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When computing the characteristic function of Cauchy distribution, we applied the Cauchy Integration theorem:

$$ \int_{C_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=\int_{-R}^{R}\frac{e^{i\alpha z}}{z^{2}+1}dz+\int_{\Gamma_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=I_{R}+J_{R} $$

We assume $\alpha>0$ and we use the curve from $(-R,0)$ to $(R,0)$ and back to $(-R,0)$ counter clockwise from the positive half plane (imaginary part $>0$) .

$C_{R}$ is the counter-clockwise contour, $\Gamma_{R}$ is the counterclockwise half circle on the upper half plane. The final result is: $$ \pi e^{-\alpha} $$

I am curious where did we use the condition $\alpha>0$ in this derivation?

I suspect it to be the choice of the integration contour. But how? Can we choose $\alpha<0$ while integrate over the same positive contour and get the same result but it is unstable for $\alpha<0$

Thanks in advance.

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    In the future, please make your questions more complete. I didn't know what $C_R$ or $\Gamma_R$ were, and the relevant steps in the calculation (pretty much everything before "The final result is:") were omitted. Luckily Google found a reference for me, otherwise I wouldn't have been able to help. Presumably the full derivation was given in your text. It would have been much more efficient to ask specific questions about the parts you did not understand.2012-02-27
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    If you hadn't assumed $\alpha>0$, the bottom line would have been $\pi e^{-|\alpha|}$.2012-02-27

2 Answers 2

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This is to ensure that $J_R\to0$ when $R\to+\infty$. Note that, if $z=R\mathrm e^{\mathrm it}$, $|\mathrm e^{\mathrm i\alpha z}|=\mathrm e^{-\alpha R\sin(t)}$ and that $\sin(t)\gt0$ for every $t$ in $(0,\pi)$, that is, on the path $\Gamma_R$. Hence, when $R\to+\infty$, $|\mathrm e^{\mathrm i\alpha z}|\to0$ if $\alpha\gt0$ but $|\mathrm e^{\mathrm i\alpha z}|\to\infty$ if $\alpha\lt0$.

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    Do you mean $|e^{i\alpha z}|=e^{-\alpha R\sin(t)}$? I understand your answer, really helps, thanks!2012-02-27
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    Absolutely, thanks for spotting this typo.2012-02-27
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    The end should be $\alpha<0$2014-01-09
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    @Danielsen Yep, thanks.2014-01-09
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Yes, the requirement $a>0$ determines the contour you choose. When $\Gamma_R$ is a half-circle in the upper half-plane and $a>0$ it can be shown that

$$\lim_{R \to \infty} \int_{\Gamma_R}\frac{e^{i\alpha z}}{z^{2}+1}dz = 0.$$

(When $a<0$ you choose $\Gamma_R$ to be a circle in the lower half-plane.) Thus you can conclude that

$$\lim_{R \to \infty} \int_{C_{R}}\frac{e^{i\alpha z}}{z^{2}+1}dz=\int_{-\infty}^{\infty}\frac{e^{i\alpha z}}{z^{2}+1}dz,$$

and the integral on the left can be calculated by the method of residues.

The full derivation can be found in this PDF.

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    Thanks, it is the $J_R \rightarrow 0 $ part that I was missing. ^_^2012-02-27