upper bound for the Meijer G function The Meijer G function $G_{2,3}^{3,0}(1|1,1;0,0,1+b)$, for $b>-1$, equals the integral 
$
\int_0^1-\log(x) \frac{e^{-1/x}}{x^{2+b}}dx.
$
Do you know any continuous function $f(b)$ satisfying $G_{2,3}^{3,0}(1|1,1;0,0,1+b)\leq f(b)$ at least in a neighborhooud of $1$? I would like to come up with a function which is less complicated than Meijer G.
 A: Special functions get less respect on this site, though I've seen Claude contribute several times.  I find this problem intriguing.  I'll work with the modification, with  $y=b+1:$
$$J(y) := \int_1^\infty dx \log{(x)} \  x^{y-1} e^{-x} \quad , \quad J(b+1)=I(b) $$
Assume $y>0.$  Then, by the integral definition of the incomplete gamma function,
$$ J(y)=\frac{d}{dy} \Gamma(y,1) .$$ Use the 'reciprocal gamma function' representation (Dig. Lib. Math. Func. 8.7.3) to get 
$$ \frac{d}{dy} \Gamma(y,1) = \frac{d}{dy} \Big(\Gamma(y)\big(1-\frac{1}{e}\sum_{n=0}^\infty \frac{1}{(y+n)!} \big)\Big)$$
$$ = \Gamma(y)\psi(y) + \frac{\Gamma(y)}{e}\sum_{n=0}^\infty \frac{1}{(n+y)!}
\big( \psi(n+y+1) - \psi(y) \big) $$
In the last line $\psi(y)$ is the digamma function.
To get a bound assume $y$ is large and expand the difference in $\psi$ functions:
$$ \psi(n+y+1) - \psi(y) \sim \frac{n+1}{y} - \frac{n(n+1)}{2y^2} + ... $$
The second term is negative, so for y large enough we know we have a bound if keeping only the first term.  Thus
$$\sum_{n=0}^\infty \frac{1}{(n+y)!}\big( \psi(n+y+1) - \psi(y) \big) \le $$
$$\frac{1}{y \ y!}\Big(1 + \frac{2}{y+1} + \frac{3}{(y+1)(y+2)} + ... \Big) \sim \frac{1}{y \ y!}\sum_{n=0}^\infty \frac{n+1}{(y+1)^n}$$
Summing the series and simplifying we get
$$J(y) \le \Gamma(y)\psi(y)+\frac{1}{e y^2}\big(1+\frac{1}{y} \big)^2 $$
This bound is very good for $y\gtrapprox 1.2,$ but diverges for $y \to 0.$  For smaller $y,$ I'd just expand the $x^y = 1+ y \ \log{x} + ...$ and get a power series in $y$ and do the integrals numerically. For an even simpler bound, one could insert the asymptotic expansion 
$$ \Gamma(y) \psi(y) \sim e^{-y} y^y \sqrt{\frac{2 \pi}{y}} \ \log{y} \quad,\quad y \to \infty $$
It should be noted that there are other expansions for the incomplete gamma function, like Temme's that involves the error function.  That attempt went nowhere for me, but it is another avenue to explore.
A: This is not an answer.
If your problem is
$$I(b)=-\int_0^1\log(x) \frac{e^{-1/x}}{x^{2+b}}\,dx$$ you could notice that $\log(I(b))$ is quite nice.
For the range $-1 \leq b \leq 3$, a quick and dirty regression
$$I(b)= \alpha + \beta\, e^{\gamma \,x}$$ gives something which is quite good
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 \alpha  & 0.11690 & 0.00243 & \{0.11211,0.12168\} \\
 \beta   & 0.10704 & 0.00077 & \{0.10553,0.10854\} \\
 \gamma  & 1.41011 & 0.00251 & \{1.40518,1.41505\} \\
\end{array}$$
