Resembling Resemblence

If you have another approach then do post it .

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Let $f(u)=\displaystyle \int_0^1 \log\Gamma(z+u) dz$

$\color{#3D99F6}{f'(u) =\displaystyle\int_0^1 \dfrac{\Gamma'(z+u)}{\Gamma(z+u)}dz = \Big[\log\Gamma(z+u)\Big]_{0}^{1} = \log\dfrac{\Gamma(u+1)}{\Gamma(u)} = \log u}$

Integrating this gives us $f(u) = f(0) + u\log u - u$ .So

$f(0) = \displaystyle\int_0^1\log\Gamma(z)dz = \frac12\int_0^1\log(\Gamma(z)\Gamma(1-z))dz$ $=\frac12 \int_0^1\log\frac{\pi}{\sin\pi z} dz$ $=\frac12\left(\log\pi - \frac{1}{\pi}\int_0^{\pi}\log \sin\theta d\theta\right) =\frac12\log(2\pi)^{**}$

This gives us

$\color{#D61F06}{f(u) = \frac12\log(2\pi) + u\log u - u}$

so,

$\begin{aligned} I = \int_0^1 f(y^3) dy = & \int_0^1 \big(\frac12\log(2\pi) + y^3 \log(y^3) - y^3\big) dy\\ = & \frac12\log(2\pi) + \left[\frac{3}{16}y^4(\log(y^4)-1) - \frac14 y^4\right]_0^1\\ = & \boxed{\color{#20A900}{\frac12\log(2\pi) - \frac{7}{16}}} \end{aligned}$

By Raabe's Formula, we get: $\int_0^1 \ln\Gamma\left(x+\alpha\right)\ dx =\frac{1}{2}\ln2\pi+\alpha \log \alpha -\alpha\quad;\quad \alpha \geq 0,$

This is a well known identity: $\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

The inner integral comes out to be: $\int_0^1 \ln\Gamma\left(x+y^3\right)\ dx=\frac{1}{2}\ln2\pi+3y^3 \ln y -y^3$

Therefore on evaluating the outer integral, we get: $\begin{aligned} \int_0^1\int_0^1 \ln\Gamma\left(x+y^3\right)\ dx\ dy&=\int_0^1 \left(\frac{1}{2}\ln2\pi+3y^3 \ln y -y^3\right)\ dy\\ &=\frac{1}{2}\ln2\pi-\frac{3}{4^2}-\frac14\\ &=\large\color{#3D99F6}{\frac{1}{2}\ln2\pi-\frac{7}{16}}.\end{aligned}$