This is the integral of the reciprocal gamma function,$$F=\int_0^\infty dx/\Gamma(x)= e+\int_0^\infty \frac{e^{-t}\ dt}{\log^2 t+\pi^2}$$ $$=e+\int_{-\infty}^{\infty}\frac{\exp(s-e^s)\ ds}{s^2+\pi^2}$$ $$=e+\frac1{2\pi i}\int_{-\infty}^{+\infty}[\frac1{s-i\pi}-\frac1{s+i\pi}]\exp(s-e^s)\ ds$$Now, defining $f(s):=\exp(s-e^s)$ and observing that $f(s+2\pi i)=f(s)$ and $f(i\pi)=-e$, consider the rectangular closed contour $\mathcal{C}=[-A,A]\cup[A,A+2\pi i]\cup[A+2\pi i,-A+2\pi i]\cup[-A+2\pi i,-A]$. The contour integral$$\frac1{2\pi i}\oint_{\mathcal C}\frac{f(s)\ ds}{s-i\pi}=f(i\pi)=-e$$and also$$=\frac1{2\pi i}\int_{-A}^A [\frac1{s-i\pi}-\frac1{s+i\pi}]f(s)\ ds+\frac1{2\pi i}(\int_{-A+2\pi i}^{-A}+\int^{A+2\pi i}_{A})\frac{f(s)\ ds}{s-i\pi}$$ Taking the $\lim_{A\to\infty}$, and as the "left branch" $\int_{-A+2\pi i}^{-A}\frac{f(s)\ ds}{s-i\pi} \to 0$, we get $$F=-\lim_{A\to\infty}\frac{e^A}{2\pi}\int_0^{2\pi}\frac{\exp(-e^A \cos\theta-ie^A \sin\theta)e^{i\theta}\ d\theta}{A+i(\theta-\pi)}$$
Question : does there exist a method, standard (Laplace, stationary phase, saddle point...) or otherwise, to make this limit explicit ?
