How to find $$\int_0^1 \int_x^1 \arctan \left( \frac{y}{x}\right)~dxdy$$ I am not looking for any full solutions just some small hints to get me started would be great.
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1Have you tried inverting the order of the integrals? – Harald Hanche-Olsen Jul 28 '15 at 09:00
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No I am unfamiliar with how to do that. – Mike H Jul 28 '15 at 09:02
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1should it be $dydx$ rather than $dxdy$? – Math-fun Jul 28 '15 at 09:05
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No it is stated as $dxdy$ that is why I am confused because then to change it is saying $x=x$ and $1=x$ – Mike H Jul 28 '15 at 09:07
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What kind of calculus course (or text) will even pose such a problem without teaching about different orders of integration first? That surprises me. – Harald Hanche-Olsen Jul 28 '15 at 09:07
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Hint: Try this https://en.wikipedia.org/wiki/Integral_of_inverse_functions – entrelac Jul 28 '15 at 09:08
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No course Harald I am learning myself. – Mike H Jul 28 '15 at 09:11
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Just realised this is a duplicate sorry. Please close the question thank you. – Mike H Jul 28 '15 at 09:12
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could you please provide the like to the duplicate stuff? Thanks! – Math-fun Jul 28 '15 at 09:40
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Suppose he is referring to this http://math.stackexchange.com/questions/1256465/find-int-limits-01-int-limits-x1-arctan-bigg-frac-yx-bigg-d?rq=1 from the Related section – Christoph Jul 28 '15 at 09:43
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HINT EDITED
There are two possible interpretations of the double integral in question over the region below
$$ \int_0^1 \left[\int_x^1 \arctan \left( \frac{y}{x}\right)~dy\ \right] dx.$$
or
$$\int_0^1 \left[\color{red}{\int_0^y} \arctan\left(\frac{y}{x}\right)\color{red}{dx}\right] dy$$
(That is, first you have to integrate with respect to $y$ since you cannot integrate with respect to $x$ if the limits depend on $x$. Or, if you want to integrate first by $x$ then the limits are $0$ and $y$.)
I've suggested the first interpretation because I knew the antiderivative of $\arctan.$
Then use the well known facts that
- $$\int \arctan (u)\ du =u\arctan(u)-\frac12\ln(u^2+1)+C$$
- If $g$ is an antiderivative of $f$ then $\frac1Ag$ is an antiderivative of $f(Au)$.
zoli
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@Math-fun: The OP unconsciously mean $dS$ denoted by $dxdy$. However, I cannot be certain about that. – zoli Jul 28 '15 at 09:47
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@Math-fun: Yes, I see your question and then his answer. Yet, I claim that $"dxdy"$ is meaningless. This should be explained somewhat better... – zoli Jul 28 '15 at 10:18
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@Math-fun: I edited my answer. I think I've found out what went wrong in the OP. It is possible to integrate first by $x$ but then the limits of integration are different. – zoli Jul 28 '15 at 11:43
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It could well be that OP wanted to have the inner integral over $y\to1$ ... – Math-fun Jul 28 '15 at 12:54