Antiderivative of the reciprocal Gamma function I have a question about the evaluation of the following function for $x>1$:
$$\Omega(x)=\int_0^x\frac{1}{\Gamma(s)}\;ds$$
In order to try to evaluate $\Omega(x)$ I used the reflection formula, the definition of the Gamma function, and formula $(105)$ here (as well as rearranging integrals) to get:
$$\int_0^x\frac{1}{\Gamma(s)}\;ds=\frac{1}{\pi}\int_0^x\Gamma(1-s)\sin\pi s\;ds=\frac{1}{\pi}\int_0^x\sin\pi s\int_0^\infty t^{-s}e^{-t}\;dt\;ds={\frac{1}{\pi}\int_0^\infty e^{-t}\int_0^x e^{-s\ln{t}}\sin\pi s\;ds\;dt}={\frac{1}{\pi}\int_0^\infty e^{-t}\frac{1}{\pi^2+\ln^2t}\left[t^{-x}(-\ln(t)\sin\pi x-\pi\cos\pi x)+\pi\right]\;dt}$$
$$\therefore\int_0^x\frac{1}{\Gamma(s)}\;ds=\int_0^\infty\frac{e^{-t}}{\pi^2+\ln^2t}\;dt-\frac{\sin\pi x}{\pi}\int_0^\infty\frac{e^{-t}\;t^{-x}\ln{t}}{\pi^2+\ln^2t}\;dt-{\cos\pi x\int_0^\infty\frac{e^{-t}\;t^{-x}}{\pi^2+\ln^2t}\;dt}\tag{1}$$
Now for values I have tested for $x\le1$ this expression appears to give the right answer. In particular for $x=1$ we get the following result (numerically correct to 6 decimal places):
$$\int_0^1\frac{1}{\Gamma(s)}\;ds=\int_0^\infty\frac{e^{-x}\left(1+\frac{1}{x}\right)}{\pi^2+\ln^2x}\;dx\tag{2}$$
and the following family of results (which are also numerically correct to a few decimal places):
$$\int_{-1}^0\frac{1}{\Gamma(s)}\;ds=\int_0^\infty\frac{e^{-x}\left(1+x\right)}{\pi^2+\ln^2x}\;dx\tag{3}$$
and in general:
$$\int_{-n}^0\frac{1}{\Gamma(s)}\;ds=\int_0^\infty\frac{e^{-x}\left[1+(-1)^{n+1}x^n\right]}{\pi^2+\ln^2x}\;dx\tag{4}$$
However larger values do not seem to work so well; it does not appear that we can get convergent results for $x>1$; in particular we cannot take a limit as $x\rightarrow\infty$ to find $\int_0^\infty\frac{1}{\Gamma(s)}\;ds$. Now I am aware from here that we have the following beautiful result:
$$\int_0^\infty\frac{1}{\Gamma(s)}\;ds=e+\int_0^\infty\frac{e^{-t}}{\pi^2+\ln^2t}\;dt\tag{5}$$
and the proof is certainly not elementary. I had been hoping to derive this result even if not rigorously for myself, but this route has not worked, although tantalizingly we do have the integral $\int_0^\infty\frac{e^{-x}}{\pi^2+\ln^2x}dx$ present. I believe that what prevents the derivation of $(1)$ extending to $x>1$ is that the integral representation of $\Gamma(s)$ may only be useful for $s\ge0$. What I am wondering is whether a simple manipulation like $(1)$ may be extended in any way to some $x>1$.

Thus my question is: Can the derivation of $(1)$ be extended to derive an expression valid for larger values of $x$ (and is there such an expression)? In particular can we use such a method to derive $(5)$ simply? If not, what is the underlying reason that it does not work? I would also like to know whether $(2)$ and $(3)$ and $(4)$ are in fact correct.
 A: I like what you were doing in (1) but my two-pennies worth observe also that for $\Re(z) < 2$
\begin{align}
\frac{1}{\Gamma(z)} &= \frac{\sin (\pi z)}{\pi z}\Gamma(1-z) \\
&= \frac{\sin (\pi (z-1))}{\pi(z-1)}\Gamma(2-z) \\
&= \frac{\sin (\pi (z-1))}{\pi(z-1)} \int_0^\infty e^{-t}t^{1-z}dt
\end{align}
Now observe that
$$2i \sin (\pi (z-1)) = e^{i \pi (z-1)}-e^{-i \pi (z-1)}$$
Whence
\begin{align}\tag{1}
\frac{1}{\Gamma(z)} &= \frac{1}{z-1}\frac{1}{2 \pi i}\int_0^\infty e^{-t}\left(e^{(z-1)(-\log(t)-i \pi)} - e^{(z-1)(-\log(t)+i \pi)}\right)
\end{align}
Now, if at the outset we wrote
$$\frac{1}{\Gamma(z)} = \sum a_n z^n$$ and then replacing $z$ by $z-1$ and dividing both sides by $z-1$ we find
$$\frac{1}{\Gamma(z)} = \frac{1}{z-1} \left(a_1 (z-1)+a_2 (z-1)^2+a_3(z-1)^3 \ldots \right)$$
Now comparing this with (1) we find that the coefficients, $a_n$ for $n \geq 1$  are represented by
\begin{align}
a_n &= \frac{1}{2 \pi i n!}\int_0^\infty e^{-t}\lim_{z \to 1}\frac{d^n}{dz^n}\left(e^{(z-1)(-\log(t)-i \pi)} - e^{(z-1)(-\log(t)+i \pi)}\right)\\
&= \frac{1}{2 \pi i n!}\int_0^\infty e^{-t}\left( (-\log(t)-i \pi)^n - (-\log(t)+i \pi)^n\right)\\
&=\frac{1}{ \pi i n!}\int_0^\infty e^{-t}\Im(-\log(t)-i \pi)^n \\
&=\frac{(-1)^n}{ \pi i n!}\int_0^\infty e^{-t}\Im(\log t-i \pi)^n
\end{align}
This integral representation of the coefficients are tricky to evaluate, but I believe there will be an asymptotic representation to evaluate them.
