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If $q(x)$ is the pdf, I can write it in terms of the characteristic function:

$$q(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-i\phi x} f_j(x,t,\phi)\, d\phi. $$

I see from the literature I'm reading that I can rewrite this as

$$q(x) = \frac{1}{2\pi}\int_{0}^{\infty} \left[e^{-i\phi x} f_j(x,t,\phi) +e^{i\phi x} f_j(x,t,-\phi)\right]\, d\phi. $$

I'm trying to arrive at this form by splitting the original integral:

$$q(x) = \frac{1}{2\pi}\left[\int_{-\infty}^{0} e^{-i\phi x} f_j(x,t,\phi) \,d\phi + \int_{0}^{\infty} e^{-i\phi x} f_j(x,t,\phi)\, d\phi\right]. $$

To get to the final form, this must mean that the integral from $-\infty$ to $0$ of the characteristic function is equal to the integral from $0$ to $\infty$ of the complex conjugate of the integrand. Is this true, and why? Thank you!

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1 Answer 1

up vote 2 down vote accepted

You don't need to take the complex conjugate. Just substitute $\psi = -\phi$ into your first integral!

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D'oh! I thought it would be something simple like that. Thank you! –  SingularityFuture Nov 28 '12 at 18:12

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