I am looking for a way to evaluate the integral
$$ \int_{-1}^{1}\left(t - 1\right)\left[\mathrm{e}^{1/\Gamma\left(t\right)} - 1\right]\mathrm{d}t $$
This integral appears to almost have a sort of symmetry about the $y$-axis that could yield a cancellation, but I have been unable to show this. Numerical integration yields a value of $−0.0001194628623602256$. I am curious if there is a way to evaluate this integral, perhaps exploiting any potential symmetry, but I would not be surprised if there is no such method given the difficulty in working with the gamma function.
I have tried substituting $t=-u$ and breaking up the integral into two parts in order to solve but have so far been unsuccessful.
You can get this kind of small numbers using the Taylor expansion of $e^\frac{1}{\Gamma(t)}$ built at $t=0$ and integrate termwise. This would be quite long; the first terms are $$e^\frac{1}{\Gamma(t)}-1=t+\left(\frac{1}{2}+\gamma \right) t^2+\left(\frac{1}{6}+\gamma +\frac{\gamma ^2}{2}-\frac{\pi ^2}{12}\right) t^3+$$ $$\frac{1}{24} \left(1+12 \gamma +24 \gamma ^2+4 \gamma ^3-2 \pi ^2-2 \gamma \pi ^2-4 \psi ^{(2)}(1)\right)t^4+O\left(t^5\right)$$
Let us admit that you are sufficiently patient to do the expansion up to $O\left(t^{p+1}\right)$. You should get $$\left( \begin{array}{cc} p & \text{result} \\ 2 & -0.05147711 \\ 4 & +0.06413149 \\ 6 & -0.00956415 \\ 8 & -0.00483854 \\ 10 & +0.00118575 \\ 12 & -0.00041961 \\ 14 & -0.00003656 \end{array} \right)$$ I give up (my computer too !).