5
$\begingroup$

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).

$\endgroup$
  • 13
    $\begingroup$ 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. $\endgroup$ – Jimmy R. Dec 17 '15 at 11:53
  • 3
    $\begingroup$ 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... $\endgroup$ – Eric S. Dec 17 '15 at 11:53
  • 1
    $\begingroup$ Should the integral read as $\int _0^1\int _0^{\frac{1}{\color{red}{y}}}\frac{x}{1+y^2}dxdy$ or $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}d\color{red}{y}d\color{red}{x}$ $\endgroup$ – Math-fun Dec 17 '15 at 12:18
  • 1
    $\begingroup$ Your bounds of integration don't make sense. In the inside integral, $x$ goes from $0$ to${}\,\ldots\ldots\,{}$what? ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 20 '15 at 16:30
  • 3
    $\begingroup$ 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. $\endgroup$ – Alex M. Dec 22 '15 at 9:37
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Yes, we can't remove variable $x$ from given double integration. $\endgroup$ – Mithlesh Upadhyay Dec 17 '15 at 12:30
  • 1
    $\begingroup$ @MithleshUpadhyay That's what I've showed you :) $\endgroup$ – Jan Dec 17 '15 at 12:31
  • 2
    $\begingroup$ then it seems like your Answer is mileading the OP. $\endgroup$ – Math-fun Dec 17 '15 at 13:40
  • 2
    $\begingroup$ 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 $\int_0^{1/x} \frac{x}{1+y^2}\,dx = \int_0^{1/x} \frac{u}{1+y^2}\,du$. $\endgroup$ – David K Dec 18 '15 at 13:51
  • 2
    $\begingroup$ @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. $\endgroup$ – Alex M. Dec 22 '15 at 11:42
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How do you change order of integral? $\endgroup$ – Mithlesh Upadhyay Dec 18 '15 at 6:39
  • $\begingroup$ I don't change it at all $\endgroup$ – Yuri Negometyanov Dec 18 '15 at 12:36
  • 1
    $\begingroup$ 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. $\endgroup$ – David K Dec 18 '15 at 13:45
  • $\begingroup$ 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 $\pi/8$. $\endgroup$ – lcv Dec 18 '15 at 14:00
  • $\begingroup$ That were intermediate calculations, sorry&thanks $\endgroup$ – Yuri Negometyanov Dec 18 '15 at 14:41
0
$\begingroup$

$$\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ \frac { 1 }{ x } }{ \frac { x }{ 1+{ y }^{ 2 } } dxdy } } \\ =\int _{ 0 }^{ \frac { 1 }{ x } }{ \int _{ 0 }^{ 1 }{ \frac { x }{ 1+{ y }^{ 2 } } dydx } } \\ =\int _{ 0 }^{ \frac { 1 }{ x } }{ x\left( arctan1 \right) dx } \\ =\frac { \pi }{ { 8x }^{ 2 } } $$

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.

$\endgroup$
  • $\begingroup$ First, you have an upper bound of $\frac 1 2$ that the OP doesn't have. Second, instead of obtaining a number, you obtain a result that depends on $x$. Your answer simply doesn't make sense. $\endgroup$ – Alex M. Dec 22 '15 at 8:53
  • $\begingroup$ @AlexM. sorry for the typo $\endgroup$ – Aditya Kumar Dec 22 '15 at 8:58
  • $\begingroup$ @AdityaKumar, you have changed order of integral :). $\endgroup$ – Mithlesh Upadhyay Dec 22 '15 at 9:36
  • $\begingroup$ @MithleshUpadhyay it is allowed to change the order of integral in multi-variable integrals. $\endgroup$ – Aditya Kumar Dec 22 '15 at 9:38
  • 1
    $\begingroup$ @AdityaKumar, math.stackexchange.com/questions/467910/… $\endgroup$ – Mithlesh Upadhyay Dec 22 '15 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.