If I have a real parameter $\Omega > 0$, and I consider the modified Bessel function of the second kind: $$ K_{i\Omega}(x) $$ What is a general formula I can use for an order $i\Omega$? If I have $x>0$, is there a simply way of writing $K_{i\Omega}(x)^{\ast}?$ in terms of $K_{i\Omega}(x)$ (or something along those lines)?
I found the following integral representation: $$ K_{\nu}(z) = \frac{\Gamma(\nu+\tfrac{1}{2})(2z)^{\nu}}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\cos(t)dt}{(t^{2}+z^{2})^{\nu+\frac{1}{2}}} $$
But I'm worried that this integral won't be valid for purely imaginary $\nu$. Blindly using the above integral, I find: $$ K_{i\Omega}(x)^{\ast} = \frac{\Gamma(\tfrac{1}{2} - i \Omega)(2z)^{-i\Omega}}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\cos(t)dt}{(t^{2}+x^{2})^{-i\Omega+\frac{1}{2}}} \ = \ K_{-i\Omega}(x) $$
So it seems that $K_{i\Omega}(x)^{\ast} = K_{-i\Omega}(x)$...is this okay?