Integrate the Functional

Calculus Level 1

A function f ( x ) f(x) satisfies the functional equation f ( tan ⁡ θ ) = sin ⁡ 2 2 θ 4 f(\tan \theta)=\frac{\sin^2 2\theta}{4} for all real θ \theta . If ∫ 0 ∞ f ( x ) d x \displaystyle \int_{0}^{\infty}f(x) \ dx is equal to π a b \dfrac{\pi^a}{b} for positive integers a a and b b , then what is the value of a + b a+b ?


The answer is 5.

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Sanjeet Raria by Sanjeet Raria , 27, India

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4 solutions

Jing Yuan Lee
Nov 18, 2016

An alternative solution which does not require you to find the function f ( x ) f(x) :

16 Helpful 0 Interesting 3 Brilliant 0 Confused

We can write f f as 4 sin ⁡ 2 ( θ ) cos ⁡ 2 ( θ ) 4 = tan ⁡ 2 ( θ ) sec ⁡ 4 ( θ ) = tan ⁡ 2 ( θ ) ( tan ⁡ 2 ( θ ) + 1 ) 2 , \dfrac{4\sin^{2}(\theta)\cos^{2}(\theta)}{4} = \dfrac{\tan^{2}(\theta)}{\sec^{4}(\theta)} = \dfrac{\tan^{2}(\theta)}{(\tan^{2}(\theta) + 1)^{2}},

and thus f ( x ) = x 2 ( x 2 + 1 ) 2 . f(x) = \dfrac{x^{2}}{(x^{2} + 1)^{2}}.

Then for the integral, we can use integration by parts, setting

u = x u = x and d v = x ( 1 + x 2 ) 2 . dv = \dfrac{x}{(1 + x^{2})^{2}}. Then d u = d x du = dx and v = − 1 2 ( x 2 + 1 ) v = -\dfrac{1}{2(x^{2} + 1)} , and thus

∫ f ( x ) d x = − x 2 ( x 2 + 1 ) + 1 2 ∫ 1 x 2 + 1 d x = − x 2 ( x 2 + 1 ) + 1 2 ∗ arctan ⁡ ( x ) , \displaystyle\int f(x) dx = -\dfrac{x}{2(x^{2} + 1)} + \dfrac{1}{2}\int \dfrac{1}{x^{2} + 1} dx = -\dfrac{x}{2(x^{2} + 1)} + \dfrac{1}{2}*\arctan(x),

which when evaluated from 0 0 to ∞ \infty comes out to

( 0 + 1 2 ∗ π 2 ) − ( 0 + 0 ) = π 4 . (0 + \dfrac{1}{2}*\dfrac{\pi}{2}) - (0 + 0) = \dfrac{\pi}{4}.

Thus a + b = 1 + 4 = 5 . a + b = 1 + 4 = \boxed{5}.

16 Helpful 0 Interesting 1 Brilliant 1 Confused

Posting Just for sake of variety :) , I did in this way sir : I = ∫ 0 ∞ f ( x ) d x P u t , x = tan ⁡ θ I = ∫ 0 π 2 f ( tan ⁡ θ ) sec ⁡ 2 θ d θ I = ∫ 0 π 2 sin ⁡ 2 2 θ 4 sec ⁡ 2 θ d θ I = ∫ 0 π 2 ( 2 sin ⁡ θ cos ⁡ θ ) 2 4 sec ⁡ 2 θ d θ I = ∫ 0 π 2 sin ⁡ 2 θ d θ I = ∫ 0 π 2 cos ⁡ 2 θ d θ 2 I = ∫ 0 π 2 d θ I = π 4 \displaystyle{{ I=\int _{ 0 }^{ \infty }{ f\left( x \right) dx } \\ Put,\quad x=\tan { \theta } \\ I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ f\left( \tan { \theta } \right) \sec ^{ 2 }{ \theta } d\theta } \\ I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \sin ^{ 2 }{ 2\theta } }{ 4 } \sec ^{ 2 }{ \theta } d\theta } \\ I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { { (2\sin { \theta } \cos { \theta } ) }^{ 2 } }{ 4 } \sec ^{ 2 }{ \theta } d\theta } \\ I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \sin ^{ 2 }{ \theta } } d\theta \\ I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cos ^{ 2 }{ \theta } d\theta } \\ 2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ d\theta } \\ \boxed { I=\cfrac { \pi }{ 4 } } }}

Deepanshu Gupta - 6 years, 3 months ago

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Nice solution. I think that your method is the one that Sanjeet had in mind when he posted the problem. :)

Brian Charlesworth - 6 years, 3 months ago

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When i was conceiving the question, my initial motive before that was to carry out direct integration from the functional equation without having to find f ( x ) f(x) at all !!

But later it seemed that i was terribly failed, as there was a loophole in the rigor. So i had to finally delete my solution and the question rendered me disenchanted as it was no more sophisticated :(

PS But I think we can still work in that way. That's why i changed the name of the problem.

Sanjeet Raria - 6 years, 3 months ago

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@Sanjeet Raria – Yes, I think that Deepanshu has solved the problem in the way that you intended. This is more elegant than my solution, but mine was fairly short as well so time-wise I don't think there was that much difference. Nice problem. :)

Brian Charlesworth - 6 years, 3 months ago

Thank you Deepanshu. You saved my time.

Sanjeet Raria - 6 years, 3 months ago

did the same ! :)

Mayank Singh - 6 years, 3 months ago

did the same !

Hasan Kassim - 6 years, 3 months ago

For even more variety,

I = ∫ 0 ∞ d z 1 + ( α z ) 2 = 1 α a r c t a n ( α z ) ∣ 0 ∞ = π 2 α ∂ I ∂ α = − 2 ∫ 0 ∞ α z 2 d z ( 1 + ( α z ) 2 ) 2 = − π 2 α 2 p u t t i n g a = 1 − 2 ∫ 0 ∞ z 2 d z ( 1 + ( z ) 2 ) 2 = − π 2 ∫ 0 ∞ z 2 d z ( 1 + ( z ) 2 ) 2 = π 4 = ∫ 0 ∞ f ( x ) d x \displaystyle I\quad =\quad \int _{ 0 }^{ \infty }{ \frac { dz }{ 1+(\alpha z)^{ 2 } } } =\frac { 1 }{ \alpha } arctan(\alpha z)\overset { \infty }{ \underset { 0 }{ | } } =\frac { \pi }{ 2\alpha } \\ \\ \frac { \partial I }{ \partial \alpha } =-2\int _{ 0 }^{ \infty }{ \frac { \alpha { z }^{ 2 }dz }{ (1+(\alpha z)^{ 2 })^{ 2 } } } =\frac { -\pi }{ 2{ \alpha }^{ 2 } } \\ \\ putting\quad a=1\\ \\ -2\quad \int _{ 0 }^{ \infty }{ \frac { { z }^{ 2 }dz }{ (1+(z)^{ 2 })^{ 2 } } } =\quad \frac { -\pi }{ 2 } \\ \int _{ 0 }^{ \infty }{ \frac { { z }^{ 2 }dz }{ (1+(z)^{ 2 })^{ 2 } } } =\frac { \pi }{ 4 } =\int _{ 0 }^{ \infty }{ f(x)dx }

Mvs Saketh - 6 years, 3 months ago

this is JEE approach

manish bhargao - 6 years, 3 months ago

hey deepanshu...hi...I am finding maths a bit difficult and don't know whether I will be able to score beyond 65% in it....(i am talking about JEE mains) what suggestions you would give me.......thnkx...

manish bhargao - 6 years, 3 months ago

No need to find the function . Could have been done directly

Prithwish Mukherjee - 2 years, 7 months ago

Relevant wiki: Integration Tricks

Using the identity sin ⁡ 2 θ = 2 tan ⁡ θ 1 + tan ⁡ 2 θ \sin 2\theta = \dfrac{2 \tan \theta}{1+\tan^2 \theta} and let x = tan ⁡ θ x = \tan \theta , then we have:

f ( x ) = 1 4 ( 2 x 1 + x 2 ) 2 = x 2 ( 1 + x 2 ) 2 f(x) = \dfrac{1}{4}\left(\dfrac{2x}{1+x^2}\right)^2 = \dfrac{x^2}{(1+x^2)^2} .

Shorter solution:

∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ x 2 ( 1 + x 2 ) 2 d x Using ∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ f ( 1 x ) x 2 d x = 1 2 ∫ 0 ∞ ( x 2 ( 1 + x 2 ) 2 + 1 x 4 ( 1 + 1 x 2 ) 2 ) d x = 1 2 ∫ 0 ∞ ( x 2 ( 1 + x 2 ) 2 + 1 ( x 2 + 1 ) 2 ) d x = 1 2 ∫ 0 ∞ 1 1 + x 2 d x = tan ⁡ − 1 x 2 ∣ 0 ∞ = π 4 \begin{aligned} \int_0^\infty f(x) \ dx & = \int_0^\infty \frac {x^2}{(1+x^2)^2} dx & \small \color{#3D99F6} \text{Using } \int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left( \frac 1x \right)} {x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac {x^2}{(1+x^2)^2} + \frac 1{x^4\left(1+\frac 1{x^2}\right)^2}\right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {x^2}{(1+x^2)^2} + \frac 1{(x^2 + 1)^2}\right) dx \\ & = \frac 12 \int_0^\infty \frac 1{1+x^2} dx \\ & = \frac {\tan^{-1}x}2 \bigg|_0^\infty \\ & = \frac \pi 4 \end{aligned}

Therefore, a + b = 1 + 4 = 5 a+b=1+4 = \boxed 5 .

Previous solution:

⇒ ∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ x 2 ( 1 + x 2 ) 2 d x Let x = tan ⁡ ϕ ⇒ d x = sec ⁡ 2 ϕ d ϕ = ∫ 0 π 2 tan ⁡ 2 ϕ sec ⁡ 2 ϕ sec ⁡ 4 ϕ d ϕ = ∫ 0 π 2 tan ⁡ 2 ϕ sec ⁡ 2 ϕ d ϕ = ∫ 0 π 2 sin ⁡ 2 ϕ cos ⁡ 2 ϕ × cos ⁡ 2 ϕ d ϕ = ∫ 0 π 2 sin ⁡ 2 ϕ d ϕ = 1 2 ∫ 0 π 2 [ 1 − cos ⁡ ( 2 ϕ ) ] d ϕ = 1 2 [ ϕ + sin ⁡ ( 2 ϕ ) 2 ] 0 π 2 = π 4 \begin{aligned} \Rightarrow \int_0^\infty f(x) \space dx & = \int_0^\infty \frac{x^2}{(1+x^2)^2} dx \quad \quad \small \color{#3D99F6}{\text{Let }x = \tan \phi \quad \Rightarrow dx = \sec^2 \phi \space d\phi} \\ & = \int_0^\frac{\pi}{2} \frac{\tan^2 \phi \sec^2 \phi}{\sec^4 \phi} d\phi \\ & = \int_0^\frac{\pi}{2} \frac{\tan^2 \phi}{\sec^2 \phi} d\phi \\ & = \int_0^\frac{\pi}{2} \frac{\sin^2 \phi}{\cos^2 \phi}\times \cos^2 \phi \space d\phi \\ & = \int_0^\frac{\pi}{2} \sin^2 \phi \space d\phi \\ & = \frac{1}{2} \int_0^\frac{\pi}{2} [1-\cos (2 \phi)] \space d\phi \\ & = \frac{1}{2}\left[\phi + \frac{\sin(2\phi)}{2}\right]_0^\frac{\pi}{2} = \frac{\pi}{4} \end{aligned}

⇒ a + b = 1 + 4 = 5 \Rightarrow a + b = 1 + 4 = \boxed{5}

4 Helpful 1 Interesting 2 Brilliant 0 Confused

you did a mistake.......in eq1 it is tan( θ \theta ) and not tan(tan( θ \theta )) ...i did exactly same as you +1....you saved my time thnx sir... you are the best !

A Former Brilliant Member - 4 years, 5 months ago

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Thanks. I have found a shorter solution. Take a look.

Chew-Seong Cheong - 2 years, 7 months ago

Solution is too long. It could have been short and compact

Prithwish Mukherjee - 2 years, 7 months ago

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Thanks. I have found a shorter solution. See the solution now.

Chew-Seong Cheong - 2 years, 7 months ago
Rohan Shinde
Dec 24, 2018

Using the identity sin ⁡ 2 θ = 2 tan ⁡ θ 1 + tan ⁡ 2 θ \sin 2\theta =\frac {2\tan \theta}{1+\tan^2\theta}

Hence f ( x ) = x 2 ( 1 + x 2 ) 2 f(x)=\frac {x^2}{(1+x^2)^2}

∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ x 2 ( 1 + x 2 ) 2 d x = 1 2 B ( 3 2 , 1 2 ) = 1 2 Γ ( 3 2 ) Γ ( 1 2 ) Γ ( 2 ) = π 4 \int_0^{\infty} f(x) dx=\int_0^{\infty} \frac {x^2}{(1+x^2)^2}dx= \frac 12 B\left(\frac 32,\frac 12\right)=\frac 12 \frac {\Gamma \left(\frac 32\right)\Gamma \left(\frac 12\right)}{\Gamma (2)}=\frac {\pi}{4}

1 Helpful 1 Interesting 1 Brilliant 1 Confused

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