一类Log-Gamma积分的一般形式

\[\Large\int_{0}^{z}x^{t}\ln\Gamma \left ( 1+x \right )\mathrm{d}x~,~z>0\, ,\, t\in N^{*}\]

---

\(\Large\mathbf{Solution:}\)
Notice that
\[\begin{align*} \int_{0}^{z}x^{t}\ln\Gamma \left ( 1+x \right )\mathrm{d}x&=\int_{0}^{z}x^{t}\ln\Big[ x\Gamma \left ( x \right ) \Big]\mathrm{d}x=\int_{0}^{z}x^{t}\ln x\mathrm{d}x+\int_{0}^{z}x^{t}\ln\Gamma \left ( x \right )\mathrm{d}x\\ &=\frac{z^{1+t}\Big[\left ( 1+t \right )\ln z-1\Big]}{\left ( 1+t \right )^{2}}+\int_{0}^{z}x^{t}\ln\Gamma \left ( x \right )\mathrm{d}x \end{align*}\]
Using the Kummer's Fourier Series of \(\displaystyle \ln \Gamma(x)\) and
\[\sum_{n=1}^{\infty }\frac{\cos\left ( 2\pi nx \right )}{2n}=-\frac{1}{2}\ln\left ( 2\sin\pi x \right )~,~\sum_{n=1}^{\infty }\frac{\sin\left ( 2\pi nx \right )}{2n}=\frac{1}{2}-x\]
we have
\[\ln\Gamma \left ( x \right )=\frac{1}{2}\ln2\pi-\frac{1}{2}\ln\left ( 2\sin\pi x \right )+\left ( \gamma +\ln2\pi \right )\left ( \frac{1}{2}-x \right )+\sum_{n=1}^{\infty }\frac{\ln n}{n\pi }\sin\left ( 2\pi nx \right )\]
Hence we have
\[\begin{align*} \int_{0}^{z}x^{t}\ln\Gamma \left ( x \right )\mathrm{d}x&=\frac{z^{t+1}\ln2\pi }{2\left ( t+1 \right )}+\frac{z^{t+1}\left ( \gamma +\ln2\pi \right )\left ( t-2zt-2z+2 \right )}{2\left ( t+1 \right )\left ( t+2 \right )}\\ &~~~-\frac{1}{2}\int_{0}^{z}x^{t}\ln\left ( 2\sin\pi x \right )\mathrm{d}x+\frac{1}{\pi }\sum_{n=1}^{\infty }\frac{\ln n}{n}\int_{0}^{z}x^{t}\sin\left ( 2\pi nx \right )\mathrm{d}x \end{align*}\]
where
\[\begin{align*} \int_{0}^{z}x^{t}\ln\left ( 2\sin\pi x \right )\mathrm{d}x&=\frac{1}{\left ( 2\pi \right )^{t+1}}\int_{0}^{2\pi z}x^{t}\ln\left ( 2\sin\frac{x}{2} \right )\mathrm{d}x\\ &=-\frac{x^{t}}{\left ( 2\pi \right )^{t+1}}\mathrm{Cl}_{2}\left ( x \right )\Bigg|_{0}^{2\pi z}+\frac{t}{\left ( 2\pi \right )^{t+1}}\int_{0}^{2\pi z} x^{t-1}\mathrm{Cl}_{2}\left ( x \right )\mathrm{d}x\\ &=-\left ( \frac{z}{2\pi } \right )^{t}\mathrm{Cl}_{2}\left ( 2\pi z \right )+\frac{t}{\left ( 2\pi \right )^{t+1}}\int_{0}^{2\pi z}x^{t-1}\sum_{k=1}^{\infty }\frac{\sin kx}{k^{2}}\mathrm{d}x\\ &=-\left ( \frac{z}{2\pi } \right )^{t}\mathrm{Cl}_{2}\left ( 2\pi z \right )+\frac{t}{\left ( 2\pi \right )^{t+1}}\sum_{k=1}^{\infty }\frac{1}{k^{2}}\int_{0}^{2\pi z}x^{t-1}\sin kx\mathrm{d}x \end{align*}\]
for the last integral,it's not hard to see that
\[\begin{align*} \int_0^{2\pi z}x^{t-1}\sin kx\,\mathrm{d}x&=\left ( t-1 \right )!\,\Bigg[~\sum_{j=0}^{\lfloor {t-1/2} \rfloor}(-1)^{j+1}\frac{x^{t-1-2j}}{k^{2j+1}(t-1-2j)!}\cos kx \,\Bigg|_0^{2\pi z}\\ &~~~+\sum_{j=0}^{\lfloor {(t-2)2} \rfloor}(-1)^{j+1}\frac{x^{t-2j-2}}{k^{2j+2}(t-2j-2)!}\sin kx\,\Bigg|_0^{2\pi z}~\Bigg] \end{align*}\]
let \(t-1=t\) and \(k=2\pi n\) we can evaluate \(\displaystyle \int_{0}^{z}x^{t}\sin\left ( 2\pi nx \right )\mathrm{d}x\) .
Now we obtain the result for the initial integral
\[\boxed{\begin{align*} &\int_{0}^{z}x^{t}\ln\Gamma \left ( 1+x \right )\mathrm{d}x=\\ &\color{blue}{\frac{z^{1+t}\left[\left ( 1+t \right )\ln z-1\right]}{\left ( 1+t \right )^{2}}+\frac{z^{t+1}\ln2\pi }{2\left ( t+1 \right )}+\frac{z^{t+1}\left ( \gamma +\ln2\pi \right )\left ( t-2zt-2z+2 \right )}{2\left ( t+1 \right )\left ( t+2 \right )}}\\ &\color{blue}{+\frac{1}{2}\left ( \frac{z}{2\pi } \right )^{t}\mathrm{Cl}_{2}\left ( 2\pi z \right )-\frac{t\left ( t-1 \right )!}{2\left ( 2\pi \right )^{t+1}}\sum_{k=1}^{\infty }\frac{1}{k^{2}}\Bigg \{\sum_{j=0}^{\lfloor {t-1/2} \rfloor}\frac{(-1)^{j+1}x^{t-1-2j}}{k^{2j+1}(t-1-2j)!}\cos kx \,\Bigg|_0^{2\pi z}}\\ &\color{blue}{+\sum_{j=0}^{\lfloor {(t-2)2} \rfloor}\frac{(-1)^{j+1}x^{t-2j-2}}{k^{2j+2}(t-2j-2)!}\sin kx\,\Bigg|_0^{2\pi z} \Bigg \}+\frac{t!}{\pi }\sum_{n=1}^{\infty }\frac{\ln n}{n}\Bigg \{ \sum_{j=0}^{\lfloor {t/2} \rfloor}\frac{(-1)^{j+1}x^{t-2j}}{\left ( 2\pi n \right )^{2j+1}(t-2j)!}\cos\left ( 2\pi nx \right ) \,\Biggr|_0^{z}}\\ &\color{blue}{+\sum_{j=0}^{\lfloor {(t-1)2} \rfloor}\frac{(-1)^{j+1}x^{t-2j-1}}{\left ( 2\pi n \right )^{2j+2}(t-2j-1)!}\sin \left ( 2\pi nx \right )\,\Biggr|_0^{z} \Bigg\}} \end{align*}}\]

---

\(\mathrm{For~example:}\)
\[\color{red}{\int_{0}^{1}x\ln\Gamma \left ( 1+x \right )\mathrm{d}x=\ln\left ( \frac{2^{\frac{1}{4}}\pi ^{\frac{1}{4}}}{\mathbf{A}e^{\frac{1}{4}}} \right )}\]
\[\color{red}{\int_{0}^{2}x^{2}\ln\Gamma \left ( 1+x \right )\mathrm{d}x=\frac{4}{3}\ln\left (\frac{8\pi }{\mathbf{A}^{3}} \right )+\frac{\zeta \left ( 3 \right )}{2\pi ^{2}}-\frac{5}{2}}\]
\[\color{red}{\int_{0}^{\frac{1}{2}}x^{2}\ln\Gamma \left ( 1+x \right )\mathrm{d}x=\ln\left ( \frac{\mathbf{A}^{\frac{1}{8}}\pi ^{\frac{1}{48}}}{2^{\frac{89}{2880}}} \right )-\frac{5}{8}\zeta '\left ( 3 \right )-\frac{3\zeta \left ( 3 \right )}{32\pi ^{2}}-\frac{3}{128}}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5440148.html