Integral and Gamma Function (3)

Let $F(a)=\int_a^{a+1}\ln\Gamma(x)\ dx,$ we have $\begin{aligned} F'(a)&=\ln\Gamma(a+1)-\ln\Gamma(a) \\&=\ln\frac{\Gamma(a+1)}{\Gamma(a)} \\&=\ln a. \end{aligned}$ Hence $\begin{aligned} F(a)&=F(0)+\int_0^a\ln x\ dx \\&=\frac{\ln(2\pi)}2+a\ln a-a. \end{aligned}$ Here is why $F(0)=\dfrac{\ln(2\pi)}2$ .

Set $a=e$ , we get $F(e)=\int_e^{e+1}\ln\Gamma(x)\ dx=\frac{\ln(2\pi)}2.$ Therefore, $A=\boxed 2$ .