# 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.

• divide the region over which you are integrating in two parts, one who has $u > v$ and the other with $v > u$. Calculate those integrals separetely and sum :-) – Ant May 26 '15 at 22:01
• @denek : Although Robert Israels answer, also proposed by "Ant" in a comment, is the right basic idea, you might also want to consider whether $s<t$ or $t<s$. See my answer below. – Michael Hardy May 26 '15 at 22:33

• …and the function being plotted is $\min(u, v)$. :) – wchargin May 27 '15 at 3:29
Hint: break it into two pieces, one with $u < v$ and the other with $u > v$.
$$\int_0^t \int_0^s \frac{\min(u,v)}{uv} \, dv \, du$$ 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 $s>t$.)