No problem, Double Integral!

Calculus Level 5

If ∫ 0 ∞ sin ⁡ 3 x x 3 d x = A \large \displaystyle \int_{0}^{\infty} \dfrac{\sin^3 x}{x^3} dx \ = \ A and ∫ 0 ∞ ( x − sin ⁡ x x 3 ) d x = m A n \large \displaystyle \int_{0}^{\infty} \left (\dfrac{x - \sin x}{x^3} \right ) dx \ = \ \dfrac{m A}{n} where m m and n n are positive relative coprimes, then what is m + n m+n ?.


The answer is 5.

Anish Harsha by Anish Harsha , 20, India

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3 solutions

Chew-Seong Cheong
Apr 23, 2016

A = ∫ 0 ∞ sin ⁡ 3 x x 3 d x As sin ⁡ 3 x = 3 sin ⁡ x − 4 sin ⁡ 3 x = 1 4 ∫ 0 ∞ 3 sin ⁡ x − sin ⁡ 3 x x 3 d x = − 1 8 [ 3 sin ⁡ x − sin ⁡ 3 x x 2 ] 0 ∞ + 3 8 ∫ 0 ∞ cos ⁡ x − cos ⁡ 3 x x 2 d x = − 3 8 [ cos ⁡ x − cos ⁡ 3 x x ] 0 ∞ − 3 8 ∫ 0 ∞ sin ⁡ x − 3 sin ⁡ 3 x x d x = − 3 8 ∫ 0 ∞ sin ⁡ x x d x + 9 8 ∫ 0 ∞ sin ⁡ 3 x 3 x d ( 3 x ) = 3 4 ∫ 0 ∞ sin ⁡ x x d x Dirichlet integral = 3 4 ⋅ π 2 = 3 π 8 \begin{aligned} A & = \int_0^\infty \frac{\color{#3D99F6}{\sin^3 x}}{x^3}\, dx \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{As }\sin 3x = 3 \sin x - 4 \sin^3 x} \\ & = \frac{1}{4} \int_0^\infty \frac{\color{#3D99F6}{3 \sin x - \sin 3x}}{x^3} \, dx \\ & = -\frac{1}{8} \left[\frac{3\sin x - \sin 3x}{x^2}\right]_0^\infty + \frac{3}{8} \int_0^\infty \frac{\cos x - \cos 3x}{x^2} dx \\ & = -\frac{3}{8} \left[\frac{\cos x - \cos 3x}{x}\right]_0^\infty - \frac{3}{8} \int_0^\infty \frac{\sin x - 3 \sin 3x}{x} dx \\ & = - \frac{3}{8} \int_0^\infty \frac{\sin x}{x} dx + \frac{9}{8} \int_0^\infty \frac{\sin \color{#3D99F6}{3x}}{\color{#3D99F6}{3x}} d(\color{#3D99F6}{3x}) \\ & = \frac{3}{4} \color{#3D99F6}{\int_0^\infty \frac{\sin x}{x} dx} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Dirichlet integral}} \\ & = \frac{3}{4} \cdot{} \color{#3D99F6}{\frac{\pi}{2}} = \frac{3 \pi}{8} \end{aligned}

m A n = ∫ 0 ∞ x − sin ⁡ x x 3 d x Integration by parts = [ − 2 x + sin ⁡ x 2 x 2 + cos ⁡ x 2 x ] 0 ∞ + 1 2 ∫ 0 ∞ sin ⁡ x x d x Dirichlet integral = 1 2 ⋅ π 2 = π 4 \begin{aligned} \frac{mA}{n} & = \int_0^\infty \frac{x-\sin x}{x^3} \, dx \quad \quad \small \color{#3D99F6}{\text{Integration by parts}} \\ & = \left[- \frac{2}{x} + \frac{\sin x}{2x^2} + \frac{\cos x}{2x} \right]_0^\infty + \frac{1}{2} \color{#3D99F6}{\int_0^\infty \frac{\sin x}{x} \, dx} \quad \quad \small \color{#3D99F6}{\text{Dirichlet integral}} \\ & = \frac{1}{2} \cdot{} \color{#3D99F6}{\frac{\pi}{2}} = \frac{\pi}{4} \end{aligned}

m A n = π 4 ⟹ 3 m π 8 n = π 4 ⟹ m n = 2 3 ⟹ m + n = 5 \begin{aligned} \frac{mA}{n} & = \frac{\pi}{4} \\ \implies \frac{3m\pi}{8n} & = \frac{\pi}{4} \\ \implies \frac{m}{n} & = \frac{2}{3} \\ \implies m + n & = \boxed{5} \end{aligned}

10 Helpful 0 Interesting 1 Brilliant 0 Confused

This should be tidied up a little, since the individual integrals such as ∫ 0 ∞ sin ⁡ x x 3 d x \int_0^\infty \frac{\sin x}{x^3}\,dx do not exist, and cannot be evaluated separately. What makes this problem work is the fact that the differences between various pairs of integrals do behave, so that, for example 3 sin ⁡ x − sin ⁡ 3 x = O ( x 3 ) x → 0 , 3\sin x - \sin3x \; = \; O(x^3) \qquad \qquad x \to 0 \;, which means that what you have written as a separate pair of terms − 3 8 [ sin ⁡ x x 2 ] 0 ∞ + 1 8 [ sin ⁡ 3 x x 2 ] 0 ∞ -\tfrac38\Big[\frac{\sin x}{x^2}\Big]_0^\infty + \tfrac18\Big[\frac{\sin 3x}{x^2}\Big]_0^\infty (neither of which exist) cancel out when considered as a single expression [ 1 8 sin ⁡ 3 x − 3 sin ⁡ x x 2 ] 0 ∞ \Big[\tfrac18\frac{\sin 3x - 3\sin x}{x^2}\Big]_0^\infty

If you wrote everything out as an integral from a a to X X , then let a → 0 a \to 0 and X → ∞ X \to \infty , then you would be OK. Otherwise you need to integrate (for example) 3 sin ⁡ x − sin ⁡ 3 x x 3 \frac{3\sin x - \sin3x}{x^3} by parts as a single integral, in which case all is well.

Mark Hennings - 5 years, 1 month ago

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does the limit x → 0 x \to 0 exist for cos ⁡ x x \frac{\cos x}{x}

Tanishq Varshney - 5 years, 1 month ago

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No. It behaves like 1 / x 1/x near 0 0 . The limit of ( 1 − cos ⁡ x ) / x (1-\cos x)/x exists and equals 0 0 , though.

Mark Hennings - 5 years, 1 month ago

I was taking:

[ sin ⁡ x x 2 ] 0 ∞ = lim ⁡ x → ∞ sin ⁡ x x 2 − lim ⁡ x → 0 sin ⁡ x x 2 By l’H o ˆ pital’s rule = 0 + 0 \begin{aligned} \left[ \dfrac{\sin x}{x^2} \right]_0^\infty & = \lim_{x \to \infty} \dfrac{\sin x}{x^2} -\color{#3D99F6}{ \lim_{x \to 0} \dfrac{\sin x}{x^2}} \quad \quad \small \color{#3D99F6}{\text{By l'Hôpital's rule}} \\ & = 0 + \color{#3D99F6}{0} \end{aligned}

Chew-Seong Cheong - 5 years, 1 month ago

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The limit of sin ⁡ x / x 2 \sin x/x^2 does not exist as x x tends to 0 0 . It behaves like 1 / x 1/x near 0 0 .

Mark Hennings - 5 years, 1 month ago

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@Mark Hennings – Thanks, I got it. lim ⁡ x → 0 sin ⁡ x x ⋅ x = 1 x \lim_{x \to 0} \dfrac{\color{#3D99F6}{\sin x}}{\color{#3D99F6}{x}\cdot{}x} = \dfrac {\color{#3D99F6}{1}}{x} .

Chew-Seong Cheong - 5 years, 1 month ago

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@Chew-Seong Cheong – Almost. The limit is not 1 / x 1/x ; the function behaves like 1 / x 1/x (think what dividing the Maclaurin series for sin ⁡ x \sin x by x 2 x^2 will look like), so there is no limit.

Mark Hennings - 5 years, 1 month ago

In the 11th step where did the the cosx/x go. Its evaluates to be infinity at x= 0

parv mor - 5 years, 1 month ago

The first integral is now fine. You need to do the same sort of thing with the second one as well, integrating x − sin ⁡ x x 3 \tfrac{x - \sin x}{x^3} by parts...

Mark Hennings - 5 years, 1 month ago

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Thanks again.

Chew-Seong Cheong - 5 years, 1 month ago

First, let's consider

sin ⁡ x = 3 sin ⁡ x 3 − 4 sin ⁡ x 3 3 \displaystyle \sin { x } =3\sin { \frac { x }{ 3 } } -4{ \sin { \frac { x }{ 3 } } }^{ 3 }

Let u = x / 3 \displaystyle u=x/3 hence,

∫ 0 ∞ x − sin ⁡ x x 3 d x = ∫ 0 ∞ 3 u − 3 sin ⁡ u + 4 sin ⁡ u 3 u 3 × 27 × 3 d u \displaystyle \int _{ 0 }^{ \infty }{ \frac { x-\sin { x } }{ { x }^{ 3 } } dx } =\int _{ 0 }^{ \infty }{ \frac { 3u-3\sin { u } +4{ \sin { u } }^{ 3 } }{ { u }^{ 3 }\times 27 } \times } 3du

∫ 0 ∞ 3 u − 3 sin ⁡ u + 4 sin ⁡ u 3 u 3 × 27 × 3 d u = ∫ 0 ∞ u − sin ⁡ u 3 u 3 d u + 4 9 ∫ 0 ∞ sin ⁡ u 3 u 3 d u \displaystyle \int _{ 0 }^{ \infty }{ \frac { 3u-3\sin { u } +4{ \sin { u } }^{ 3 } }{ { u }^{ 3 }\times 27 } \times } 3du=\int _{ 0 }^{ \infty }{ \frac { u-\sin { u } }{ 3{ u }^{ 3 } } } du+\frac { 4 }{ 9 }\int _{ 0 }^{ \infty }{ \frac { { \sin { u } }^{ 3 } }{ { u }^{ 3 } } du }

∫ 0 ∞ x − sin ⁡ x x 3 d x = I = I 3 + 4 9 A \displaystyle \int _{ 0 }^{ \infty }{ \frac { x-\sin { x } }{ { x }^{ 3 } } dx } =I=\frac { I }{ 3 } +\frac { 4 }{ 9 } A

2 I 3 = 4 9 A \displaystyle \frac { 2I }{ 3 } =\frac { 4 }{ 9 } A

I = 2 3 A = m n A \displaystyle I=\frac { 2 }{ 3 } A=\frac { m }{ n } A

m + n = 5 \displaystyle m+n=\boxed{5}

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Parv Mor
Apr 25, 2016

Can anyone tell whats wrong in this.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Well, you don't know whether ∫ s i n ( x ) x 3 \int \frac{sin(x)}{x^{3}} exist or not and in fact it doesn't so you cannot equate this one to the other one

คลุง แจ็ค - 5 years, 1 month ago

You can separate integrals only if both are defined

roja rani - 3 years, 5 months ago

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