Double integral with minimum and maximum How can I compute the following integral:
$$\int_0^t\int_0^se^{u \land r-u\lor r}drdu$$
I know how to solve it when there is only one integral, but the double integral it is making it harder.
 A: With the assumption that $u \land r=\min\{u,r\}$ and $u\lor r = \max \{u,r\}$ we have two cases: $t>s$ or $t<S$. We define $f(u,r)=\frac{e^{\min\{u,r\}}}{e^{\max \{u,r\}}}$
First consider $t<s$ in this 
$$\int_0^t\int_0^sf(u,r)drdu = \int_0^tdu\int_0^u \frac{e^r}{e^u}dr+\int_t^sdu\int_0^t \frac{e^r}{e^u}dr+\int_0^tdu\int_u^t \frac{e^u}{e^r}dr$$
Now if we consider $t>s$ the we have
$$\int_0^t\int_0^sf(u,r)drdu =\int_0^sdr\int_r^s \frac{e^r}{e^u}du+\int_0^sdr\int_0^r \frac{e^u}{e^r}du+\int_s^tdr\int_0^s \frac{e^u}{e^r}du$$
Now we have to evaluate each part. Here we go for the first one
$$\int_0^t\int_0^sf(u,r)drdu = \left(u+e^{-u}\right)\Bigl|_0^t + \left(1-e^t \right)e^{-u} \Bigl|_t^s+\left(u-e^ue^{-t}\right)\Bigl|_0^t=2t-2+2e^{-t} + (e^t-1)(e^{-t}-e^{-s})$$
For the second one we have
$$\int_0^t\int_0^sf(u,r)drdu = \left( -e^{-s}e^r-r\right)\Bigl|_0^s+\left(r+e^{-r} \right)\Bigl|_0^s+\left( e^{-r}(1-e^{s})\right)\Bigl|_s^t=(e^{-t}-e^{-s})(1-e^s)$$
In each integral you can change the order of integration, thus each first and second cases have 8 representations, but the answer must be the same.
