I was going through the solid state book by Philip Phillips. I came across an integral similar to:

$$\int_{0}^{\beta}d\tau d\tau^{'}e^{-E_c|\tau-\tau^{'}|}$$

where $\beta E_c >> 1$.

I am not able to solve this integral. I am not sure how to deal with the | | sign occurring in the exponent of e. Can anyone please help?

  • $\begingroup$ what are the limits for $\tau$, $\tau'$ ? $\endgroup$ – lineage Dec 4 '19 at 9:14
  • $\begingroup$ it is 0 to $\beta$ for both of them $\endgroup$ – physu Dec 4 '19 at 9:15
  • $\begingroup$ is $\beta\gt 0$? $\endgroup$ – lineage Dec 4 '19 at 9:17
  • 1
    $\begingroup$ the integral is in a square so try integrating in two regions $\tau\gt\tau'$ and $\tau\lt\tau'$ $\endgroup$ – lineage Dec 4 '19 at 10:01
  • $\begingroup$ $\approx\beta/E_c$ $\endgroup$ – lineage Dec 4 '19 at 10:02

Divide the region of integration into two triangular regions I and II. The diagonal is $\tau=\tau'$. The original integral is the sum of these two sub-integrals.

fig 1

In region I

here $\tau\gt\tau'$, therefore $$ \begin{align} I_I &=\int_{0}^{\beta}d\tau\int_{0}^{\tau} d\tau^{'}e^{-E_c(\tau-\tau^{'})}\\ &=\frac{e^{-E_c \beta}+\beta E_c-1}{E_c^2} \end{align} $$

In region 2

here $\tau\lt\tau'$, therefore

$$ \begin{align} I_{II} &=\int_{0}^{\beta}d\tau\int_{\tau}^{\beta} d\tau^{'}e^{+E_c(\tau-\tau^{'})}\\ &=\frac{e^{-E_c \beta}+\beta E_c-1}{E_c^2}\\ &=I_I \end{align} $$

therefore the original integral

$$I=I_I+I_{II}=\frac{2(e^{-E_c \beta}+\beta E_c-1)}{E_c^2}$$

which under $E_c\beta \gg 1$ reduces to $$2\frac{\beta}{E_c}$$

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