Proof of an identity connecting a generalized Hypergeometric function to exponential integral. In this post one can find a proof of the following identity 
$$ \sum^{\infty}_{m=1} \frac{1}{m} \frac{(-x)^m}{m!} = Ei(-x)-\ln(x)-\gamma, \qquad (1)$$
for $x>0$. On the other hand, it is easy to see that 
$$ \sum^{\infty}_{m=1} \frac{1}{m} \frac{(-x)^m}{m!} = -x \cdot \ _2F_2(1,1;2,2;-x). \qquad (2) $$
For a definition of the generalized hypergeometric function, see this page. Comparing (1) and (2) implies that we should have the following identity
$$  -x \cdot \ _2F_2(1,1;2,2;-x) = Ei(-x)-\ln(x)-\gamma, \qquad x>0. \qquad (3) $$
I want to establish (3) independently, more precisely from the integral representation of generalized hypergeometric function which can be found here. From the equation given in DLMF and the identity $\Gamma(z+1)=z\Gamma(z)$ we can write
$$ _2F_2(1,1;2,2;-x) = \frac{1}{2\pi i} \int_{C} \frac{\Gamma(-s)}{(1+s)^2}x^sds, $$
where $C$, as suggested by DLMF, is any contour which separates the poles of $\Gamma(-s)$ from those of $\Gamma(1+s)$. I choose $$C := \Sigma_{-,\varepsilon} \cup \Sigma_{+,\varepsilon} \cup C_{\varepsilon},$$ where $\Sigma_{-,\varepsilon}$ starts at $-\infty-i\varepsilon$, continues to the right along $y=-\varepsilon$ to reach $-1-\varepsilon - i \varepsilon$, $\Sigma_{+,\varepsilon}$ starts at $-1-\varepsilon + i \varepsilon$  and continues to the left along $y=\varepsilon$ all the way to $-\infty + i \varepsilon$, and $C_{\varepsilon}$ encircles $-1$ in a ccw way and connects $-1-\varepsilon - i \varepsilon$ to $-1-\varepsilon + i \varepsilon$. Since the integrand is analytic on $(-\infty,-1-\varepsilon)$, the integral over $C$ can be replaced by the integral over the positively oriented circle $C'_{\varepsilon}$ of radius $\varepsilon$ centered at $-1$ (by deforming the  $\Sigma_{-,\varepsilon} $ and $ \Sigma_{+,\varepsilon}$ to $(-\infty,-1-\varepsilon)$ and the corresponding integrals get cancelled). Thus by a residue calculation we have 
$$ _2F_2(1,1;2,2;-x) = \frac{1}{2\pi i} \int_{C'_{\varepsilon}} \frac{\Gamma(-s)}{(1+s)^2}x^sds = \frac{d}{ds} \left( \Gamma(-s)x^s \right) \bigg|_{s=-1} = $$
$$ -\Gamma'(-s)x^s + \ln(x) \Gamma(-s)x^s \bigg|_{s=-1}  = \frac{\gamma + \ln(x)}{x}, $$
using $\Gamma'(1)=-\gamma$. Hence
$$ -x \cdot  _2F_2(1,1;2,2;-x) = - \ln(x) - \gamma. $$
So I have not shown (3), as I am missing $Ei(-x)$ on the right hand side. I suspect that I am missing something when I am collapsing the integrals on $\Sigma_{-,\varepsilon} $ and $ \Sigma_{+,\varepsilon}$ but I do not see it... I appreciate any help in advance!
 A: Basically, your mistake is the choice of $C$. The integrals over $\Sigma_{-,\varepsilon}$ and $\Sigma_{+,\varepsilon}$ both diverge, and this breaks the rest of the above. It's important to understand the "how and when" of $\int_C\Gamma(-s)F(s)\,ds$. Surely, if $F(s)$ is regular at $s\in\mathbb{Z}_{\geqslant 0}$, and $C_n$ (is simple positively oriented and) encircles $s=0, \ldots, s=n$ but not other singularities of the integrand, then $$\int_{C_n}\Gamma(-s)F(s)\,ds=2\pi i\sum_{k=0}^{n}(-1)^{k-1}F(k)/k!$$ (by the residue theorem). But if we're using unbounded contours, we must justify validity of the underlying $n\to\infty$. Returning to your formula $$_2 F_2(1,1;2,2;-x)=\frac{1}{2\pi i}\int_C\frac{\Gamma(-s)}{(1+s)^2}x^s\,ds,$$
we see that it holds for a "reflected" version of "your" $C$ (going from $+\infty-i\varepsilon$ to $-i\varepsilon$, encircling $0$ clockwise, and continuing from $i\varepsilon$ to $+\infty+i\varepsilon$), by forming $C_n$ in $\{s:\Re s\leqslant n+1/2\}$, say, and noting that the integral over the part of $\Re s=n+1/2$ tends to $0$ with $n\to\infty$.
Similarly, it holds for $C$ the line $\Re s=-1/2$ (say); this time we use "large rectangles". To prove $(1)$, we move $C$ to the left of $s=-1$. Or, alternatively, we can leave $C$ as is, but replace $s$ by $s-1$, which, in view of $\Gamma(1-s)=-s\Gamma(-s)$, suggests to consider
$$\frac{1}{2\pi i}\int_C\frac{\Gamma(-s)}{s}x^s\,ds=\gamma+\ln x+\sum_{n=1}^{\infty}\frac{(-x)^n}{n\cdot n!}$$
(shown exactly the way you did). On the other hand, writing $x^s/s=-\int_x^\infty y^{s-1}\,dy$,
$$\frac{1}{2\pi i}\int_C\frac{\Gamma(-s)}{s}x^s\,ds=-\int_x^\infty\underbrace{\frac{1}{2\pi i}\int_C\Gamma(-s)y^s\,ds}_{=e^{-y}\text{ by the above}}\,\frac{dy}{y}.$$
