Do we have any methods for evaluating $$\int_1^{\infty} \frac{1}{\Gamma(s)} \,ds$$? I thought about perhaps rewriting as $$\int_1^{\infty} \frac{\Gamma(1-s)}{\Gamma(1-s) \Gamma(s)} \,ds$$
$$=\frac{1}{\pi} \int_1^{\infty} \Gamma(1-s) \sin(\pi s) \,ds $$
But I'm not too sure if this is all that useful. Thoughts?
$$\frac{1}{\pi}\int_{1}^\infty\Gamma(1-s)\sin(\pi s)ds=\frac{1}{\pi}\int_{0}^\infty e^{-x}\int_1^\infty \sin(\pi s)e^{-s\log(x)}dsdx=-\frac{1}{\pi}\int_0^\infty e^{-x}\frac{ \frac{\pi}{x}}{\log^2(x)+\pi^2}dx=-\int_0^\infty \frac{e^{-x}}{x}\frac{dx}{\log^2(x)+\pi^2}$$ which looks similar to the Fransen-Robinson Constant and has no known closed form.
If we want to evaluate this integral, series expansion are not too bad.
Around $s=0$ $$ \frac{1}{\Gamma(s)}=s+\gamma s^2+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) s^3+\frac{1}{12} s^4 \left(2 \gamma ^3-\gamma \pi ^2-2 \psi ^{(2)}(1)\right)+O\left(s^5\right)\tag 1$$ Around $s=1$ $$ \frac{1}{\Gamma(s)}=1+\gamma (s-1)+\left(\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) (s-1)^2+\frac{1}{12} (s-1)^3 \left(2 \gamma ^3-\gamma \pi ^2-2 \psi ^{(2)}(1)\right)+\frac{(s-1)^4 \left(60 \gamma ^4-60 \gamma ^2 \pi ^2+\pi ^4-240 \gamma \psi ^{(2)}(1)\right)}{1440}+O\left((s-1)^5\right)\tag 2$$ As a function of the order of the expansions, we would get (in decimal representation) $$\left( \begin{array}{ccc} n & (1) & (2) \\ 1 & 0.500000 & 0.711392 \\ 2 & 0.692405 & 0.492766 \\ 3 & 0.528436 & 0.503267 \\ 4 & 0.520035 & 0.536575 \\ 5 & 0.547792 & 0.543607 \\ 6 & 0.541763 & 0.542233 \\ 7 & 0.540561 & 0.541331 \\ 8 & 0.541363 & 0.541201 \\ 9 & 0.541246 & 0.541223 \\ 10 & 0.541227 & 0.541234 \end{array} \right)$$ for an exact value $\approx 0.541236$.
You can express in terms of Nu Function:
$\int_1^{\infty}\dfrac{1}{\Gamma(s)}~ds=\int_0^{\infty}\dfrac{1}{\Gamma(s+1)}~ds=\nu(1)$
