Writing an Integral in Different Form?

Question

Is it possible to write $\int_{-\infty}^\infty e^{-iqu}\left( \frac{A}{\sqrt{1-iau}}+\frac{B}{\sqrt{1-ibu}}\right)^Ndu$ in this form $\int\frac{e^{-iu}}{(\sqrt{1-iau})^n(\sqrt{1-ibu})^m}du$

Answer 1

Raskolnikov is absolutely right: $$\int e^{-\mathrm{i} qu} \left( \frac{A}{\sqrt{1-\mathrm{i}au}} + \frac{B}{\sqrt{1-\mathrm{i}bu}} \right)^N du = \sum_{k=0}^N \binom{N}{k} \int e^{-\mathrm{i} qu} \left[ \frac{A}{\sqrt{1-\mathrm{i}au}}\right]^{N-k} \left[ \frac{B}{\sqrt{1-\mathrm{i}bu}} \right]^{k} du$$

See here.
Sincerely

Robert

Answer 2

In view of your recent question on MathOverflow, this new question is regarding (inversion of) characteristic functions. Let's begin with the characteristic function corresponding to the right integral. Suppose that $W \sim {\rm N}(0,1)$. Then $W^2$ has characteristic function of chi-square with $1$ degree of freedom: $$ {\rm E}[e^{iuW^2 } ] = \frac{1}{{\sqrt {1 - 2iu} }}. $$ From this it follows straightforwardly that the characteristic function of the sum $$ {\frac{a}{2}} \sum\limits_{i = 1}^n W_i^2 + {\frac{b}{2}} \sum\limits_{i = 1}^m {\tilde W}_i^2, $$ where $W_i$ and ${\tilde W}_i$ are independent copies of $W$, is given by $$ {\rm E}\bigg[\exp \bigg(i\frac{{au}}{2}\sum\limits_{i = 1}^n {W_i^2 } \big)\bigg]{\rm E}\bigg[\exp \bigg(i\frac{{bu}}{2}\sum\limits_{i = 1}^m {{\tilde W}_i^2 } \bigg)\bigg] = \frac{1}{{(\sqrt {1 - iau} )^n }}\frac{1}{{(\sqrt {1 - ibu} )^m }}. $$ As for the characteristic function corresponding to the left integral, define random variable $T$ as follows. With probability $A$, $T = (a/2)W_1^2$, and with probability $B(=1-A)$, $T = (b/2)W_2^2$, where $W_1$ and $W_2$ are independent ${\rm N}(0,1)$ rv's. Then, the characteristic function of $T$ is given by $$ A{\rm E}\bigg[\exp \bigg(i\frac{{au}}{2}W_1^2 \bigg)\bigg] + B{\rm E}\bigg[\exp \bigg(i\frac{{bu}}{2}W_1^2 \bigg)\bigg] = \frac{A}{{\sqrt {1 - iau} }} + \frac{B}{{\sqrt {1 - ibu} }}. $$ Thus, the characteristic function of the sum of $N$ independent copies of $T$ is given by $$ \bigg(\frac{A}{{\sqrt {1 - iau} }} + \frac{B}{{\sqrt {1 - ibu} }}\bigg)^N. $$