# How to simplify complex integration?

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). – anon May 5 '12 at 15:03
sure, i can add derivation as – Readon Shaw May 5 '12 at 15:11

Since you hinted at the probabilistic origin of the integral in question, it is likely that $k$ is actually negative, since the characteristic function of the $\Gamma(k,\theta)$ random variable is $\phi_X(t) = \left(1- i t \theta\right)^{-k}$. Then, for $U_1$ and $U_2$ independent uniformly distributed random variables on intervals $(0,d_1)$ and $(0,d_2)$, and both independent on $X$, and for r.v. $Z = X + U_1 + U_2$ we get: $$\phi_Z(t) = \phi_{X}(t) \phi_{U_1}(t) \phi_{U_2}(t) = \exp\left(i \frac{d_1+d_2}{2} t\right) \operatorname{sinc}\left(\frac{ d_1 t}{2} \right) \operatorname{sinc}\left(\frac{ d_2 t}{2} \right) \left(1 - i t \theta\right)^{-k}$$ Now, you would like to compute the probability density function $f_Z(z)$ by means of the inverse Fourier transform: $$f_Z(z) = \frac{1}{2 \pi} \int_{-\infty}^\infty \mathrm{e}^{-i t z} \phi_Z(t) \mathrm{d} t$$ The you seem to be interested in $f_Z(d_2)$. The answer will depend on whether $d_1 < d_2$, or $d_1 > d_2$.