Sunday, 3 May 2015

calculus - Integrate inti0nftyfractanh(x)x3fracoperatornamesech(x)x2dx




I would like to integrate:



0tanh(x)x3sech(x)x2dx
I'm not sure where I found this integral, but I have a feeling I wrote it down because its solution. I want to say it's related to the Zeta Function, but I'm not sure. I've managed to rewrite it as:
n=1(1)n+1(2ln(2n1n1)+4nln(n12n1)+2n2ln(nn1)+4nln(2)2n2ln(2)+1)



Above follows by writing the hyperbolic functions in terms of exponential functions and then using series. Then I used differentiating under the integral.



This makes me think otherwise about the Zeta Function/having a closed form for the original integral. I would appreciate any help in solving this.



Answer



Start by splitting the integral into two convergent parts:
I=0tanhxx+xxsechxx3dx=0tanhxxx3dx+01sechxx2dx
I1=0tanhxxx3dxIBP=120sech2x1x2dx
IBP=0tanhxsech2xxdx=7ζ(3)π2



I2=01sechxx2dxIBP=0tanhxsechxxdx=4Gπ







I2=0tanhxsechxxdxx=lnt=21(t21)(t2+1)lntdtt=1x=0x21(x2+1)2lnxdx
Above the two integrals were averaged after the reciprocal subtitution was done.



Now we will use Feynman's trick alongside beta function: I(a)=0xa1(x2+1)2lnxdxI(a)=0xa(x2+1)2dx=12(1a2)πsin(π(a+1)2)
I(0)=0I2=π4201asinπ(a+1)2da=2ππ20tsintdt=4Gπ
I1 can be found here.


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