double integral $\int_0^t \int_0^s \frac{\min(u,v)}{uv} \, dv \, du$ I want to calculate the double integral: 
$$\int_0^t \int_0^s \frac{\min(u,v)}{uv} \, dv \, du$$
I don't know how to o that even if it seems simple.
Thanks in advance for your help
 A: Hint: break it into two pieces, one with $u < v$ and the other with $u > v$.
A: Someone changed the function to be plotted from the original function, but the below figure is for the equation as currently stated.

A: $$
\int_0^t \int_0^s \frac{\min(u,v)}{uv} \, dv \, du
$$
The suggestion given by "Ant" and by Robert Israel will work. I would add this: What you get may depend on whether $s<t$ or $t<s$.  To divide the region into two parts as suggested, you need to look at that.
Suppose $s<t$.  Then the region where $u\le v$ is where $0\le u\le v\le s$, and the region where $u\ge v$ is where $0\le v\le\min\{u,s\}$.
(And a similar thing happens when $s>t$.)
A: You didn't mention $s<t$ or $t<s$ ? Let's consider $t<s$, and since we can alter the order of differential operator in the integral we change $dvdu$ to $dudv$; otherwise, the result will diverge. Surely the alter one is given in the original problem.
Now, $$min(u,v)= \begin{cases} 
      u & 0<u\leq v \\
      v & v\leq u\leq s 
   \end{cases} $$
Now the integral becomes,
$$\int_{0}^{t} \int_{0}^{s} \frac{min(u,v)}{uv} \,du\,dv$$
$$=\int_{0}^{t}\{\int_{0}^{v}\frac{u}{uv}du+\int_{v}^{s}\frac{v}{uv}du\}\,dv$$
$$=\int_{0}^{t}(\frac{1}{v}(v-0)+ln(\frac{s}{v}))dv$$
$$=\int_{0}^{t}(1+ln(s)-ln(v))dv$$
$$=(v+v.ln(s)-(v.ln(v)-v)\Big|_0^t$$
$$=t+t.ln(s)-(t. ln(t)-t)$$
$$=t(2+ln(\frac{s}{t}))$$
If you have a question about how to write the integration's second line, please refer to the link: how to solve double integral of a min function
