \\ &=& \lim_{r \rightarrow \infty} \int_{-t}^{0} dy\left\{ \exp\left[-\frac{1}{2} \left(-r+iy\right)^2\right] - \exp\left[-\frac{1}{2} \left(r+iy\right)^2\right]\right\} + \sqrt{2 \pi} \\ &=& 2 i \lim_{r \rightarrow \infty} e^{-r^2/2} \int_{-t}^{0} dy\left\{ e^{y^2/2} \sin \left(r y\right)\right\} + \sqrt{2 \pi} \\ &=& \sqrt{2 \pi} \end{eqnarray} $