Definite Gamma function integral Curiosity Question
It's very well known that 

$$\int_a^{a+1} \ln\Gamma(x)\ dx = \frac{1}{2}\ln(2\pi) - a - \ln(a) - (a+1)\ln\Gamma(a) + (a+1)\ln\Gamma(1+a)$$

Clearly provided that $\Gamma(a)\geq 0$ and $\Gamma(1+a) \geq 0$.
If in the above result the variable $a\in\mathbb{N}$, then the result simplifies into

$$\int_a^{a+1} \ln\Gamma(x)\ dx = \frac{1}{2}\ln(2\pi) + a(\ln(a)-1) ~~~~~~~;~ a\in\mathbb{N}$$

Now these two ones are very cool integrals involving the logarithm of the Gamma function.
Question
Is there a closed form (provided all the existence conditions and so on) of the following?

$$\int_a^{a+1} \Gamma(x)\ dx \tag 1$$

Some ideas
If $a$ is very large, then I except a use of one of the many approximations available (Menes, Ramanujan, Windschitl et cetera) to be legit. For example, adopting Menes one:
$$\Gamma(z)\approx \sqrt{\frac{2\pi}{z}}\left(\frac{1}{e}\left(z + \frac{1}{12z - \frac{1}{10z}}\right)\right)^z \to \sqrt{\frac{2\pi}{z}}e^{-z}\left(\frac{10 z}{120 z^2-1}+z\right)^z$$
And applying the Binomial theorem we gain
$$\sqrt{\frac{2\pi}{z}}e^{-z} z^z \sum_{k = 0}^z \left(\frac{10}{120z^2 + 1}\right)^z$$
Where now we are forced to only take $z\in\mathbb{N}$, fair enough for my purpose.
Observing the series can be turned into a geometric series, we can writ it again as
$$\sqrt{\frac{2\pi}{z}}e^{-z} z^z \frac{-10^{z+1} \left(\frac{1}{120 z^2+1}\right)^z+120 z^2+1}{120 z^2-9}$$
That is, 
$$\frac{\sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}} \left(-10^{z+1} \left(120 z^2+1\right)^{-z}+120 z^2+1\right)}{3 \left(40 z^2-3\right)}$$
This is surely a hell to integrate
But with numerical methods, whilst for small $a$ the integration of the latter form and the integration of $\Gamma(x)$ merely fails, when choosing larger numbers I get:
$$\int_{10}^{11} \Gamma(x)\ dx \approx 1.407\cdot 10^6$$
$$\int_{10}^{11} \frac{\sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}} \left(-10^{z+1} \left(120 z^2+1\right)^{-z}+120 z^2+1\right)}{3 \left(40 z^2-3\right)}\ dz \approx 1.397\cdot 10^6$$
Or again:
$$\int_{30}^{31} \Gamma(x)\ dx \approx 7.521\cdot 10^{31}$$
$$\int_{30}^{31} \frac{\sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}} \left(-10^{z+1} \left(120 z^2+1\right)^{-z}+120 z^2+1\right)}{3 \left(40 z^2-3\right)}\ dz \approx 7.501\cdot 10^{31}$$
Corollary
I am perfectly aware of the fact that this is a very hard treatment for such an integral, and that there are lots of easier and faster ways to get some numerical great result.
So apart from this, back to the original question: is there some closed for the integral in $(1)$? Even approximate one, but simple (read: hand-doable).
 A: A good approximation for the proposed integral is
$$ I(a) = \int_a^{a+1}\,\Gamma(x) dx \sim \frac{a-1-\epsilon}{\log{(a-\epsilon)}}\,\Gamma(a) \, , \, \epsilon \approx 0.12 $$
The form with $\epsilon=0$ can be shown to be a first order approximation to that obtained with the saddle point strategy, and the given  $\epsilon=0.12$ is an empirical constant that gives agreement of about 3 digits for all $a>2$ and not just in the limit $a \to \infty$.
By differentiation with respect to $a$ (I actually did it first through operator methods) one can see that 
$$ I(a) = \int_0^{\infty}\,e^{-t}t^{a-1}\big(\frac{t-1}{\log{t}}\big) dt \,. $$
The saddle point strategy, to first order, says to put the $e^{-t}t^{a}$ into an exponential $\exp(-h(t)) = \exp(-(t-\log{a})).$ Differentiate $h(t)$ with respect to $t$ and set the expression to zero to find the saddle point $t=a.$  The 'slowly varying' function $g(t),$ with  $g(t)=(t-1)/\log(t)$ for our case, is pulled through the integral sign and evaluated at the saddle point.  Instead of going through the saddle point machinery (which for g(t) = 1) many will recognize as the manner in which to obtain the Stirling approximation to the $\Gamma$ function, we stop.  We realize that what we have left after pulling the 'slowly varying' function through IS the gamma function.
