# Solving a two variable integration [closed]

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?

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

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 $$2\frac{\beta}{E_c}$$