I would like to integrate:
∫∞0tanh(x)x3−sech(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(2n−1n−1)+4nln(n−12n−1)+2n2ln(nn−1)+4nln(2)−2n−2ln(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=∫∞0tanhx−x+x−xsechxx3dx=∫∞0tanhx−xx3dx+∫∞01−sechxx2dx
I1=∫∞0tanhx−xx3dxIBP=12∫∞0sech2x−1x2dx
IBP=−∫∞0tanhxsech2xxdx=−7ζ(3)π2
I2=∫∞01−sechxx2dxIBP=∫∞0tanhxsechxxdx=4Gπ
I2=∫∞0tanhxsechxxdxx=lnt=2∫∞1(t2−1)(t2+1)lntdtt=1x=∫∞0x2−1(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)=∫∞0xa−1(x2+1)2lnxdx⇒I′(a)=∫∞0xa(x2+1)2dx=12(1−a2)πsin(π(a+1)2)
I(0)=0⇒I2=π4∫201−asinπ(a+1)2da=2π∫π20tsintdt=4Gπ
I1 can be found here.
No comments:
Post a Comment