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$

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

Solution 1:

Let $\displaystyle\;\;f(u) = \int_0^1 \log\Gamma(z+u) \,dz,\;\;$ we have:

$$f'(u) = \int_0^1 \frac{\Gamma'(z+u)}{\Gamma(z+u)}\,dz = \Big[\log\Gamma(z+u)\Big]_{z=0}^1 = \log\frac{\Gamma(u+1)}{\Gamma(u)} = \log u$$

Integrate this gives us $f(u) = f(0) + u\log u - u$. Now

$$f(0) = \int_0^1\log\Gamma(z)dz = \frac12\int_0^1\log(\Gamma(z)\Gamma(1-z)) \, dz =\frac12 \int_0^1\log\frac{\pi}{\sin\pi z} \, dz\\ =\frac12\left(\log\pi - \frac{1}{\pi}\int_0^{\pi}\log \sin\theta \, d\theta\right) \stackrel{\color{blue}{[1]}}{=} \frac12\log(2\pi) $$ This gives us $$f(u) = \frac12\log(2\pi) + u\log u - u$$ and hence

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

Notes

• $\color{blue}{[1]}$ We are using the result $$\frac 1 \pi \int_0^\pi \log\sin\theta \, d\theta = -\log 2,$$ For a proof, see answers of this question.

Solution 2:

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}