The Gamma Function is defined as the following integral: $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}\, \text{d}x$$
For $z \in \mathbb{C}$, and $\text{Re}(z)$ is not a non-negative integer. If we let $z = \sigma + it$, we can expand out the integral definition and and express this function as a complex number:
$$\Gamma(\sigma + it) = \int_0^\infty x^{\sigma-1 +it}e^{-x}\, \text{d}x \\= \int_0^\infty x^{\sigma-1}e^{i(t\log x)}e^{-x}\, \text{d}x \\ \implies \Gamma (\sigma +it) = \int_0^\infty x^{\sigma-1}e^{-x}\cos(t\log x) \, \text{d}x + i \int_0^\infty x^{\sigma-1}e^{-x}\sin(t\log x) \, \text{d}x $$
Therefore, we can express the real and imaginary parts of the Gamma function as:
$$ \text{Re}(\Gamma(z)) = \int_0^\infty x^{\sigma-1}e^{-x}\cos(t\log x) \, \text{d}x \\ \text{Im}(\Gamma(z)) = \int_0^\infty x^{\sigma-1}e^{-x}\sin(t\log x) \, \text{d}x $$
While I know of fast fixed-precision algorithms for computing the Gamma function for all real numbers (excluding the negative integers), I’m not familiar with any that can compute these complex parts quickly.
Are there any fast algorithms out there that can compute these integrals quickly?
What’s more, these integrals look very familiar to the original integral definition of the Gamma function. Is it possible to express them in terms of that original definition?
The real and imaginary parts of $\Gamma(s)$ where $s=\sigma+i t\land \sigma,t\in\mathbb{R}$ can be derived from the Laplace transforms
$$\Re(\Gamma(\sigma+i t))=\mathcal{L}_x\left[x^{\sigma -1} \cos(t \log(x))\right](1)=\mathcal{L}_x\left[\frac{1}{2} \left(x^{\sigma+i t-1}+x^{\sigma-i t-1}\right)\right](1)$$ $$=\int_0^{\infty } \frac{1}{2} \left(x^{\sigma +i t-1}+x^{\sigma -i t-1}\right) e^{-x} \,dx=\frac{1}{2}\, (\Gamma(\sigma+i t)+\Gamma(\sigma-i t))\tag{1}$$
and
$$\Im(\Gamma(\sigma+i t))=\mathcal{L}_x\left[x^{\sigma -1} \sin(t \log(x))\right](1)=\mathcal{L}_x\left[-\frac{i}{2} \left(x^{\sigma +i t-1}-x^{\sigma -i t-1}\right)\right](1)$$ $$=\int\limits_0^{\infty} -\frac{i }{2} \left(x^{\sigma +i t-1}-x^{\sigma -i t-1}\right) e^{-x}\,dx=-\frac{i}{2}\, (\Gamma(\sigma+i t)-\Gamma (\sigma-i t))\tag{2}$$
which are equivalent to the Mellin transforms
$$\Re(\Gamma(\sigma+i t))=\frac{1}{2} \left(\mathcal{M}_x\left[e^{-x}\right](\sigma+i t)+\mathcal{M}_x\left[e^{-x}\right](\sigma-i t)\right)$$ $$=\frac{1}{2} \left(\int\limits_0^{\infty} e^{-x} x^{\sigma+i t-1}\,dx+\int\limits_0^{\infty} e^{-x} x^{\sigma-i t-1}\,dx\right)=\frac{1}{2} (\Gamma(\sigma +i t)+\Gamma(\sigma-i t))\tag{3}$$
and
$$\Im(\Gamma(\sigma+i t))=-\frac{i}{2} \left(\mathcal{M}_x\left[e^{-x}\right](\sigma+i t)-\mathcal{M}_x\left[e^{-x}\right](\sigma-i t)\right)$$ $$=-\frac{i}{2} \left(\int_0^{\infty} e^{-x} x^{\sigma+i t-1}\,dx-\int_0^{\infty} e^{-x} x^{\sigma-i t-1}\,dx\right)=-\frac{i}{2} (\Gamma(\sigma+i t)-\Gamma(\sigma-i t))\tag{4}$$
and these results also follow from
$$\Re(\Gamma(s))=\Re\left(\Gamma\left(s^*\right)\right)\tag{5}$$
and
$$\Im(\Gamma(s))=-\Im\left(\Gamma\left(s^*\right)\right)\tag{6}$$
where $s^*$ is the complex conjugate of $s$.
There are many formulas for evaluating $\Gamma(s)$ at the Gamma page of the Wolfram function site.