Need to help to solve an integral from quantum conductance

[skarthikselvan]

Hi All,

While going through quantum conductnace derivation, I got stuck by an integral.

I = ((e*(h/2pi))/(m*pi)) Integral from 0 to infinity

dt / ( exp (((((h/2pi)^2 * t)/(2*m)) - Ef) / kT)) + 1)

The answer is
I = (2e/h)*kT*ln(1 + exp(Ef/kT))

I am stuggling to find the answer. Any help will be appreciated.

Thanks
Karthikselvan

 

[VietDao29]

Hi All,

While going through quantum conductnace derivation, I got stuck by an integral.

I = ((e*(h/2pi))/(m*pi)) Integral from 0 to infinity

dt / ( exp (((((h/2pi)^2 * t)/(2*m)) - Ef) / kT)) + 1)

Cannot really get what you are asking here...
Are you trying to evaluate:
$$\mathop{\int} \limits_{0} ^ {\infty} \frac{dt}{e ^ { \frac{ \frac{ \left( \frac{h}{2 \pi} \right) ^ 2 t}{2m} - Ef}{kT} + 1}}$$?
If that's what you mean, then what does the part before it: I = ((e*(h/2pi))/(m*pi)) have to do here? Is it also an integral? Or a constant?
Or do you mean:
Are you trying to evaluate:
$$I = \frac{e ^ {\frac{h}{2 \pi}}}{2m} \mathop{\int} \limits_{0} ^ {\infty} \frac{dt}{e ^ {\frac{\frac{ \left( \frac{h}{2 \pi} \right) ^ 2 t}{2m} - Ef}{kT} + 1}}$$?

 

[Data]

I think $e$ is the electronic charge (hence the exp notation later on), it looks like he's trying to calculate a charge density (the integrand is the Fermi-Dirac distribution for average number of fermions in a given state).

To evaluate

$$\int_0^\infty \frac{1}{e^{\frac{ax-b}{c}} + 1}dx$$

where a,b,c > 0, just sub

$$u = e^{\frac{ax-b}{c}}.$$

Then $\frac{c}{au}du = dx$, and when $x=0$ you get $u = e^{-b/c}$, so the integral is

$$\int_{e^{-b/c}}^\infty \frac{c}{au(u+1)}du.$$

You can split the integrand up by partial fractions:

$$\frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1},$$

and I think you should be able to finish from there (you'll need to simplify a bit after integrating to get to the form of the answer you posted above, but it's right)

 

[skarthikselvan]

Thanks Data. I am able to get the answer after following your suggestions.