# The value of double integral $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$?

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).

• 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
• 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
• 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 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
• 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 9:37

\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}

• Yes, we can't remove variable $x$ from given double integration. – Mithlesh Upadhyay Dec 17 '15 at 12:30
• @MithleshUpadhyay That's what I've showed you :) – Jan Dec 17 '15 at 12:31
• then it seems like your Answer is mileading the OP. – Math-fun Dec 17 '15 at 13:40
• 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
• @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

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.

• How do you change order of integral? – Mithlesh Upadhyay Dec 18 '15 at 6:39
• I don't change it at all – Yuri Negometyanov Dec 18 '15 at 12:36
• 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

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

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.

• 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. – Alex M. Dec 22 '15 at 8:53
• @AlexM. sorry for the typo – Aditya Kumar Dec 22 '15 at 8:58
• @AdityaKumar, you have changed order of integral :). – Mithlesh Upadhyay Dec 22 '15 at 9:36
• @MithleshUpadhyay it is allowed to change the order of integral in multi-variable integrals. – Aditya Kumar Dec 22 '15 at 9:38