The real and imaginary parts of the Incomplete Gamma function for second argument being purely imaginary Dear all,
I am looking for explicit (at least more explicit than the original expression) for
1) Re$(\Gamma(a, i\omega))$ 
as well as
2) Im$(\Gamma(a, i\omega)),$
where i Re and Im denote the real and imaginary part, and $\Gamma(a, i\omega)$ is the Incomplete Gamma function with the arguments $a$ and $i\omega$. The letter $i$ denotes imaginary unit, $a>0$ is a real number and $\omega$ is also a real number.
I would presume that 1) and 2) could be written as functions of $\Gamma(a, \omega)$, however by application of change of variables in the integral definition of the Incomplete Gamma function did not succeed for me.
Thank you in advance for any kind of input.
 A: The usual integral representation gives you
$$\Gamma(a,i\omega) = \Gamma(a) - (i\omega)^a \int_0^1 t^{a-1} e^{-i\omega t}\ dt$$
I assume you have no trouble writing $(i\omega)^a$ in terms of its real and imaginary parts.
As for the integral, Maple says
$$ \eqalign{\int_0^1 t^{a-1} \cos(\omega t)\ dt &= 
{\frac { \left( {w}^{-a-1/2}\sin \left( w \right) -{w}^{-a+1/2}\cos \left( w
 \right)  \right) {S_1} \left( a+1/2,1/2,w \right) }{a}}\cr &-{\frac {{w}^{-a+1/2}{S_1} \left( a+3/2,3/2,w \right) \sin
 \left( w \right) }{ \left( 2+a \right) a}}\cr &+{\frac { \left( 2
+a \right) \cos \left( w \right) +w\sin \left( w \right) }{ \left( 2+a
 \right) a}}
\cr}
$$
$$ \eqalign{\int_0^1 t^{a-1} \sin(\omega t)\ dt &=-{\frac {{w}^{-a+1/2}\sin \left( w \right) {S_1} \left( a+1/2,3
/2,w \right) }{  a+1 }}\cr+&{\frac { \left( {w}^{-a
+1/2}\cos \left( w \right) -{w}^{-a-1/2}\sin \left( w \right)  \right) {S_1} \left( a+3/2,1/2,w \right) }{ \left( a+1 \right) a}
}\cr+&{\frac { \left( a+1 \right) \sin \left( w \right) -\cos \left( w
 \right) w}{ \left( a+1 \right) a}}
\cr}$$ 
where $S_1$ is the Lommel S1 function.
