Bernoulli Numbers, Euler-Maclauring Formula - Math Methods class

kde2520
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Homework Statement


The Bloch-Gruneissen approximation for the resistance on a monovalent metal is

\rho=C(T^{5}/\Theta^{6})\int^{\Theta/T}_{0}\frac{x^{5}dx}{(e^{x}-1)(1-e^{-x})}

(a)For T->\infty, show that \rho=(C/4)(T/\Theta^{2})

(b)For T->0, show that \rho=5!\zeta(5)C\frac{T^{5}}{\Theta^{6}}


Homework Equations


The section is on Bernoulli numbers and the Euler-Maclaurin Formula. Several definitions including x/(e^x-1)=sum->(Bn*x^n)/n!, Bernoulli Polynomials, Reimann-Zeta function, etc.


The Attempt at a Solution


For part (a) I see that as T->infinity the upper integration limit goes to zero, thus I may approximate the integrand giving (as the integrand) x^5/[(x+x^2/2!+x^3/3!+...)(-x+x^/2!-x^3/3!+...)]. Can I just multiply this out, simplify, and integrate term by term? If so, over what integration limits?

For part (b) the upper limit goes to infinity so I'm guessing I need to do the integral by substituting some definition of the Bernoulli Numbers?...

Help...

PS - Sorry if the equations are unclear. I'm new to LaTex. Help with that would be appreciated too.
 
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kde2520 said:
Can I just multiply this out, simplify, and integrate term by term?

Yep! (You have a sign wrong though ...)
kde2520 said:
If so, over what integration limits?

Same as before, you just think of the upper limit as a small parameter.
kde2520 said:
For part (b) the upper limit goes to infinity so I'm guessing I need to do the integral by substituting some definition of the Bernoulli Numbers?...

Try expanding everything but the x^5 in powers of e^{-x}.

As for LaTeX, just put the beginning and ending tex and /tex commands around the whole equation:

\rho=C(T^{5}/\Theta^{6})\int^{\Theta/T}_{0}\frac{x^{5}dx}{(e^{x}-1)(1-e^{-x})}
 
I guess you mean the sign is wrong in the expansion of e^{-x}? What is it's expansion?
 
Avodyne said:
Try expanding everything but the x^5 in powers of e^{-x}.

Isn't that what we did for part (a)?

Thanks for the LaTex tip!
 
I mean an expansion like

{1\over 1-e^{-x}}=\sum_{n=0}^\infty (e^{-x})^n
 
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