[Largeint_{0}^{pi } heta ln anfrac{ heta }{2}mathrm{d} heta
]
(Largemathbf{Solution:})
显然
[int_{0}^{pi } heta ln anfrac{ heta }{2}mathrm{d} heta =4int_{0}^{pi /2}xln an xmathrm{d}x
]
利用 (mathbf{Lobachevskiy}) 函数的定义
[mathrm{L}left ( x ight )=-int_{0}^{x}lncos xmathrm{d}x~,~ ~ ~ ~ ~ -frac{pi }{2}leq xleq frac{pi }{2} ]
所以
[egin{align*}
int_{0}^{pi /2}xln an xmathrm{d}x &= xleft [ mathrm{L}left ( x
ight )+mathrm{L}left ( frac{pi }{2}-x
ight )
ight ]_{0}^{pi /2}-int_{0}^{pi /2}left [ mathrm{L}left ( x
ight )+mathrm{L}left ( frac{pi }{2}-x
ight )
ight ]mathrm{d}x\
&= left ( frac{pi }{2}
ight )^{2}ln 2-2int_{0}^{pi /2}mathrm{L}left ( x
ight )mathrm{d}x
end{align*}]
再利用
[mathrm{L}left ( x
ight )=xln 2-frac{1}{2}sum_{k=1}^{infty }frac{left ( -1
ight )^{k-1}}{k^{2}}sin 2kx
]
就能计算得
[egin{align*}
int_{0}^{pi /2}mathrm{L}left ( x
ight )mathrm{d}x&=frac{1}{2}left ( frac{pi }{2}
ight )^{2}ln 2-frac{1}{2}sum_{k=1}^{infty }frac{left ( -1
ight )^{k-1}}{k^{2}}int_{0}^{pi /2}sin 2kxmathrm{d}x \
&= frac{pi ^{2}}{8}ln 2-frac{1}{2}sum_{k=1}^{infty }frac{1}{left ( 2k-1
ight )^{3}}
end{align*}]
所以
[egin{align*}
int_{0}^{pi/2}xln an xmathrm{d}x &=frac{pi ^{2}}{4}ln 2-2left [ frac{pi ^{2}}{8}ln 2-frac{1}{2}sum_{k=1}^{infty }frac{1}{left ( 2k-1
ight )^{3}}
ight ] \
&=sum_{k=1}^{infty }frac{1}{left ( 2k-1
ight )^{3}}\
&=sum_{k=1}^{infty } frac{1}{k^{3}}-sum_{k=1}^{infty }frac{1}{left ( 2k
ight )^{3}}=frac{7}{8}zeta left ( 3
ight )
end{align*}]
亦即
[Largeoxed{displaystyle int_{0}^{pi } heta ln anfrac{ heta }{2}mathrm{d} heta=color{blue}{frac{7}{2}zeta left ( 3
ight )}}
]
另外还可以得到
[int_{0}^{pi } heta ln anfrac{ heta }{2}mathrm{d} heta=sum_{n=1}^{infty }frac{1}{n^{2}}left [ psi left ( n+frac{1}{2}
ight )-psi left ( frac{1}{2}
ight )
ight ]=frac{7}{2}zeta left ( 3
ight )
]
或
[Largecolor{purple}{sum_{n=1}^{infty }frac{1}{n^{2}}psi left ( n+frac{1}{2}
ight )=frac{7}{2}zeta left ( 3
ight )-left ( gamma +2ln 2
ight )frac{pi ^{2}}{6}}
]