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

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    $\begingroup$ 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 :-) $\endgroup$ – Ant May 26 '15 at 22:01
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    $\begingroup$ @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. $\endgroup$ – Michael Hardy May 26 '15 at 22:33

Someone changed the function to be plotted from the original function, but the below figure is for the equation as currently stated.

enter image description here

  • $\begingroup$ …and the function being plotted is $\min(u, v)$. :) $\endgroup$ – 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 $s>t$.)


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