Integral Integral9

Calculus Level 3

0 1 ( π csc ( π k ) ) 2 d k = 1 a \large \int_0^{\infty } \frac{1}{\left(\pi \csc \left(\frac{\pi }{k}\right)\right)^2} \, dk=\frac{1}{a}

where a a is a positive integer. Submit a a .


The answer is 2.

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2 solutions

Chew-Seong Cheong
Dec 25, 2017

I = 0 1 π 2 csc 2 π k d k = 1 π 2 0 sin 2 π k d k By integration by part = k sin 2 π k π 2 0 1 π 2 0 2 π sin π k cos π k k d k Let x = 1 k d x = d k k 2 = 0 + 2 π 0 sin x cos x x d x = 2 π 0 sin 2 x 2 x d x Let u = 2 x d u = 2 d x = 1 π 0 sin u u d u = 1 π Si ( ) where Si ( x ) is the sine integral. = 1 π × π 2 = 1 2 \begin{aligned} I & = \int_0^\infty \frac 1{\pi^2\csc^2 \frac \pi k}dk \\ & = \frac 1{\pi^2} \int_0^\infty \sin^2 \frac \pi k \ dk & \small \color{#3D99F6} \text{By integration by part} \\ & = \frac {k\sin^2 \frac \pi k}{\pi^2} \bigg|_0^\infty - \color{#3D99F6} \frac 1{\pi^2} \int_0^\infty \frac {-2\pi \sin \frac \pi k \cos \frac \pi k}k d k & \small \color{#3D99F6} \text{Let }x = \frac 1k \implies dx = - \frac {dk}{k^2} \\ & = 0 + \color{#3D99F6} \frac 2\pi \int_0^\infty \frac {\sin x \cos x}x dx \\ & = \frac 2\pi \int_0^\infty \frac {\sin 2x}{2x} dx & \small \color{#3D99F6} \text{Let }u = 2x \implies du = 2\ dx \\ & = \frac 1\pi \color{#3D99F6} \int_0^\infty \frac {\sin u}{u} du \\ & = \frac 1\pi \color{#3D99F6} \text{Si}(\infty) & \small \color{#3D99F6} \text{where }\text{Si}(x) \text{ is the sine integral.} \\ & = \frac 1\pi \times \color{#3D99F6} \frac \pi 2 \\ & = \frac 12 \end{aligned}

Therefore, a = 2 a=\boxed{2} .

. .
Mar 6, 2021

We know that a 0 , then a must be 2 . \text { We know that } a \ne 0, \text { then } a \text { must be \boxed { 2 } .}

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