一个包含arctan与arctanh的积分

[Largeint_0^1frac{arctan x \,operatorname{arctanh} x\, ln x}{x}mathrm{d}x=frac{pi^2}{16}mathbf{G}-frac{7pi}{32}zeta(3) ]


(Large mathrm{mathbf{Proof:}})

Let (n=0,1,2,cdots), We define (I,I_{1,n},I_{2,n}) and (I_n) as follows:

[egin{align*}I &= int_0^1 frac{ln(x) an^{-1}(x) anh^{-1}(x)}{x}mathrm{d}x~~,~~I_{1,n} =int_0^1 x^{2n}ln(x)ln(1-x); mathrm{d}x \ I_{2,n}&=int_0^1 x^{2n}ln(x)ln(1+x); mathrm{d}x~~,~~I_n = int_0^1 x^{2n} ln(x) anh^{-1}(x); mathrm{d}x end{align*} ]

Part I : Evaluation of (I_{1,n}), (I_{2,n}) and (I_n)

[egin{align*} I_{1,n} &= int_0^1 x^{2n}ln(x)ln(1-x) mathrm{d}x= int_0^1 x^{2n}ln(x)left(-sum_{j=1}^infty frac{x^j}{j} ight) mathrm{d}x\ &= -sum_{j=1}^infty frac{1}{j} int_0^1 x^{2n+j}ln(x) mathrm{d}x= sum_{j=1}^infty frac{1}{jleft(2n+1+j ight)^2}\ &= frac{1}{(2n+1)^2}sum_{j=1}^infty left(frac{1}{j}-frac{1}{j+2n+1} ight)-frac{1}{(2n+1)}sum_{j=1}^infty frac{1}{(j+2n+1)^2}\ &= frac{gamma +psi_0(2n+2)}{(2n+1)^2}-frac{psi_1(2n+2)}{2n+1} ag{1} end{align*}]

Similarly,

[egin{align*} I_{2,n}&=int_0^1 x^{2n}ln(x)ln(1+x) mathrm{d}x = int_0^1 x^{2n}ln(x)left(sum_{j=1}^inftyfrac{(-1)^{j+1}x^j}{j} ight) mathrm{d}x\&= sum_{j=1}^infty frac{(-1)^{j+1}}{j}int_0^1 x^{2n+j}ln(x) mathrm{d}x = sum_{j=1}^infty frac{(-1)^{j}}{jleft(2n+1+j ight)^2} \ &= frac{1}{(2n+1)^2}sum_{j=1}^infty frac{(-1)^j}{j}-frac{1}{(2n+1)^2}sum_{j=1}^inftyfrac{(-1)^j}{j+2n+1}-frac{1}{2n+1}sum_{j=1}^infty frac{(-1)^j}{(j+2n+1)^2} \ &= -frac{ln(2)}{(2n+1)^2}+frac{psi_0left( n+dfrac{3}{2} ight)-psi_0(n+1)}{2(2n+1)^2}+frac{psi_1(n+1)-psi_1left(n+dfrac{3}{2} ight)}{4(2n+1)} end{align*} ]

We can make some simplifications using the following identities:

[egin{align*} psi_0left(n+frac{3}{2} ight) &= 2psi_0(2n+2)-psi_0(n+1)-2ln(2) \ psi_1left( n+frac{3}{2} ight) &= 4psi_1(2n+2)-psi_1(n+1) end{align*} ]

So, (I_{2,n}) can be written as:

[egin{align*} I_{2,n}&= -frac{2ln(2)}{(2n+1)^2}+frac{psi_0(2n+2)-psi_0(n+1)}{(2n+1)^2}+frac{2psi_1(n+1)-4psi_1(2n+2)}{4(2n+1)} ag{2} end{align*} ]

Also note that (displaystyle I_n=frac{I_{2,n}-I_{1,n}}{2}). Therefore,

[egin{align*} I_n=-frac{ln(2)}{(2n+1)^2}-frac{gamma+psi_0(n+1)}{2(2n+1)^2}+frac{psi_1(n+1)}{4(2n+1)} ag{3} end{align*}]

Part II : Expressing (I) in terms of Euler Sums

[egin{align*} I &= int_0^1 frac{ln(x) an^{-1}(x) anh^{-1}(x)}{x}mathrm{d}x= int_0^1 frac{ln(x) anh^{-1}(x)}{x}left(sum_{n=0}^infty frac{(-1)^n}{2n+1}x^{2n+1} ight)mathrm{d}x \ &= sum_{n=0}^infty frac{(-1)^n}{2n+1} int_0^1 x^{2n}ln(x) anh^{-1}(x) mathrm{d}x= sum_{n=0}^infty frac{(-1)^n}{2n+1} I_n \ &= sum_{n=0}^infty frac{(-1)^n}{2n+1}left(-frac{ln(2)}{(2n+1)^2}-frac{gamma+psi_0(n+1)}{2(2n+1)^2}+frac{psi_1(n+1)}{4(2n+1)} ight) \ &= -ln(2)frac{pi^3}{32}-frac{1}{2}sum_{n=0}^infty frac{(-1)^n}{(2n+1)^3}left( gamma+psi_0(n+1) ight)+frac{1}{4}sum_{n=0}^infty frac{(-1)^n psi_1(n+1)}{(2n+1)^2} end{align*} ]

Let us denote the euler sums by (E_1) and (E_2):

[E_1 = sum_{n=0}^infty frac{(-1)^n}{(2n+1)^3}left( gamma+psi_0(n+1) ight)~~,~~E_2 = sum_{n=0}^infty frac{(-1)^n psi_1(n+1)}{(2n+1)^2} ]

Part III : Evaluation of (E_1)
We use the FS-contour method. Let (displaystyle f(z)=frac{picsc(pi z) left(gamma+psi_0(-z) ight)}{(2z+1)^3}). Then the sum of all residues of (f(z)) is zero.
The sum of the residues at the negative integers is equal to:

[sum_{n=1}^{infty} ext{Res}_{z=-n}f(z) = sum_{n=1}^infty frac{(-1)^{n-1} left(gamma+psi_0(n) ight)}{(2n-1)^3}= E_1 ]

At (z=-dfrac{1}{2}), the residue is

[ ext{Res}_{z=-1/2}f(z) = frac{pi^3}{8}ln(2)+frac{7pi}{8}zeta(3) ]

The sum of the residues at the positive integers is:

[sum_{n=0}^{infty} ext{Res}_{z=n}f(z) = sum_{n=0}^infty left(-6frac{(-1)^n}{(2n+1)^4}+frac{(-1)^n H_n}{(2n+1)^3} ight)= -6eta(4)+E_1 ]

Therefore,

[E_1+frac{pi^3}{8}ln(2)+frac{7pi}{8}zeta(3)+E_1-6eta(4) = 0 implies E_1 = oxed{3eta(4)-dfrac{7pi}{16}zeta(3)-dfrac{pi^3}{16}ln(2)} ]

Part IV : Evaluation of (E_2)
This time we use FS contour method to the function (displaystyle g(z)=frac{picsc(pi z)psi_1(-z)}{(2z+1)^2}).
The sum of the residues at the negative integers is:

[sum_{n=1}^infty ext{Res}_{z=-n}g(z) =-sum_{n=1}^infty frac{(-1)^{n-1}psi_1(n)}{(2n-1)^2} = -E_2 ]

The residue at (z=-dfrac{1}{2}) is :

[ ext{Res}_{z=-1/2}g(z)=-frac{7pi}{2}zeta(3) ]

The sum of the residues at the positive integers is:

[egin{align*} sum_{n=0}^infty ext{Res}_{z=n}g(z) &= sum_{n=0}^infty left(12frac{(-1)^n}{(2n+1)^4}+frac{pi^2 (-1)^n}{2(2n+1)^2} -frac{(-1)^npsi_1(n+1)}{(2n+1)^2} ight) \ &= 12eta(4)+frac{pi^2}{2}mathbf{G}-E_2 end{align*} ]

The sum of all the residues is zero. Therefore,

[-E_2-frac{7pi}{2}zeta(3)+12eta(4)+frac{pi^2}{2}mathbf{G}-E_2 = 0 implies E_2 = oxed{6eta(4)-dfrac{7pi}{4}zeta(3)+dfrac{pi^2}{4}mathbf{G}} ]

Part V : The Final Answer

[egin{align*} I &= -frac{pi^3 ln(2)}{32}-frac{E_1}{2}+frac{E_2}{4} \ &= -frac{pi^3 ln(2)}{32}-frac{1}{2}left(3eta(4)-frac{7pi}{16}zeta(3)-frac{pi^3}{16}ln(2) ight)+frac{1}{4}left( 6eta(4)-frac{7pi}{4}zeta(3)+frac{pi^2}{4}mathbf{G} ight) \ &=Largeoxed{color{blue}{dfrac{pi^2}{16}mathbf{G}-dfrac{7pizeta(3)}{32}}} end{align*} ]

原文地址:https://www.cnblogs.com/Renascence-5/p/5432396.html