Bài toán.Chứng minh rằng :
$ln\Gamma (x)=\frac{1}{2}ln(2\pi )+\frac{1}{2}\sum_{k=1}^{\infty }\frac{cos2k\pi x}{k}+\sum_{k=1}^{\infty }\frac{(\gamma +ln2\pi k)sin2k\pi x }{\pi k}, (0< x< 1)$, trong đó $\gamma$ là hằng số Euler
Lời giải. Để thực hiện được phép chứng minh trên ta cần đến các kết quả quen thuộc sau :
$\Gamma (1-x)\Gamma (x)=\frac{\pi }{sin(\pi x)}$ ; $\int_{0}^{\pi }ln(sinx)dx=-\pi ln2$ và công thức
$\Gamma (x)=\frac{e^{-\gamma x}}{x}\prod_{k=1}^{\infty }e^{\frac{x}{k}}\left ( 1+\frac{x}{k} \right )^{-1}$
Ta có:
$a_{0}=2\int_{0}^{1}ln\Gamma (x)dx=2\int_{1}^{0}ln\Gamma (1-x)d(1-x)=2\int_{0}^{1}ln\Gamma (1-x)dx$
$\Rightarrow a_{0}=\int_{0}^{1}ln\left ( \frac{\pi }{sin\pi x} \right )dx=ln\pi -\int_{0}^{1}ln(sin\pi x)dx=ln(2\pi )$
$a_{k}=2\int_{0}^{1}ln\Gamma (x)cos2k\pi xdx=2\int_{0}^{1}ln\Gamma (1-x)cos2k\pi xdx$
$\Rightarrow a_{k}=\int_{0}^{1}ln\left ( \frac{\pi }{sin\pi x} \right )cos2k\pi xdx=-\int_{0}^{1}ln(sin\pi x)cos2k\pi xdx$$=-\frac{1}{\pi }\int_{0}^{\pi }ln(sinx)cos2kxdx=-ln(sinx)\frac{sin2k x}{2k\pi } _{0}^{\pi } +\frac{1}{2k\pi }\int_{0}^{\pi }\frac{sin2k xcosx}{sinx}dx$=$\frac{1}{2k\pi }\int_{0}^{\pi }\frac{sin2kxcosxdx}{sinx}$
Đặt $I_{k}=\int_{0}^{\pi }\frac{sin2kxcosxdx}{sinx}\Rightarrow I_{k+1}=\int_{0}^{\pi }\frac{sin2(k+1)xcosxdx}{sinx}$. Do đó $I_{k+1}-I_{k}=2\int_{0}^{\pi }\frac{cos(2k+1)xsinxcosxdx}{sinx}=2\int_{0}^{\pi }cos(2k+1)xcosxdx=0$ nên $I_{k}=2\int_{0}^{\pi }cos^{2}xdx=\pi \Rightarrow a_{k}=\frac{1}{2k\pi }\times \pi =\frac{1}{2k}$
Theo một kết quả quen thuộc:
$\Gamma (x)=\frac{e^{-\gamma x}}{x}\prod_{k=1}^{\infty }e^{\frac{x}{k}}\left ( 1+\frac{x}{k} \right )^{-1}\Rightarrow ln\Gamma (x)=-(\gamma x+lnx)+\sum_{k=1}^{\infty }\left \{ \frac{x}{k}-ln\left ( 1+\frac{x}{k} \right ) \right \}$$\Rightarrow ln\Gamma (x)sin2k\pi x=-(\gamma x+lnx)sin2k\pi x+\sum_{k=1}^{\infty }\left \{ \frac{x}{k}-ln\left ( 1+\frac{x}{k} \right ) \right \}sin2k\pi x$.
trong đó :
$\int_{0}^{1}(-\gamma x-lnx)sin2k\pi xdx=\frac{\gamma xcos2k\pi x}{2k\pi }_{0}^{1}-\frac{\gamma }{2k\pi }\int_{0}^{1}cos2k\pi xdx+\frac{lnxcos2k\pi x}{2k\pi }_{0}^{1}-\frac{1}{2k\pi }\int_{0}^{1}\frac{cos2k\pi xdx}{x}=\frac{\gamma }{2k\pi }+\left [ \frac{lnxcos2k\pi x-Ci(2k\pi x)}{2k\pi } \right ]_{0}^{1}$$\frac{\gamma -Ci(2k\pi )}{2k\pi }-lim_{\epsilon \rightarrow \infty }\frac{ln\epsilon -Ci(2k\pi \epsilon )}{2k\pi }=\frac{\gamma -Ci(2k\pi )}{2k\pi }-lim_{\epsilon \rightarrow \infty }\frac{ln\epsilon -ln(2k\pi )-ln\epsilon -\gamma +O(1)}{2k\pi }$$=\frac{2\gamma +ln(2k\pi )-Ci(2k\pi )}{2k\pi }$
$\int_{0}^{1}\sum_{k=1}^{\infty }\left \{ \frac{x}{k}-ln(1+\frac{x}{k}) \right \}sin2k\pi xdx=-\frac{1}{2k\pi }\sum_{n=1}^{\infty }\left \{ \frac{1}{n}+Ci(2k\pi n+2k\pi )-Ci(2k\pi n)-ln\left ( 1+\frac{1}{n} \right ) \right \}$
$=-\frac{1}{2k\pi }\left [ \sum_{n=1}^{\infty } \left \{ \frac{1}{n}-ln\left ( 1+\frac{1}{n} \right ) \right \}+lim_{N\rightarrow \infty }\left ( \sum_{n=2}^{N+1}Ci(2k\pi n)-\sum_{n=1}^{N}Ci(2k\pi n) \right )\right ]$
$=-\frac{1}{2k\pi }\left [ \gamma +lim_{N\rightarrow \infty } \left ( Ci(2k\pi x+2k\pi N) -Ci(2k\pi )\right )\right ]=\frac{Ci(2k\pi )-\gamma }{2k\pi }$.
Do đó
$\int_{0}^{1}ln\Gamma (x)sin2k\pi xdx=\frac{\gamma +ln(2k\pi )}{2k\pi }\Rightarrow b_{k}=\frac{\gamma +ln(2k\pi )}{k\pi }$
Bài viết đã được chỉnh sửa nội dung bởi sinh vien: 06-05-2015 - 07:58
