Minimizing Integral

$\begin{aligned} \frac d{da}\left(\int_a^{a+1}\ln\Gamma(x)\ dx\right)&=\ln\Gamma(a+1)-\ln\Gamma(a) \\&=\ln\frac{\Gamma(a+1)}{\Gamma(a)} \\&=\ln a \end{aligned}$

Therefore, the integeral has a minimum value when $\ln a=0$ , that is, $a=\boxed 1$ .

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For the Bonus question:

According to the solution of the problem Integeral and Gamma Function (3) , the minimum value is $\int_1^2\ln\Gamma(x)\ dx=\frac{\ln(2\pi)}2-1$