The real and imaginary parts of the Incomplete Gamma function for second argument being purely imaginary

Question

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.

Answer 1

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.