奇怪的唇膏(奇怪的唇膏Large奇怪的唇膏int_0^1奇怪的唇膏frac{ )奇怪的唇膏arctan x奇怪的唇膏，奇怪的唇膏operatorname ) arctan x ) 奇怪的唇膏mathrm{d}x=奇怪的唇膏frac{奇怪的唇膏pi^2}{16}奇怪的唇膏mathbf{G}-奇怪的唇膏frac{7奇怪的唇膏pi}{32}
奇怪的唇膏(奇怪的唇膏Large奇怪的唇膏mathrm{ (奇怪的唇膏mathbf ) proof: ) ) ) ) ) ) )。
Let奇怪的唇膏(n=0，1，2，奇怪的唇膏cdots奇怪的唇膏)、We define奇怪的唇膏(I，I_{1，n}，I_{2，n}奇怪的唇膏)
奇怪的唇膏(奇怪的唇膏begin(aligndtdqz ) I=奇怪的唇膏int_0^1奇怪的唇膏frac{ )奇怪的唇膏ln(x )奇怪的唇膏tan ^ {-1 } 奇怪的唇膏mathrm{d}x奇怪的唇膏I_{2，n}=奇怪的唇膏int_0^1 x^{2n}奇怪的唇膏ln(x )奇怪的唇膏ln ) 1x )奇怪的唇膏~~I_n=奇怪唇膏int_0^1 x^{2n}奇怪唇膏ln(x )奇怪唇膏tanh^{-1}(x ) x )奇怪唇膏奇怪唇膏mathrm{d}x
Part I : Evaluation of 搞怪的唇膏(I_{1,n}搞怪的唇膏), 搞怪的唇膏(I_{2,n}搞怪的唇膏) and 搞怪的唇膏(I_n搞怪的唇膏)
奇怪的唇膏[奇怪的唇膏begin{aligndtdqz} I_{1， n}=奇怪唇膏int_0^1 x^{2n}奇怪唇膏ln(x )奇怪唇膏ln(1-x )奇怪唇膏mathrm ) x=奇怪唇膏int_0^1 x^{2n} 奇怪的唇膏ln ) x )奇怪的唇膏奇怪的唇膏right (奇怪的唇膏mathrm ) d ) x奇怪的唇膏=-奇怪的唇膏sum_(j=1) ^奇怪的唇膏infty奇怪的莉莉奇怪的唇膏^奇怪的唇膏infty奇怪的唇膏FRAC{1}{2n1j奇怪的唇膏left(2n1j奇怪的唇膏right )2}奇怪的唇膏=奇怪的唇膏FRAC(1) ) ^ -奇怪的唇膏frac{1}{j 2n 1}奇怪的唇膏right(-奇怪的唇膏FRAC{1}{(2n1) }奇怪的唇膏sum_{j=1}^奇怪的唇膏infty 奇怪的唇膏=奇怪的唇膏FRAC (奇怪的唇膏gamma奇怪的唇膏psi_0(2n2 ) (2n1 ) ^2) -奇怪的唇膏FRAC )奇怪的唇膏psi_1(2n2 )
Similarly，
奇怪的唇膏[奇怪的唇膏begin{aligndtdqz} I_{2，n}=奇怪的唇膏int_0^1 x^{2n}奇怪的唇膏ln(x )奇怪的唇膏ln ) 1x 奇怪的唇膏ln ) x )奇怪的唇膏(x ) j )奇怪的唇膏right )奇怪的唇膏mathrm ) d ) x奇怪的唇膏=奇怪的唇膏sum_{j=1}^奇怪的唇膏奇怪的唇膏x=奇怪的唇膏sum_{j=1}^奇怪的唇膏infty奇怪的唇膏FRAC{(-1 ) ^{j}}{j奇怪的唇膏left(2n1j奇怪的唇膏right (奇怪的唇膏(infty )奇怪的唇膏FRAC ) ) j ) -奇怪的唇膏FRAC ) )2n1 )2)奇怪的唇膏sum_{j=1}^奇怪的唇膏infty
唇膏 &= -搞怪的唇膏frac{搞怪的唇膏ln(2)}{(2n+1)^2}+搞怪的唇膏frac{搞怪的唇膏psi_0搞怪的唇膏left( n+搞怪的唇膏dfrac{3}{2}搞怪的唇膏right)-搞怪的唇膏psi_0(n+1)}{2(2n+1)^2}+搞怪的唇膏frac{搞怪的唇膏psi_1(n+1)-搞怪的唇膏psi_1搞怪的唇膏left(n+搞怪的唇膏dfrac{3}{2} 搞怪的唇膏right)}{4(2n+1)} 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]We can make some simplifications using the following identities:
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} 搞怪的唇膏psi_0搞怪的唇膏left(n+搞怪的唇膏frac{3}{2} 搞怪的唇膏right) &= 2搞怪的唇膏psi_0(2n+2)-搞怪的唇膏psi_0(n+1)-2搞怪的唇膏ln(2) 搞怪的唇膏搞怪的唇膏 搞怪的唇膏psi_1搞怪的唇膏left( n+搞怪的唇膏frac{3}{2}搞怪的唇膏right) &= 4搞怪的唇膏psi_1(2n+2)-搞怪的唇膏psi_1(n+1) 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]
So, 搞怪的唇膏(I_{2,n}搞怪的唇膏) can be written as:
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} I_{2,n}&= -搞怪的唇膏frac{2搞怪的唇膏ln(2)}{(2n+1)^2}+搞怪的唇膏frac{搞怪的唇膏psi_0(2n+2)-搞怪的唇膏psi_0(n+1)}{(2n+1)^2}+搞怪的唇膏frac{2搞怪的唇膏psi_1(n+1)-4搞怪的唇膏psi_1(2n+2)}{4(2n+1)} 搞怪的唇膏tag{2} 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]
Also note that 搞怪的唇膏(搞怪的唇膏displaystyle I_n=搞怪的唇膏frac{I_{2,n}-I_{1,n}}{2}搞怪的唇膏). Therefore,
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} I_n=-搞怪的唇膏frac{搞怪的唇膏ln(2)}{(2n+1)^2}-搞怪的唇膏frac{搞怪的唇膏gamma+搞怪的唇膏psi_0(n+1)}{2(2n+1)^2}+搞怪的唇膏frac{搞怪的唇膏psi_1(n+1)}{4(2n+1)}搞怪的唇膏tag{3} 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]
Part II : Expressing 搞怪的唇膏(I搞怪的唇膏) in terms of Euler Sums
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} I &= 搞怪的唇膏int_0^1 搞怪的唇膏frac{搞怪的唇膏ln(x)搞怪的唇膏tan^{-1}(x)搞怪的唇膏tanh^{-1}(x)}{x}搞怪的唇膏mathrm{d}x= 搞怪的唇膏int_0^1 搞怪的唇膏frac{搞怪的唇膏ln(x)搞怪的唇膏tanh^{-1}(x)}{x}搞怪的唇膏left(搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{2n+1}x^{2n+1} 搞怪的唇膏right)搞怪的唇膏mathrm{d}x 搞怪的唇膏搞怪的唇膏 &= 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{2n+1} 搞怪的唇膏int_0^1 x^{2n}搞怪的唇膏ln(x)搞怪的唇膏tanh^{-1}(x) 搞怪的唇膏mathrm{d}x= 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{2n+1} I_n 搞怪的唇膏搞怪的唇膏 &= 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{2n+1}搞怪的唇膏left(-搞怪的唇膏frac{搞怪的唇膏ln(2)}{(2n+1)^2}-搞怪的唇膏frac{搞怪的唇膏gamma+搞怪的唇膏psi_0(n+1)}{2(2n+1)^2}+搞怪的唇膏frac{搞怪的唇膏psi_1(n+1)}{4(2n+1)} 搞怪的唇膏right) 搞怪的唇膏搞怪的唇膏 &= -搞怪的唇膏ln(2)搞怪的唇膏frac{搞怪的唇膏pi^3}{32}-搞怪的唇膏frac{1}{2}搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{(2n+1)^3}搞怪的唇膏left( 搞怪的唇膏gamma+搞怪的唇膏psi_0(n+1)搞怪的唇膏right)+搞怪的唇膏frac{1}{4}搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n 搞怪的唇膏psi_1(n+1)}{(2n+1)^2} 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]
Let us denote the euler sums by 搞怪的唇膏(E_1搞怪的唇膏) and 搞怪的唇膏(E_2搞怪的唇膏):
搞怪的唇膏[E_1 = 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n}{(2n+1)^3}搞怪的唇膏left( 搞怪的唇膏gamma+搞怪的唇膏psi_0(n+1)搞怪的唇膏right)~~,~~E_2 = 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^n 搞怪的唇膏psi_1(n+1)}{(2n+1)^2}搞怪的唇膏]
Part III : Evaluation of 搞怪的唇膏(E_1搞怪的唇膏)
We use the FS-contour method. Let 搞怪的唇膏(搞怪的唇膏displaystyle f(z)=搞怪的唇膏frac{搞怪的唇膏pi搞怪的唇膏csc(搞怪的唇膏pi z) 搞怪的唇膏left(搞怪的唇膏gamma+搞怪的唇膏psi_0(-z) 搞怪的唇膏right)}{(2z+1)^3}搞怪的唇膏). Then the sum of all residues of 搞怪的唇膏(f(z)搞怪的唇膏) is zero.
The sum of the residues at the negative integers is equal to:
搞怪的唇膏[搞怪的唇膏sum_{n=1}^{搞怪的唇膏infty}搞怪的唇膏text{Res}_{z=-n}f(z) = 搞怪的唇膏sum_{n=1}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^{n-1} 搞怪的唇膏left(搞怪的唇膏gamma+搞怪的唇膏psi_0(n) 搞怪的唇膏right)}{(2n-1)^3}= E_1 搞怪的唇膏]
At 搞怪的唇膏(z=-搞怪的唇膏dfrac{1}{2}搞怪的唇膏), the residue is
搞怪的唇膏[搞怪的唇膏text{Res}_{z=-1/2}f(z) = 搞怪的唇膏frac{搞怪的唇膏pi^3}{8}搞怪的唇膏ln(2)+搞怪的唇膏frac{7搞怪的唇膏pi}{8}搞怪的唇膏zeta(3)搞怪的唇膏]
The sum of the residues at the positive integers is:
搞怪的唇膏[搞怪的唇膏sum_{n=0}^{搞怪的唇膏infty}搞怪的唇膏text{Res}_{z=n}f(z) = 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏left(-6搞怪的唇膏frac{(-1)^n}{(2n+1)^4}+搞怪的唇膏frac{(-1)^n H_n}{(2n+1)^3} 搞怪的唇膏right)= -6搞怪的唇膏beta(4)+E_1搞怪的唇膏]
Therefore,
搞怪的唇膏[E_1+搞怪的唇膏frac{搞怪的唇膏pi^3}{8}搞怪的唇膏ln(2)+搞怪的唇膏frac{7搞怪的唇膏pi}{8}搞怪的唇膏zeta(3)+E_1-6搞怪的唇膏beta(4) = 0 搞怪的唇膏implies E_1 = 搞怪的唇膏boxed{3搞怪的唇膏beta(4)-搞怪的唇膏dfrac{7搞怪的唇膏pi}{16}搞怪的唇膏zeta(3)-搞怪的唇膏dfrac{搞怪的唇膏pi^3}{16}搞怪的唇膏ln(2)} 搞怪的唇膏]
Part IV : Evaluation of 搞怪的唇膏(E_2搞怪的唇膏)
This time we use FS contour method to the function 搞怪的唇膏(搞怪的唇膏displaystyle g(z)=搞怪的唇膏frac{搞怪的唇膏pi搞怪的唇膏csc(搞怪的唇膏pi z)搞怪的唇膏psi_1(-z)}{(2z+1)^2}搞怪的唇膏).
The sum of the residues at the negative integers is:
搞怪的唇膏[搞怪的唇膏sum_{n=1}^搞怪的唇膏infty 搞怪的唇膏text{Res}_{z=-n}g(z) =-搞怪的唇膏sum_{n=1}^搞怪的唇膏infty 搞怪的唇膏frac{(-1)^{n-1}搞怪的唇膏psi_1(n)}{(2n-1)^2} = -E_2 搞怪的唇膏]
The residue at 搞怪的唇膏(z=-搞怪的唇膏dfrac{1}{2}搞怪的唇膏) is :
搞怪的唇膏[搞怪的唇膏text{Res}_{z=-1/2}g(z)=-搞怪的唇膏frac{7搞怪的唇膏pi}{2}搞怪的唇膏zeta(3)搞怪的唇膏]
The sum of the residues at the positive integers is:
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏text{Res}_{z=n}g(z) &= 搞怪的唇膏sum_{n=0}^搞怪的唇膏infty 搞怪的唇膏left(12搞怪的唇膏frac{(-1)^n}{(2n+1)^4}+搞怪的唇膏frac{搞怪的唇膏pi^2 (-1)^n}{2(2n+1)^2} -搞怪的唇膏frac{(-1)^n搞怪的唇膏psi_1(n+1)}{(2n+1)^2}搞怪的唇膏right) 搞怪的唇膏搞怪的唇膏 &= 12搞怪的唇膏beta(4)+搞怪的唇膏frac{搞怪的唇膏pi^2}{2}搞怪的唇膏mathbf{G}-E_2 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]
The sum of all the residues is zero. Therefore,
搞怪的唇膏[-E_2-搞怪的唇膏frac{7搞怪的唇膏pi}{2}搞怪的唇膏zeta(3)+12搞怪的唇膏beta(4)+搞怪的唇膏frac{搞怪的唇膏pi^2}{2}搞怪的唇膏mathbf{G}-E_2 = 0 搞怪的唇膏implies E_2 = 搞怪的唇膏boxed{6搞怪的唇膏beta(4)-搞怪的唇膏dfrac{7搞怪的唇膏pi}{4}搞怪的唇膏zeta(3)+搞怪的唇膏dfrac{搞怪的唇膏pi^2}{4}搞怪的唇膏mathbf{G}}搞怪的唇膏]
Part V : The Final Answer
搞怪的唇膏[搞怪的唇膏begin{aligndtdqz} I &= -搞怪的唇膏frac{搞怪的唇膏pi^3 搞怪的唇膏ln(2)}{32}-搞怪的唇膏frac{E_1}{2}+搞怪的唇膏frac{E_2}{4} 搞怪的唇膏搞怪的唇膏 &= -搞怪的唇膏frac{搞怪的唇膏pi^3 搞怪的唇膏ln(2)}{32}-搞怪的唇膏frac{1}{2}搞怪的唇膏left(3搞怪的唇膏beta(4)-搞怪的唇膏frac{7搞怪的唇膏pi}{16}搞怪的唇膏zeta(3)-搞怪的唇膏frac{搞怪的唇膏pi^3}{16}搞怪的唇膏ln(2) 搞怪的唇膏right)+搞怪的唇膏frac{1}{4}搞怪的唇膏left( 6搞怪的唇膏beta(4)-搞怪的唇膏frac{7搞怪的唇膏pi}{4}搞怪的唇膏zeta(3)+搞怪的唇膏frac{搞怪的唇膏pi^2}{4}搞怪的唇膏mathbf{G}搞怪的唇膏right) 搞怪的唇膏搞怪的唇膏 &=搞怪的唇膏Large搞怪的唇膏boxed{搞怪的唇膏color{blue}{搞怪的唇膏dfrac{搞怪的唇膏pi^2}{16}搞怪的唇膏mathbf{G}-搞怪的唇膏dfrac{7搞怪的唇膏pi搞怪的唇膏zeta(3)}{32}}} 搞怪的唇膏end{aligndtdqz}搞怪的唇膏]