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Hi,

I am reading "An Introduction of Solid State Physics" from Ibach Lüth and don't understand the integration process.

They write $$\sigma=\frac{e^2}{8\pi^3 \hbar}

\int df_{E}dE \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k}) \delta(E-E_F)

$$

$$

= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k})

$$

But why does ## k ## NOT become ##k_F## after the Integration of the Delta-function?

I would think that the integral becomes immedately after the dE-integration:

$$

= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k_F})}{v(\bf{k_F})} \tau(\bf{k_F})

$$

THX

Abbi

I also add the pages

I am reading "An Introduction of Solid State Physics" from Ibach Lüth and don't understand the integration process.

They write $$\sigma=\frac{e^2}{8\pi^3 \hbar}

\int df_{E}dE \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k}) \delta(E-E_F)

$$

$$

= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k})

$$

But why does ## k ## NOT become ##k_F## after the Integration of the Delta-function?

I would think that the integral becomes immedately after the dE-integration:

$$

= \int_{E=E_F}^{}df_{E} \frac{v^2_x(\bf{k_F})}{v(\bf{k_F})} \tau(\bf{k_F})

$$

THX

Abbi

I also add the pages

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