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Given double integral is :

$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$$

My attempt :

We can't solve since variable $x$ can't remove by limits, but if we change order of integration, then

$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx$$

$$\implies\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx = \frac{1}{2}$$

Can you explain in formal way, please?

Edit : This question was from competitive exam GATE. The link is given below on comments by Alex M. and Martin Sleziak(Thanks).

Mithlesh Upadhyay
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    The first expression says that y goes from 0 to 1 (outer integral) which is ok and x goes from 0 to 1/x (inner integral) which makes no sense. – Jimmy R. Dec 17 '15 at 11:53
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    With $\int_0^x{\frac{x}{1+y^2}dx}$, do you mean $\int_0^x{\frac{s}{1+y^2}ds}$? Otherwise your variable of integration also appears in the interval over which you integrate... – Eric S. Dec 17 '15 at 11:53
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    Should the integral read as $\int _0^1\int _0^{\frac{1}{\color{red}{y}}}\frac{x}{1+y^2}dxdy$ or $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}d\color{red}{y}d\color{red}{x}$ – Math-fun Dec 17 '15 at 12:18
  • Your problem contains a mistake, as Eric S. has pointed out: you cannot have $x$ simultaneously as an integration variable and as an integration endpoint, and yet this is what happens in your inner integral. As it is formulated now, your problem is meaningless. – Alex M. Dec 20 '15 at 15:39
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    Your bounds of integration don't make sense. In the inside integral, $x$ goes from $0$ to${},\ldots\ldots,{}$what? ${}\qquad{}$ – Michael Hardy Dec 20 '15 at 16:30
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    As the OP mentioned in chat, the problem was taken from here. (It would probably be better to add this information to the question.) – Martin Sleziak Dec 22 '15 at 08:12
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    Indeed, this is problem 2.6 of the CS section of the 1993 GATE. Remarkably, there is a very similar one (again, problem 2.6) in the EC section of the 1993 GATE - this one having a lower integration bound of $x$ instead of $0$ in the inner integral. Both problems are mistaken, most probably it should have been $\Bbb d y \Bbb d x$ instead of $\Bbb d x \Bbb d y$, in which case it seems that the OP knows what to do. – Alex M. Dec 22 '15 at 09:37

3 Answers3

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$$\begin{align}\int_{0}^{1}\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\text{d}y&= \int_{0}^{1}\left(\int_{0}^{\frac{1}{x}}\frac{x}{1+y^2}\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{1}{1+y^2}\int_{0}^{\frac{1}{x}}x\space\text{d}x\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\left[x^2\right]_{0}^{\frac{1}{x}}}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\left(\frac{1}{x}\right)^2-0^2}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\left(\frac{\frac{1}{x^2}}{2\left(1+y^2\right)}\right)\text{d}y\\&=\int_{0}^{1}\frac{1}{2x^2(1+y^2)}\text{d}y\\&=\frac{1}{2x^2}\int_{0}^{1}\frac{1}{1+y^2}\text{d}y\\&=\frac{\left[\arctan(y)\right]_{0}^{1}}{2x^2}\\&=\frac{\arctan(1)-\arctan(0)}{2x^2}\\&=\frac{\frac{\pi}{4}-0}{2x^2}\\&=\frac{\frac{\pi}{4}}{2x^2}\\&=\frac{\pi}{8x^2}\end{align}$$

ParaH2
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Jan Eerland
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    Yes, we can't remove variable $x$ from given double integration. – Mithlesh Upadhyay Dec 17 '15 at 12:30
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    @MithleshUpadhyay That's what I've showed you :) – Jan Eerland Dec 17 '15 at 12:31
  • I think Riemannian way: when we change $x$ the integral limits must be fixed :-) otherwise I can't use general integration rules, isn't it like that? – Math-fun Dec 17 '15 at 12:34
  • @Math-fun Yes it is! – Jan Eerland Dec 17 '15 at 12:35
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    then it seems like your Answer is mileading the OP. – Math-fun Dec 17 '15 at 13:40
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    Technically I think the answer is correct (at least up to the first expression without $dx$; I didn't check further). In a definite integral, $dx$ binds all occurrences of $x$ inside the integral, but not in the "start" and "stop" points of the integration, so $\int_0^{1/x} \frac{x}{1+y^2},dx = \int_0^{1/x} \frac{u}{1+y^2},du$. – David K Dec 18 '15 at 13:51
  • As it is now, the question is mistaken and consequently so is this answer: you cannot have $x$ simultaneously as an integration variable and as an integration endpoint, yet this is what happens in the inner integral. – Alex M. Dec 20 '15 at 15:41
  • @AlexM.: as the question is written, this answer is correct. The $\frac1x$ in the limit of integration is not bound by the $x$ in the inner integration and is therefore free. That is, we can change the name of the inner variable of integration to get $$\int_0^1\int_0^{1/x}\frac{z}{1+y^2},\mathrm{d}z,\mathrm{d}y$$ The question is written confusingly, but it is not "mistaken". However, the intended question may have been different. – robjohn Dec 22 '15 at 11:27
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    @robjohn: See my latest comment under the original question (it is currently the last one there): the question comes from an exam, where it was given in this form presumably because of a typographical error. – Alex M. Dec 22 '15 at 11:42
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Right way is: $$\int\limits_0^1x\:dx\int\limits_0^\dfrac1x \dfrac{dy}{1+y^2} = \int\limits_0^1\arctan y\:\Biggl.\Biggr|_0^{\dfrac1x} x\:dx =\int\limits_0^1x\arctan \dfrac1x\:dx = \int\limits_0^1\left(\dfrac{\pi}2-\arctan x\right)x\:dx =$$$$ \left.\dfrac{\pi}2\dfrac{x^2}2\right|_0^1 -\int\limits_0^1\arctan x\: d\dfrac{x^2}2 = \dfrac{\pi}4-\left.\dfrac{x^2}2\arctan x\right|_0^1 +{\dfrac12\int\limits_0^1\dfrac{x^2}{1+x^2}\:dx} =$$$$ \dfrac{\pi}4-\dfrac{\pi}8 + \dfrac12\int\limits_0^1\left(1 - \dfrac1{1+x^2}\right)dx = \dfrac{\pi}8+\left.\dfrac12(x-\arctan x)\right|_0^1=\dfrac12$$

To change order of integral, you build the region of integration in the chart, where you can see that the region of integration is made up of a square and curved trapezoid, so: $$\int\limits_0^1x\:dx\int\limits_0^\dfrac1x \dfrac1{1+y^2}\:dy = \int\limits_0^1 \dfrac1{1+y^2}\:dy\int\limits_0^1 x\:dx + \int\limits_1^\infty \dfrac1{1+y^2}\:dy\int\limits_0^\dfrac1y x\:dx = $$$$ \arctan y\:\Biggl.\Biggr|_0^1\cdot\left.\dfrac{x^2}2\right|_0^1 + \int\limits_1^\infty \left.\dfrac{x^2}2\right|_0^\dfrac1y\dfrac1{1+y^2}dy = \dfrac{\pi}8 + \dfrac12\int\limits_1^\infty \dfrac1{1+y^2}\dfrac1{y^2}\:dy = $$$$\dfrac{\pi}8+\dfrac12\int\limits_1^\infty \left(\dfrac1{y^2}-\dfrac1{1+y^2}\right)\:dy = \dfrac{\pi}8+\dfrac12\left(-\dfrac1y-\arctan y\right)\Biggr.\Biggr|_1^\infty = \dfrac12$$

And this way of calculation does not seem more simple.

Yuri Negometyanov
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  • How do you change order of integral? – Mithlesh Upadhyay Dec 18 '15 at 06:39
  • I don't change it at all – Yuri Negometyanov Dec 18 '15 at 12:36
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    I think the usual interpretation of $\iint f,dx,dy$ is that the inner integral is $\int f,dx$. In this answer it seems to me the inner integral is taken to be $\int f,dy$, a change in the order of integration. – David K Dec 18 '15 at 13:45
  • Your formula for "changing the order of integration" is incorrect. You are leaving out a square $(x,y) \in[0,1]\times [0,1]$ which contributes a factor $\pi/8$. – lcv Dec 18 '15 at 14:00
  • That were intermediate calculations, sorry&thanks – Yuri Negometyanov Dec 18 '15 at 14:41
  • @lcv It is easy to see, because the answer is known – Yuri Negometyanov Dec 18 '15 at 14:45
  • You are welcome – lcv Dec 18 '15 at 15:38
  • As it is now, the question is mistaken and consequently so is this answer: you cannot have $x$ simultaneously as an integration variable and as an integration endpoint, yet this is what happens in the OP's inner integral. You have silently changed $\Bbb d x \Bbb d y$ into $\Bbb d y \Bbb d x$ right from the beginning, and this is why you don't encounter this problem, but in doing so you are not answering the original question. – Alex M. Dec 20 '15 at 15:44
  • @Alex M. Sorry, but I am not obliged to answer questions clearly erroneous. – Yuri Negometyanov Dec 20 '15 at 15:55
  • @Alex M. There are two independent answers. The first doesn't use the change order of integral at all, the second accompanied by explanations... I am waiting for an apology – Yuri Negometyanov Dec 20 '15 at 16:22
  • The present situation is pretty funny: the OP asks a question that cannot be answered (because it is wrong), while you answer a question that nobody asked. Technically, this should be flagged as "not an answer" (even though from a mathematical point of view it is correct). What happened is that the OP asked a wrong question, you corrected it and then answered this new question. That's fine - but it not the OP's question. I suggest to wait for the OP to come up and clarify his question. In the meantime, note that the comment of David K. and the one of Eric S. have spotted the same mistake. – Alex M. Dec 20 '15 at 16:56
  • @Alex M I see that OP has a wrong attempt and asks to explain. So I try to show another ways. Is it funny? – Yuri Negometyanov Dec 20 '15 at 17:14
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$$\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \frac { 1 }{ x } }{ \frac { x }{ 1+{ y }^{ 2 } } dxdy } } \\ =\int _{ 0 }^{ \frac { 1 }{ x } }{ \int _{ 0 }^{ 1 }{ \frac { x }{ 1+{ y }^{ 2 } } dydx } } \\ =\int _{ 0 }^{ \frac { 1 }{ x } }{ x\left( arctan1 \right) dx } \\ =\frac { \pi }{ { 8x }^{ 2 } } $$

Now Mithlesh your method of approach is wrong as the limits of a multivariable integral are associated with the appropriate variable itself. You can not change the limits pertaining to a variable.

Aditya Kumar
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