Registro de rayos Gamma integral

Question

¿Cuál es la constante de $\phi$ en la evaluación \begin{align} \int_{0}^{1/4} \ln\Gamma\left( t + \frac{1}{4}\right) \, dt = \frac{1}{8} \ln(\phi) \end{align} y la constante de $\theta$ en el valor \begin{align} \int_{0}^{1/2} \ln\Gamma\left( t + \frac{1}{2}\right) \, dt = \frac{1}{24} \ln(\theta) \hspace{5mm} ? \end{align}

Answer 1

La primera y la segunda integral (de la que hablaré respectivamente llamar a $\mathcal{I}_1$$\mathcal{I}_2$) puede escribirse como \begin{align} \mathcal{I}_1=&\int^\frac{1}{2}_0\ln{\Gamma(x)}\ {\rm d}x-\int^\frac{1}{4}_0\ln{\Gamma(x)}\ {\rm d}x\\ \mathcal{I}_2=&\int^1_0\ln{\Gamma(x)}\ {\rm d}x-\int^\frac{1}{2}_0\ln{\Gamma(x)}\ {\rm d}x\\ \end{align} Empezar con el infinito representación de los productos de ${\Gamma(z)}$. $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod^\infty_{k=1}e^\frac{z}{k}\left(1+\frac{z}{k}\right)^{-1}$$ Tomando logaritmos obtenemos \begin{align} \ln{\Gamma(z)} =&-\gamma z-\ln{z}+\sum^\infty_{k=1}\left\{\frac{z}{k}-\ln\left(1+\frac{z}{k}\right)\right\}\\ =&-\gamma z-\ln{z}+\sum^\infty_{k=1}\left\{\frac{z}{k}+\sum^\infty_{m=1}\frac{(-1)^m}{m}\frac{z^m}{k^m}\right\}\\ =&-\gamma z-\ln{z}+\sum^\infty_{k=1}\sum^\infty_{m=2}\frac{(-1)^m}{m}\frac{z^m}{k^m}\\ =&-\gamma z-\ln{z}+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}z^m \end{align} La integración de $0$ $z$rendimientos \begin{align} \int^z_0\ln{\Gamma(x)}\ {\rm d}x =&-\frac{\gamma z^2}{2}-z\ln{z}+z+\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m}z^{m+1}-\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m+1}z^{m+1}\\ =&-\frac{\gamma z^2}{2}-z\ln{z}+z+z\ln{\Gamma(z)}+\gamma z^2+z\ln{z}-\sum^\infty_{m=2}\frac{(-1)^m\zeta(m)}{m+1}z^{m+1}\\ =&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{m=3}\frac{(-1)^{m-1}\zeta(m-1)}{m}z^m\\ =&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{m=3}\sum^\infty_{k=1}\frac{(-1)^{m-1}}{m}\frac{z^m}{k^{m-1}}\\ =&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\sum^\infty_{k=1}\left\{k\ln\left(1+\frac{z}{k}\right)-z+\frac{z^2}{2k}\right\}\\ =&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\ln\prod^\infty_{k=1}e^{\frac{z^2}{2k}-z}\left(1+\frac{z}{k}\right)^k\\ =&\frac{\gamma z^2}{2}+z+z\ln{\Gamma(z)}-\ln{G(z+1)}+\frac{z}{2}\ln(2\pi)-\frac{z^2}{2}-\frac{\gamma z^2}{2}-\frac{z}{2}\\ =&\frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi)+z\ln{\Gamma(z)}-\ln{G(z+1)} \end{align} donde el infinito representación de los productos de $G(z+1)$ $$G(z+1)=\left(\sqrt{2\pi}\right)^{z}e^{-\frac{(1+\gamma)z^2+z}{2}}\prod^\infty_{k=1}e^{\frac{z^2}{2k}-z}\left(1+\frac{z}{k}\right)^k$$ fue utilizado en la segunda y última línea. Por lo tanto, tenemos \begin{align} \color{blue}{\mathcal{I}_1=}&\color{blue}{\frac{1}{32}+\frac{1}{8}\ln\left(\frac{2}{\pi}\right)+\frac{3}{4}\ln{\Gamma\left(\frac{1}{4}\right)}-\ln{G\left(\frac{1}{2}\right)}+\ln{G\left(\frac{1}{4}\right)}}\\ \color{blue}{\mathcal{I}_2=}&\color{blue}{-\frac{1}{8}+\frac{1}{4}\ln{2}+\frac{1}{2}\ln{\pi}+\ln{G\left(\frac{1}{2}\right)}} \end{align} La duplicación de la fórmula para el Barnes G los estados que \begin{align} \ln{G(2z)}=&-\frac{1}{4}+\left(2z^2-3z+\frac{11}{12}\right)\ln{2}-\left(z-\frac{1}{2}\right)\ln{\pi}+3\ln{A}+\ln{\Gamma(z)}\\&+2\ln{G(z)}+2\ln{G\left(z+\frac{1}{2}\right)} \end{align} Dejando $z=\frac{1}{2}$ rendimientos $$\color{blue}{\ln{G\left(\frac{1}{2}\right)}=\frac{1}{8}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}-\frac{3}{2}\ln{A}}$$ La reflexión de la fórmula para el Barnes G establece que (este es bastante trivial para probar) $$\ln{G(1-z)}=z\ln\left(\frac{\sin(\pi z)}{\pi}\right)+\ln{\Gamma(z)}+\frac{1}{2\pi}{\rm Cl}_2(2\pi z)+\ln{G(z)}$$ Dejando $z=\frac{1}{4}$, $$\ln{G\left(\frac{3}{4}\right)}=-\frac{1}{8}\ln{2}-\frac{1}{4}\ln{\pi}+\frac{\mathbf{G}}{2\pi}+\ln{\Gamma\left(\frac{1}{4}\right)}+\ln{G\left(\frac{1}{4}\right)}$$ y dejando $z=\frac{1}{4}$ en la duplicación de la fórmula, \begin{align} \frac{1}{8}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}-\frac{3}{2}\ln{A} =&-\frac{1}{4}+\frac{1}{24}\ln{2}-\frac{1}{4}\ln{\pi}+\frac{\mathbf{G}}{\pi}+3\ln{A}\\ &+3\ln{\Gamma\left(\frac{1}{4}\right)}+4\ln{G\left(\frac{1}{4}\right)} \end{align} Reorganización, $$\color{blue}{\ln{G\left(\frac{1}{4}\right)}=\frac{3}{32}-\frac{\mathbf{G}}{4\pi}-\frac{9}{8}\ln{A}-\frac{3}{4}\ln{\Gamma\left(\frac{1}{4}\right)}}$$ El desplume estos valores en $\mathcal{I}_1$$\mathcal{I}_2$, obtenemos \begin{align} \color{red}{\mathcal{I}_1=}&\color{red}{\frac{1}{4}\ln\left(\sqrt[3]{2}\sqrt{\pi A^3}\right)-\frac{\mathbf{G}}{4\pi}}\\ \color{red}{\mathcal{I}_2=}&\color{red}{\frac{1}{2}\ln\left(\frac{2^\frac{7}{12}\sqrt{\pi}}{A^3}\right)} \end{align}

Answer 2

Para dar un enfoque diferente utilizando otro método:

Desde el Weierstrass-representación tenemos: $$ \frac{1}{\Gamma(x)}=x\cdot e^{\gamma x}\cdot\prod_{k=1}^\infty \left(1+\frac{x}{k}\right)\cdot e^{-\frac{x}{k}} \ffi \ln(\Gamma(x))=-\ln(x)-\gamma x+\sum_{k=1}^{\infty} \frac{x}{k}-\ln\left(1+\frac{x}{k}\right) $$ Y por lo tanto: $$ \int_0^x \ln(\Gamma(t))dt=\int_0^x\ln(t)-\gamma t+\sum_{k=1}^{\infty} \frac{t}{k}-\ln\left(1+\frac{t}{k}\right)dt=-\int_0^x\ln(t)dt-\int_0^x\gamma t \espacio dt+\sum_{k=1}^{\infty} \left[\int_0^x\frac{t}{k}\espacio dt-\int_0^x\ln\left(1+\frac{t}{k}\right)dt\right] $$ Ahora, tenemos: $$ \int_0^x\ln(t)dt=x\ln(x)-x $$ $$ \int_0^x\gamma t \espacio dt=\frac{\gamma}{2}x^2 $$ $$ \int_0^x\frac{t}{k}\espacio dt=\frac{x^2}{2k} $$ $$ \int_0^x\ln\left(1+\frac{t}{k}\right)dt=(k+x)\cdot\left(\ln\left(1+\frac{x}{k}\right)-1\right)+k=\ln\left(\frac{(k+x)^{k+x}}{k^{k+x}}\right)-x $$ Por lo tanto, podemos escribir: $$ \int_0^x \ln(\Gamma(t))dt=-x\ln(x)+x-\frac{\gamma}{2}x^2+\sum_{k=1}^{\infty} \left[\frac{x^2}{2k}-\ln\left(\frac{(k+x)^{k+x}}{k^{k+x}}\right)+x\right] $$ Para calcular la suma, podemos echar un vistazo a las sumas parciales: $$ S_n=\sum_{k=1}^{n} \left[\frac{x^2}{2k}-\ln\left(\frac{(k+x)^{k+x}}{k^{k+x}}\right)+x\right]=\frac{x^2}{2}H_n-\ln\left(\prod_{k=1}^n\frac{(k+x)^{k+x}}{k^{k+x}}\right)+nx=\frac{x^2}{2}H_n-\ln\left(\frac{H(n+x)}{H(x)\cdot H(n)\cdot(n!)^x}\right)+nx $$ Donde $H(x)$ denota la generalizada hyperfactorial dado por $H(n)=\prod_{k=1}^n k^k$. A partir de la definición de Glaisher-Kinkelin $A$ constante, el número de Euler $\gamma$ y Stirlings aproximación tenemos: $$ A=\lim_{n\to\infty}\frac{1^1\cdot 2^2\cdots n^n}{n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}\ffi \lim_{n\to\infty}\frac{H(n)}{A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}=1\iff\lim_{n\to\infty}\ln\left(\frac{H(n)}{A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}}\right)=0 $$ $$ \gamma=\lim_{n\to\infty}\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}-\ln(n)\ffi \lim_{n\to\infty}\left(H_n-\ln(n)-\gamma\right)=0 $$ $$ 1=\lim_{n\to\infty}\frac{n!}{n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}}\ffi\lim_{n\to\infty}\ln\left(\frac{n!}{n^{n+\frac{1}{2}}\cdot e^{-n}\cdot\sqrt{2\pi}}\right)=0 $$ Y por lo tanto: $$ S=\sum_{k=1}^{\infty} \left[\frac{x^2}{2k}-\ln\left(\frac{(k+x)^{k+x}}{k^{k+x}}\right)+x\right]=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\left(\frac{x^2}{2}H_n-\ln\left(\frac{H(n+x)}{H(x)\cdot H(n)\cdot(n!)^x}\right)+nx\right)=\lim_{n\to\infty}\left(\frac{x^2}{2}(\ln(n)+\gamma)-\ln\left(\frac{(n+x)^{\frac{n^2}{2}+nx+\frac{x^2}{2}+\frac{n}{2}+\frac{x}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}-\frac{nx}{2}-\frac{x^2}{4}}\cdot}{H(x)\cdot A\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}\cdot n^{nx+\frac{x}{2}}\cdot e^{-nx}\cdot \left(\sqrt{2\pi}\right)^{x}}\right)+nx\right)=\lim_{n\to\infty}\left(\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})+\ln\left(\frac{n^{\frac{x^2}{2}}\cdot n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}}\cdot n^{nx+\frac{x}{2}}\cdot e^{-nx}\cdot e^{nx}}{(n+x)^{\frac{n^2}{2}+nx+\frac{x^2}{2}+\frac{n}{2}+\frac{x}{2}+\frac{1}{12}}\cdot e^{-\frac{n^2}{4}-\frac{nx}{2}-\frac{x^2}{4}}}\right)\right)=\lim_{n\to\infty}\left(\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})+\ln\left(\frac{e^{\frac{nx}{2}+\frac{x^2}{4}}}{\left(1+\frac{x}{n}\right)^{\frac{n^2}{2}+nx+\frac{x^2}{2}+\frac{n}{2}+\frac{x}{2}+\frac{1}{12}}}\right)\right) $$ Ahora tenemos: $$ \left(1+\frac{x}{n}\right)^{\frac{n^2}{2}}=\exp\left(\frac{n^2}{2}\cdot\ln\left(1+\frac{x}{n}\right)\right)=\exp\left(\frac{n^2}{2}\cdot\left(\frac{x}{n}-\frac{x^2}{2n^2}+O\left(\frac{1}{n^3}\right)\right)\right)=\exp\left(\frac{xn}{2}-\frac{x^2}{4}+O\left(\frac{1}{n}\right)\right)\to\lim_{n\to\infty}\frac{e^{\frac{xn}{2}}}{\left(1+\frac{x}{n}\right)^{\frac{n^2}{2}}}=e^{\frac{x^2}{4}} $$ Esto produce: $$ S=\lim_{n\to\infty}\left(\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})+\ln\left(\frac{e^{\frac{nx}{2}}\cdot e^{\frac{x^2}{4}}}{\izquierda(1+\frac{x}{n}\right)^{\frac{n^2}{2}}\cdot \left(1+\frac{x}{n}\right)^{nx+\frac{n}{2}}\cdot \left(1+\frac{x}{n}\right)^{\frac{x^2}{2}+\frac{x}{2}+\frac{1}{12}}}\right)\right)=\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})+\ln\left(\frac{e^{\frac{x^2}{4}}\cdot e^{\frac{x^2}{4}}}{e^{x^2+\frac{x}{2}}}\right)=\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})-\frac{x^2}{2}-\frac{x}{2} $$ Y por lo tanto: $$ \int_0^x \ln(\Gamma(t))dt=-x\ln(x)+x-\frac{\gamma}{2}x^2+\frac{x^2}{2}\gamma+\ln(H(x))+x\ln(\sqrt{2\pi})-\frac{x^2}{2}-\frac{x}{2}=-x\ln(x)+\frac{x}{2}+\ln(H(x))+x\ln(\sqrt{2\pi})-\frac{x^2}{2} $$ Así que para responder a la pregunta: $$ \int_{0}^{\frac{1}{4}} \ln\left(\Gamma\left(t+\frac{1}{4}\right)\right)dt=\int_{\frac{1}{4}}^{\frac{1}{2}} \ln(\Gamma(t))dt=\int_{0}^{\frac{1}{2}} \ln(\Gamma(t))dt-\int_{0}^{\frac{1}{4}} \ln(\Gamma(t))dt=-\frac{1}{2}\ln\left(\frac{1}{2}\right)+\frac{\left(\frac{1}{2}\right)}{2}+\ln\left(H\left(\frac{1}{2}\right)\right)+\frac{1}{2}\ln\left(\sqrt{2\pi}\right)-\frac{\left(\frac{1}{2}\right)^2}{2}+\frac{1}{4}\ln\left(\frac{1}{4}\right)-\frac{\left(\frac{1}{4}\right)}{2}-\ln\left(H\left(\frac{1}{4}\right)\right)-\frac{1}{4}\ln\left(\sqrt{2\pi}\right)+\frac{\left(\frac{1}{4}\right)^2}{2}=\frac{1}{32}+\ln\left(\frac{H\left(\frac{1}{2}\right)}{H\left(\frac{1}{4}\right)}\right)+\frac{1}{8}\ln\left(2\pi\right)=\frac{1}{8}\ln\left(e^{\frac{1}{4}}\cdot2\pi\cdot\frac{H\left(\frac{1}{2}\right)^8}{H\left(\frac{1}{4}\right)^8}\right) $$ Y $$ \int_{0}^{\frac{1}{2}} \ln\left(\Gamma\left(t+\frac{1}{2}\right)\right)dt=\int_{0}^{1} \ln(\Gamma(t))dt-\int_{0}^{\frac{1}{2}} \ln(\Gamma(t))dt=-1\ln(1)+\frac{1}{2}+\ln\left(H(1)\right)+1\ln\left(\sqrt{2\pi}\right)-\frac{1^2}{2}+\frac{1}{2}\ln\left(\frac{1}{2}\right)-\frac{\left(\frac{1}{2}\right)}{2}-\ln\left(H\left(\frac{1}{2}\right)\right)-\frac{1}{2}\ln\left(\sqrt{2\pi}\right)+\frac{\left(\frac{1}{2}\right)^2}{2}=\frac{1}{4}\ln\left(2\pi\right)-\frac{1}{2}\ln(2)-\frac{1}{8}-\ln\left( H\left(\frac{1}{2}\right)\right)=\frac{1}{24}\ln\left(\frac{(2\pi)^6}{2^{12}\cdot e^3 \cdot H\left(\frac{1}{2}\right)^{24}}\right) $$ Ahora usted puede utilizar identidades como $H\left(\frac{1}{2}\right)=\frac{A^{\frac{3}{2}}}{2^{\frac{13}{24}}\cdot e^{\frac{1}{8}}}$ a transformar estos resultados.

Answer 3

Consideremos $$ \int \ln\Gamma\left( t + \right) \, dt = -(a+t) \text{log$\Gamma $}(a+t)+\psi ^{(-2)}(a+t)+(a+t) \log (\Gamma (a+t))$$ So,$$\int_0^a \ln\Gamma\left( t + a\right) \, dt =a \text{log$\Gamma $}(a)-2 a \text{log$\Gamma $}(2)-\psi ^{(-2)}(a)+\psi ^{(-2)}(2)-a \log (\Gamma (a))+2 a \log (\Gamma (2))$$ So, according to a CAS, for $a=\frac{1}{4}$, the result of the integral from $0$ to $un$ is $$I=\frac{1}{24} \left(\log \left(4 \pi ^3 A^9\right)-\frac{6 C}{\pi }\right)$$ where $Un$ is Glaisher's constant and $C$ catalán del número.

Para $a=\frac{1}{2}$, el resultado de la integral a partir de $0$ $a$$$I=\frac{1}{24} \log \left(\frac{128 \pi ^6}{A^{36}}\right)$$