The following problem:$$\text{PV}\, \int_0^\infty (-1)^{\lfloor x\rfloor}\ln\left(\frac{\sqrt{x}\,e^x\,\Gamma(x)}{\sqrt{2\pi}\,x^x}\right)\,\mathrm{d}x=\frac{7\,\zeta(3)}{4\,\pi^2}$$
where $\text{PV}$ is the principle value and $\lfloor.\rfloor$ is the floor function, was proposed by my friend, Khalef Ruhemi. I see that the argument of the logarithm is similar to Stirling's approximation but have no idea how to use this fact, so I wanted to share it with you hoping to see some solutions.
Denotes $$\mathscr{I}: = \mathcal{P.V.}\int_0^\infty (-1)^{\lfloor x\rfloor}\ln\left(\frac{\sqrt{x}e^x\Gamma(x)}{\sqrt{2\pi}x^x}\right)\mathrm{d}x$$ And $$ \mathscr{I}=\sum_{n=0}^{\infty} (-1)^n\lim_{\epsilon \to 0} \int_{n+\epsilon}^{n+1-\epsilon}\ln\left(\frac{\sqrt{x}e^x\Gamma(x)}{\sqrt{2\pi}x^x}\right)\mathrm{d}x $$ But we have $$ \int_{n}^{n+1}\ln\left(\frac{\sqrt{x}e^x\Gamma(x)}{\sqrt{2\pi}x^x}\right)\mathrm{d}x =\frac{n}{2} \ln n-\frac{n}{2}\ln(1+n) +\frac{n^2}{2}\ln n-\frac{n^2}{2}\ln(1+n) +\frac{n}{2}+\frac{1}{4} $$ Plug back to the original integral and I obtain: $$ \mathscr{I}=\frac{1}{4}+\sum_{n=1}^{\infty}(-1)^n\left(\frac{n}{2} \ln n-\frac{n}{2}\ln(1+n) +\frac{n^2}{2}\ln n-\frac{n^2}{2}\ln(1+n) +\frac{n}{2}+\frac{1}{4}\right) $$ From Analytic continuation,the last series can express in terms Riemann zeta function. $$ \sum_{n=1}^{\infty} (-1)^nn^s=\zeta(-s)(2^{s+1}-1) $$ And use the special value $$ \zeta'(-2)=-\frac{\zeta(3)}{4\pi^2} $$ The result follows.