TITLE: How prove this integral $ \int\limits_0^1 \int\limits_0^1 \ln\Gamma(x+y^3)\,dx\,dy =-\frac 7 {16}+\frac{1}{2}\ln 2\pi$
QUESTION [12 upvotes]: show that  

$$ I=\int_0^1 \int_0^1 \ln\Gamma(x+y^3) \, dx \, dy =-\frac 7 {16} + \frac 1 2 \ln 2\pi$$

where $$\Gamma(a)=\int_0^\infty x^{a-1}e^{-x} \, dx$$
then
$$I=\int_0^1 \int_0^1 \ln\left(\int_0^\infty t^{x+y^3-1}e^{-t} \, dt\right) \, dx \, dy$$
Then I can't works,Thank you

REPLY [7 votes]: Using
$$
\int_0^1 \ln\Gamma\left(x+\alpha\right)\ dx
=\frac{1}{2}\ln2\pi+\alpha \log \alpha -\alpha\quad;\quad \alpha \geq 0,
$$
and
$$
\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots
$$
then
$$
\int_0^1 \ln\Gamma\left(x+y^3\right)\ dx=\frac{1}{2}\ln2\pi+3y^3 \ln y -y^3
$$
and
\begin{align}
\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{blue}{\frac{1}{2}\ln2\pi-\frac{7}{16}}.\tag{Q.E.D.}
\end{align}