Integral of Absolute Difference of Piecewise Functions

Question

I'm trying to determine an analytic formula for the Wasserstein-1 distance between two Laplace distributions.

Since the distriutions are in 1D, I can use the relationship between the W1 distance and the $L_1$ difference of the cumulative density functions:

$$ \mathbb{W}(\mathrm{L}(\mu_1,b_1),\mathrm{L}(\mu_2,b_2)) = \int_{x\in\mathbb{R}} |c_1(x)-c_2(x)| $$

where $c_i(x) = \frac{1}{2}e^{\frac{x-\mu_i}{b_i}}$ when $x \leq \mu_i$ and $1-\frac{1}{2}e^{-\frac{x-\mu_i}{b_i}}$ otherwise (sorry I can't write this as a piecewise function; the SE editor thinks it's code and won't let me publish the question).

We can tell already that this is a simple integral in the sense that it breaks down into sums of exponential integrals. I was hoping Mathematica could spare me from filling in the details, so I wrote the following code:

f1[x_] := Piecewise[{{0.5*Exp[(x - \[Mu]1)/b1], x < \[Mu]1}, {1 - 0.5*Exp[-(x - \[Mu]1)/b1], x > \[Mu]1}}]
f2[x_] := Piecewise[{{0.5*Exp[(x - \[Mu]2)/b2], x < \[Mu]2}, {1 - 0.5*Exp[-(x - \[Mu]2)/b2], x > \[Mu]2}}]
Integrate[Abs[f1[x] - f2[x]], {x, -Infinity, Infinity}, Assumptions -> {b1 > 0, b2 > 0,Element[\[Mu]1, Reals], Element[\[Mu]2, Reals]}]

Unfortunately, this code doesn't work (it just returns the integral statement).

Is there a better way to formulate this integral?

Answer 1

Four parameters are too much for Integrate. It works well, when specifying the parameters:

\[Mu]1 = -1; b1 = 1; b2 = 1/Pi; \[Mu]2 = 2; 
Integrate[RealAbs[Piecewise[{{1/2*Exp[(x - \[Mu]1)/b1], 
x <= \[Mu]1}, {1 - 1/2*Exp[-(x - \[Mu]1)/b1], x > \[Mu]1}}] - 
Piecewise[{{1/2*Exp[(x - \[Mu]2)/b2], x <= \[Mu]2}, {1 - 1/2*Exp[-(x - \[Mu]2)/b2], 
x > \[Mu]2}}]], {x, -Infinity, Infinity}]

$$ \frac{e^{-\frac{3 \pi }{\pi -1}} \left(-1+\pi +3 e^{\frac{3 \pi }{\pi -1}} \pi \right)}{\pi }$$