Double integral of exponential with absolute variables

How would I compute:
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http://latex.codecogs.com/png.latex?\tiny&space;\dpi{200}&space;\iint_{-T}^{T}e^{-\alpha&space;|t_1-t_2|}dt_1dt_2


Thanks

Assume that α is nonzero.
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Assuming that both t₁, t₂ are in [-T, T], we need to split this region into two pieces to remove the absolute value sign: (i) t₁ ≥ t₂, and (ii) t₁ < t₂

So, ∫∫ e^(-α|t₁ - t₂|) dt₁ dt₂
= ∫(t₂ = -T to T) ∫(t₁ = -T to t₂) e^(-α * -(t₁ - t₂)) dt₁ dt₂
+ ∫(t₂ = -T to T) ∫(t₁ = t₂ to T) e^(-α(t₁ - t₂)) dt₁ dt₂

= ∫(t₂ = -T to T) (1/α)e^(α(t₁ - t₂)) {for t₁ = -T to t₂} dt₂
+ ∫(t₂ = -T to T) (-1/α)e^(-α(t₁ - t₂)){for t₁ = t₂ to T} dt₂

= ∫(t₂ = -T to T) (1/α) [1 - e^(α(-T - t₂))] dt₂ + ∫(t₂ = -T to T) (-1/α) [e^(-α(T - t₂)) - 1] dt₂
= (1/α) ∫(t₂ = -T to T) [2 + e^(-αT) (e^(αt₂) - e^(-αt₂))] dt₂
= (1/α) ∫(t₂ = -T to T) [2 + 2e^(-αT) sinh(αt₂)] dt₂
= (1/α) ∫(t₂ = -T to T) [2 + 0] dt₂, since the second term is odd
= 4T/α.

I hope this helps!