Change of order of integration of a triple integral Consider $$ I = \int_0^{\omega}\int_0^{\alpha}\int_0^{\alpha}F(\beta){\tilde{F}(\gamma)}e^{i\beta t}e^{-i\gamma t}R(\alpha)d\beta d\gamma d\alpha$$
In this triple integral,I want to bring about, a change of order of integration, where in I take integration with respect to $\alpha$ inside most and integration with respect to other variables outside. Appreciate some help in this regard.
Here $\tilde{F}(\gamma)$ is the complex conjugate of $F(\gamma)$.
My approach yielded me this result, which I am not sure about :
$$I = \int_0^{\omega}\int_0^{\gamma}F(\beta){\tilde{F}(\gamma)}e^{i(\beta-\gamma )t}\int_0^{2(\omega-\gamma)}R(\alpha)d\alpha d\beta d\gamma $$

What you have is wrong. 
Hint. The original integral is over the domain consisting of all triples $(\alpha,\beta,\gamma)$ for which
$$0 \leq \beta, \gamma \leq \alpha \leq \omega.$$
To change the order, note that this is equivalent to picking $\beta$ and $\gamma$ anywhere in $[0,\omega]$, and then picking $\alpha$ in the interval $[\max(\beta,\gamma),\omega]$.
Edit: It might be convenient to divide the domain into two parts, one where $\beta \leq \gamma$, and one where $\gamma \leq \beta$.
