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I found the following identity while I was working on a recent question of mine:
For $z\in\mathbb{C}$, $$G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=\gamma\ln{z}+\frac12\ln^2(z)-z\,_3F_3(1,1,1;2,2,2;-z)+\frac{\gamma^2}2+\frac{\pi^2}{12}.\tag{1}$$ WolframAlpha confirms it, but I couldn't find the formula stated anywhere online from my searches. Has this result already been discovered? If so, could someone point me in the right direction? I've made a proof for the identity, which I have included below. There's likely a more elegant, succinct proof for it, but this is what I could muster using my current knowledge.

Proof: Taking the derivative gives us $\frac{d}{dz}G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=-\frac{\Gamma(0,z)}z$, from which we can state $$-\int\frac{\Gamma(0,z)}z dz=G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)+C.\tag{2}$$ Since $\Gamma(0,z)=E_1(z)$, we can use the series expansion for $E_1(z)$, which converges for all complex $z$, to rewrite $(2)$ as $$\begin{align}\int\frac{\gamma+\ln{z}}z dz-\int\sum_{k=1}^\infty\frac{(-z)^{k-1}}{k\,k!}dz&=G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)+C\\ \gamma\ln{z}+\frac12\ln^2(z)-z\sum_{k=1}^\infty\frac{(-z)^{k-1}}{k^2\,k!}+C_0&=G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)+C\\ \gamma\ln{z}+\frac12\ln^2(z)-z\,_3F_3(1,1,1;2,2,2;-z)&=G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)+C^*,\tag{3}\label{3}\end{align}$$ where $C_0$ is a second constant of integration and $C^*=C-C_0$. In order to maintain both equivalence and continuity in $(3)$, $C^* $ can only hold one value--we must therefore solve for $C^*$. One method of doing so involves evaluating $\displaystyle\int_1^\infty\frac{-\Gamma(0,z)}z dz$. Because $z$ is positive along the given interval, this can be equivalently expressed as $$\int_1^\infty\frac{-\Gamma(0,z)}z dz=\int_1^\infty\frac{\mathrm{Ei}(-z)}z dz.\tag{4}$$ Repeated application of integration by parts yields $$\int_1^\infty\frac{\mathrm{Ei}(-z)}z dz=-\int_1^\infty\frac{\ln{z}}{ze^z}dz=\int_1^\infty\frac{\ln{z}}{ze^z}dz-\int_1^\infty\frac{\ln^2(z)}{e^z}dz,$$ from which we can infer that $$\begin{align}-\int_1^\infty\frac{\ln{z}}{ze^z}dz=-\frac12\int_1^\infty\frac{\ln^2(z)}{e^z}dz&=\frac12\int_0^1\frac{\ln^2(z)}{e^z}dz-\frac12\int_0^\infty\frac{\ln^2(z)}{e^z}dz\\&=\frac12\int_0^1\frac{\ln^2(z)}{e^z}dz-\frac{\gamma^2}2-\frac{\pi^2}{12}\tag{5}\end{align}$$ (see here). After integration by parts and manipulation of series, we get $$\begin{align}\frac12\int_0^1\frac{\ln^2(z)}{e^z}dz&=-\frac12\left[\frac{\ln^2(z)}{e^z}\right]_0^1+\int_0^1\frac{\ln{z}}{ze^z}dz\\&=-\frac12\left[\frac{\ln^2(z)}{e^z}\right]_0^1+\int_0^1\sum_{k=0}^\infty\frac{(-1)^k z^{k-1}\ln{z}}{k!}dz\\&=\int_0^1\sum_{k=1}^\infty\frac{(-1)^k z^{k-1}\ln{z}}{k!}dz\\&=\left[\sum_{k=1}^\infty\frac{(-z)^k\ln{z}}{k\,k!}\right]_0^1+\int_0^1\sum_{k=1}^\infty\frac{(-z)^{k-1}}{k\,k!}dz\\&=\left[z\,_3F_3(1,1,1;2,2,2;-z)\right]_0^1\\&={}_3F_3(1,1,1;2,2,2;-1).\tag{6}\end{align}$$ Putting together everything we have so far, we arrive at $$\int_1^\infty\frac{-\Gamma(0,z)}z dz={}_3F_3(1,1,1;2,2,2;-1)-\frac{\gamma^2}2-\frac{\pi^2}{12}.\tag{7}$$ Notice that $-{}_3F_3(1,1,1;2,2,2;-1)$ is the value obtained from the left-hand side of $\eqref{3}$ when $z=1$. Using both this observation and the first fundamental theorem of calculus, we can deduce that $$\lim_{z\to\infty}\left(\gamma\ln{z}+\frac12\ln^2(z)-z\,_3F_3(1,1,1;2,2,2;-z)\right)=-\frac{\gamma^2}2-\frac{\pi^2}{12}.$$ Furthermore, $$\lim_{z\to\infty}G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=-C^*-\frac{\gamma^2}2-\frac{\pi^2}{12}.\tag{8}$$ From this identity, we know that $$\int_z^\infty\frac{\Gamma(0,x)}xdx=\int_z^\infty\frac{1}xG_{1,2}^{2,0}\left(x\left|\begin{array}{c}1\\0,0\\\end{array}\right.\right)dx=G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right).\tag{9}$$ This implies $$\lim_{z\to\infty}G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=\lim_{z\to\infty}\left[\int_z^\infty\frac{1}xG_{1,2}^{2,0}\left(x\left|\begin{array}{c}1\\0,0\\\end{array}\right.\right)dx\right]=0,\tag{10}$$ and we may now conclude with $$C^*=-\frac{\gamma^2}2-\frac{\pi^2}{12}.\tag{11}$$ Using $(11)$, we can rewrite $\eqref{3}$ as the identity $$G_{2,3}^{3,0}\left(z\left|\begin{array}{c}1,1\\0,0,0\\\end{array}\right.\right)=\gamma\ln{z}+\frac12\ln^2(z)-z\,_3F_3(1,1,1;2,2,2;-z)+\frac{\gamma^2}2+\frac{\pi^2}{12}.\tag{12}$$ Q.E.D.

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