Good day to everyone.
In my scientific research I've got stuck with a contour integration problem.
I would like to evaluate the following integral:
$$I=\int_0^{\infty } \frac{e^{\frac{\alpha \mathrm i \Omega \beta }{1+\Omega ^2\sigma ^2}}}{\left(1+\Omega ^2\sigma ^2\right)^L\Omega } \, d\Omega \tag{1}$$
Where $\alpha, \beta, \sigma \in \mathbb{R}$ and $\sigma>0$.
First of all I've changed the variable to $x=\sigma\Omega$ and turned the exponent part in partial fractions with $A,B$ depending only on $\alpha, \beta, \sigma$, treating them as constants.
$$I=\int_0^{\infty } \frac{e^{\frac{A}{1+\mathrm i x }+\frac{B}{1\mathrm i x }}}{(1+\mathrm i x )^L(1\mathrm i x )^Lx } \, dx \tag{1*}$$
Then performed linear fractional transformation (Möbius transformation):
$$
\begin{cases}
x =\mathrm i\frac{p1}{p+1} \\ p=\frac{1\mathrm i x }{1+\mathrm i x }
\end{cases} \tag{2}
$$
$$I=\frac{e^{\frac{a}{2}}}{2^{2L1}}\int _{\gamma }\frac{(1+p)^{2L1}}{(1p)p^L}e^{\frac{B}{2}p+\frac{A}{2}\frac{1}{p}}dp\tag{3}$$
where $\gamma$ is and arc in complex pplane, running counterclockwise from $1$ to $1$ (is that correct?).
Then using the binomial theorem for $(1+p)^{2L1}$ I get:
$$I=\frac{e^{\frac{a}{2}}}{2^{2L1}}\int _{\gamma }\frac{\sum_{k=0}^{2L1} {{2L1} \choose k}p^k}{(1p)p^L}e^{\frac{B}{2}p+\frac{A}{2}\frac{1}{p}}dp\tag{4}$$
$$ I=\sum_{k=0}^{2L1} {{2L1} \choose k}\frac{e^{\frac{a}{2}}}{2^{2L1}}\underbrace{\int _{\gamma }\frac{1}{(1p)p^{Lk}}e^{\frac{B}{2}p+\frac{A}{2}\frac{1}{p}}dp}_{I_1}\tag{4*}
$$
So integral $I_1$ is of primary interest.
Form Connections between the Generalized Marcum QFunction and a class of Hypergeometric Functions by D. MoralesJimenez, F. J. LopezMartinez, E. MartosNaya, J. F. Paris, A. Lozano I know that
$$Q_m(a,b)e^{\frac{A^2+B^2}{2}}=\oint_\Gamma \frac{1}{(1p)p^{m}}\exp\left(\frac{1}{2}\left({B p+\frac{A}{p}}\right)\right)dp \tag{5}$$
where $Q_m(a,b)$ is the generalized Marcum Qfunction and $\Gamma$ is "any closed contour enclosing the singularity at $p = 0$ (in a counterclockwise direction) and no other singularities of the integrand".
So it would be perfect if $\gamma=\Gamma$ then what I have is just a sum of Marcum Qfunctions. But it does not.
Ar first I thought about closing the contour in a manner like that
But then realized that I'm not able to cope with integrals alond the real axe (the blue one's).
So, is there any way to cope with this integral? And what about the singularity at $p=1$?
Any hints would be of great help.
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1$\begingroup$ isn't your integral logarithmically divergent at $\Omega=0$? $\endgroup$ – Carlo Beenakker Sep 26 '13 at 7:01