- #1
Littlepig
- 99
- 0
I there.
I'm currently using this kind of integrals, with n even, and I couldn't found anything in internet for calculate this.
Let From the book I'm studying Ashcroft/Mermin, Solid State Physics, Append C, it says that
[tex]a_{n}=\int_{-\infty }^{\infty } \frac{x^n
e^x}{\left(e^x+1\right)^2} \, dx[/tex]
can, by elementary operations, be written as
[tex]a_{n}=1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}-\frac{1}{4^{2n}}+...[/tex] and so can be written with the zeta function:
[tex]a_{n}=(2-\frac{1}{2^{2(n+1)}}) \zeta(2n)[/tex]
and, [tex]\zeta(2n)=2^{2n-1}\frac{\pi^{2n}}{(2n)!}B_{n}[/tex] where B_n are the bernoulli numbers.
Well, aren't any easier way, using integration in complex plane? Can you give me an ideia of where can I find a resolution to this?
Thanks,
littlepig
I'm currently using this kind of integrals, with n even, and I couldn't found anything in internet for calculate this.
Let From the book I'm studying Ashcroft/Mermin, Solid State Physics, Append C, it says that
[tex]a_{n}=\int_{-\infty }^{\infty } \frac{x^n
e^x}{\left(e^x+1\right)^2} \, dx[/tex]
can, by elementary operations, be written as
[tex]a_{n}=1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}-\frac{1}{4^{2n}}+...[/tex] and so can be written with the zeta function:
[tex]a_{n}=(2-\frac{1}{2^{2(n+1)}}) \zeta(2n)[/tex]
and, [tex]\zeta(2n)=2^{2n-1}\frac{\pi^{2n}}{(2n)!}B_{n}[/tex] where B_n are the bernoulli numbers.
Well, aren't any easier way, using integration in complex plane? Can you give me an ideia of where can I find a resolution to this?
Thanks,
littlepig