I came up with an unusual integral at the bottom of the page when tried to find a closed form for this factorial-related integral:
$$\boxed{\int_0^1 \Gamma(t+1) \text{ d}t = 0.92274\ldots}$$
Here is what I did:
$$ \boxed{\textbf{I}=\int_0^1 \Gamma(t+1) \text{ d}t} \\ t\in(0,1) \Rightarrow \Gamma(t+1)= \int_0^1 (-\ln(s))^t \text{ d}s \\ =\int_0^1\int_0^1 (-\ln(s))^t\text{ d}s \text{ d}t \\ =\int_0^1\int_0^1(-\ln(s))^t\text{ d}t \text{ d}s \\ =\int_0^1\frac{-\ln(s)-1}{\ln(-\ln(s))}\text{ d}s, \\ \textbf{Substitution: }u=-\ln(s), \\ =\int_\infty^0 \frac{u-1}{-e^u\ln(u)}\text{ d}u \\ =\int_0^\infty \frac{u-1}{e^u\ln(u)} \text{ d}u $$
How can I continue now? Although this integral doesnt seem to be very complicated, i could not come up with a closed form or series representation. Neither did Wolframalpha.
Does anyone have an idea?