Imaginary Factorial?

We want $|\Gamma(1 + 2i)|$ . Now $\big|\Gamma(1+2i)\big|^2 \; = \; \Gamma(1 + 2i)\Gamma(1-2i) \; = \; \Gamma(2)B(1+2i,1-2i) \; = \; B(1+2i,1-2i)$ and hence $\big|\Gamma(1+2i)\big|^2 \; = \; \int_0^\infty \frac{u^{2i}}{(1+u)^2}\,du$ If we integrate $f(u) = \frac{u^{2i}}{(u+1)^2}$ around the "keyhole contour" $\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4$ consisting of the segments (for $0 < \epsilon < 1 < R$ ):

• $\gamma_1$ , the straight line segment from $\epsilon$ to $R$ just above the positive real axis,
• $\gamma_2$ , the circular arc $u = Re^{i\theta}$ for $0 \le \theta \le 2\pi$ ,
• $\gamma_3$ , the straight line segment from $\epsilon e^{2\pi i}$ to $Re^{2\pi i}$ just below the positive real axis,
• $\gamma_4$ , the circular arc $u = \epsilon e^{i\theta}$ for $0 \le \theta \le 2\pi$ ,

where we have cut the complex plane along the positive real axis, so that $0 \le \mathrm{Arg}\,u \le 2\pi$ for all $u$ , then $\int_{\gamma_1}f(u)\,du \; = \; \int_\epsilon^R \frac{u^{2i}}{(u+1)^2}\,du \qquad \int_{\gamma_3} f(u)\,du \; =\; e^{-4\pi}\int_\epsilon^R \frac{u^{2i}}{(u+1)^2}\,du$ while $\int_{\gamma_2} f(u)\,du \; =\; O(R^{-1}) \qquad \qquad \int_{\gamma_4}f(u)\,du \; = \; O(\epsilon)$ as $R \to \infty$ and $\epsilon \to 0$ respectively. Hence, letting $R \to \infty$ and $\epsilon \to 0+$ , $(1 - e^{-4\pi})\int_0^\infty \frac{u^{2i}}{(u+1)^2}\,du \; =\; \left(\int_{\gamma_1} + \int_{\gamma_2} - \int_{\gamma_3} - \int_{\gamma_4}\right)f(u)\,du \; = \; 2\pi i\mathrm{Res}_{u=-1} f(u) \; = \; 4\pi e^{-2\pi}$ so that $\big|\Gamma(1+2i)\big|^2 \; =\; \frac{4\pi e^{-2\pi}}{1 - e^{-4\pi}} \; = \; 2\pi \mathrm{cosech}2\pi \; = \; \frac{4\pi e^{2\pi}}{e^{4\pi}-1}$ and hence $\big| \Gamma(1+2i)\big| \; = \; 2e^{\pi} \sqrt{\frac{\pi}{e^{4\pi}-1}}$ making the answer $2 + 4 + 1 = \boxed{7}$

Start with:

$\big|\Gamma(1+2i)\big|^2 = \Gamma(1 - 2i)\Gamma(1+2i)$

Use the Recursive relation of the Gamma Function :

$\displaystyle =(-2i) \Gamma(- 2i)\Gamma(1+2i)$

Apply Euler's Reflection Formula of the Gamma Function :

$\displaystyle = (-2i) \frac{\pi}{\sin (\pi (-2i)) } = \frac{2\pi i}{\sin (2\pi i) }$

Recall the Trigonometric-Hyperbolic Functions relations:

$\displaystyle = \frac{2\pi i}{i \sinh (2\pi ) } = \frac{4\pi }{ e^{2\pi} - e^{-2\pi} } = 4e^{2\pi} \frac{\pi}{e^{4\pi} -1}$

Hence:

$\displaystyle \big|\Gamma(1+2i)\big| = \sqrt{4e^{2\pi} \frac{\pi}{e^{4\pi} -1}} = \boxed{ 2e^{\pi} \sqrt{\frac{\pi}{e^{4\pi} -1}}}$

Making the answer: $2+4+1 = \boxed{7}$