How to calculate $\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du $ I am trying to calculate $\int_0^2\int_0^2 e^{-|t-u|} \,dt \,du $ 
I solved the first integral to obtain
$$\int_0^2 e^{-|t-u|} \, dt = \left|\frac{e^{-|t-u|}}{u}\right|_0^2 $$
$$ \Longrightarrow \int_0^2 e^{-|t-u|} \, dt = \frac{e^{-|2-u|}}{u} - \frac{e^u}{u} $$
Putting the answer above into the outer integral, I now have
$$ \int_0^2\int_0^2 e^{-|t-u|} = \int_0^2\left[\frac{e^{-|2-u|}}{u} - \frac{e^u}{u} \right]\, du $$
I tried separating the integrals and using integration by parts, but I could not get an answer.
Please can you help me with tips on how to solve this integral or show me where I'm wrong?
 A: The square $[0, 2] \times [0, 2]$ can be divided into two triangles, the first where $t>u,$ the second where $t<u$. On both of these you can drop the absolute values, and by changing the variable to $v=t-u,$ $w=t+u$ easily integrate.
A: You are trying to integrate over a square from $(0,0)$ to $(2,2)$. Above the diagonal, you have $t<u$ and below the diagonal we have $t>u$ (assuming $t$ on horizontal and $u$ on vertical)
To integrate over this square, what we can do is first fix $t$, and then integrate over $u$, then integrate over $t$
Suppose we look at a particular value of $t$
For $u < t$, $|t-u| = t-u$
$u>t$, $|t-u| = u-t$
Hence we can re-write the integral as
$$I = \int_0^2 \left(\int_0^t \exp(u-t))du  + \int_t^2 \exp(t-u)du\right)dt$$
Can you proceed with this?
A: Per the symmetry with respect to the line $u-t=0$
$$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du 
= 2\int_0^2\int_0^t e^{-|t-u|} du \>dt = 2\int_0^2\int_0^t e^{-(t-u)} du \>dt =2 (1+e^{-2})
$$
A: Thank you. I have seen my error. I made a mistake in the integration
Following @Quanto 's guide,
$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du 
= 2\int_0^2\int_0^t e^{-|t-u|} du \>dt = 2\int_0^2\int_0^t e^{-(t-u)} du \>dt =2 (1+e^{-2})$
$\int_0^t e^{(u-t)}du = \big|\frac{e^{(u-t)}}{1}\big|_0^t$ 
$\int_0^t e^{(u-t)}du = e^{t-t} - e^{-t} = 1 - e^{-t}$

Solving the outer integral gives
$2\int_0^2(1 - e^{-t})dt = 2\big[ t + e^{-t}\big]_0^2$ 
$2\int_0^2(1 - e^{-t})dt = 2\big[ (2 + e^{-2})-(0 + e^0)\big]_0^2$ 
$2\int_0^2(1 - e^{-t})dt = 2\big[ 2 + e^{-2} - 1\big]$ 
$2\int_0^2(1 - e^{-t})dt = 2(e^{-2} + 1)$ 
$=>\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = 2(e^{-2} + 1)$ 

Also, following @DhanviSreenivasan 's solution,
$\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = \int_0^2 \left(\int_0^t e^{(u-t)})du  + \int_t^2 e^{(t-u)}du\right)dt$

$\int_0^t e^{(u-t)}du  + \int_t^2 e^{(t-u)}du = \big[ \frac{e^{u-t}}{1}\big]_0^t - \big[\frac{e^{t-u}}{1}\big]_t^2$

$\int_0^t e^{(u-t)}du  + \int_t^2 e^{(t-u)}du = (e^0 - e^{-t}) - (e^{t-2} - e^0) = 2 - e^{t-2} - e^{-t}$

Applying the outer integral, 
$\int_0^2 (2 - e^{t-2} - e^{-t})dt = \big[2t-e^{t-2}+e^{-t}\big]_0^2$ 
$\int_0^2 (2 - e^{t-2} - e^{-t})dt = 2(1+e^{-2})$ 
$=>\int_0^2\int_0^2 e^{-|t-u|} \, dt \, du = 2(e^{-2} + 1)$ 
Thanks to those who helped me.
