Cómo probar esta integral$ \int\limits_0^1{\int\limits_0^1{\ln\Gamma\left({x+{y^3}}\right)}}dxdy =-\frac{7}{{16}}+\frac{1}{2}\ln 2\pi$

Question

muestra esa

PS

donde$$ I=\int\limits_0^1{\int\limits_0^1{\ln\Gamma\left({x+{y^3}}\right)}}dxdy =-\frac{7}{{16}}+\frac{1}{2}\ln 2\pi$ $

entonces$$\Gamma{(a)}=\int_{0}^{\infty}x^{a-1}e^{-x}dx$ $ entonces no puedo trabajar, gracias

Answer 1

Deje $\displaystyle\;\;f(u) = \int_0^1 \log\Gamma(z+u) dz,\;\;$ tenemos:

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

Integrar esto nos da $f(u) = f(0) + u\log u - u$. Ahora

$$f(0) = \int_0^1\log\Gamma(z)dz = \frac12\int_0^1\log(\Gamma(z)\Gamma(1-z))dz =\frac12 \int_0^1\log\frac{\pi}{\sin\pi z} dz =\frac12\left(\log\pi \frac{1}{\pi}\int_0^{\pi}\log \sin\theta d\theta\right) \stackrel{\color{blue}{[1]}}{=} \frac12\log(2\pi) $$ Esto nos da $$f(u) = \frac12\log(2\pi) + u\log u - u$$ y por lo tanto

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

Notas

• $\color{blue}{[1]}$ Estamos utilizando el resultado $$\frac{1}{\pi}\int_0^{\pi}\log\sin\theta d\theta = -\log 2,$$ Para una prueba, vea las respuestas de esta pregunta.

Answer 2

Usando $$ \ int_0 ^ 1 \ ln \ Gamma \ left (x + \ alpha \ right) \ dx = \ frac {1} {2} \ ln2 \ pi + \ alpha \ log \ alpha - \ alpha \ quad; \ quad \ alpha \ geq 0, $$ y $$ \ int_0 ^ 1 x ^ \ alpha \ ln ^ nx \ dx = \ frac {(- 1) ^ nn!} {(\ alpha +1) ^ {n +1}} , \ qquad \ text {para} \ n = 0,1,2, \ ldots $$ luego $$ \ int_0 ^ 1 \ ln \ Gamma \ left (x + y ^ 3 \ right) \ dx = \ frac {1 } {2} \ ln2 \ pi +3y ^ 3 \ ln y -y ^ 3 $$ y \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}