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The integration function is very complicated, as below, where $\theta, k \in \mathbb{R}_{+}$ $$\int_{-\infty }^{\infty }\frac{{e}^{-i\,t\,x}\,\left(1- {e}^{200\,i\,t}\right) \,{\left( 1-i\,t\,\theta\right) }^{k}\,\left(1-{e}^{i\,t\,x}\right)}{{t}^{2}}dt$$ There ara so many multiplier in the integration, and result should be formed as a function to $x$, and $\theta,k$ as it's parameters. I have no idea about that now.

btw, this is not a home work. it is originate from my question about extreme value of my custom distrbution which is a sum of gamma distribution and uniform distribution, described here.

Above integration is from probablity theory. I multiply characteristic functions of gamma distribution and uniform distribution, and inverse the result function to get distribution function of new distribution. $\theta$ and $k$ have to be estimated from samples.

I checked another inverse method to convert characteristc function to distribution function the integration can also be expressed as below. $$\int_{0}^{\infty }\frac{\left( \left( \mathrm{cos}\left( 200\,t\right) -1\right) \,\mathrm{sin}\left( k\,\mathrm{atan}\left( t\,\theta\right) \right) -\mathrm{sin}\left( 200\,t\right) \,\mathrm{cos}\left( k\,\mathrm{atan}\left( t\,\theta\right) \right) \right) \,{\left( {t}^{2}\,{\theta}^{2}+1\right) }^{\frac{k}{2}}\,\mathrm{sin}\left( t\,x\right) }{{t}^{2}}dt$$+$$\int_{0}^{\infty }\frac{\left( \left( 1-\mathrm{cos}\left( 200\,t\right) \right) \,\mathrm{cos}\left( k\,\mathrm{atan}\left( t\,\theta\right) \right) -\mathrm{sin}\left( 200\,t\right) \,\mathrm{sin}\left( k\,\mathrm{atan}\left( t\,\theta\right) \right) \right) \,{\left( {t}^{2}\,{\theta}^{2}+1\right) }^{\frac{k}{2}}\,\mathrm{cos}\left( t\,x\right) }{{t}^{2}}dt$$

it is a little bit more complex than the one before.

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    Can you confirm or deny if that middle term is actually supposed to be $(1-e^{it\theta})^k$? Otherwise for $k$ not even that large the integral will diverge (and have no principal value either).2012-05-05
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    sure, i can add derivation as2012-05-05

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