## Problema do mês :: Problem of the month #4. (Integral impróprio :: Improper integral). Resolução :: Solution

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Problema

Prove ou infirme: $\pi =\displaystyle\int_{0}^{\infty }\dfrac{\cos x-\cos 3x}{x^{2}}dx$

Solução de Prof. Paulo Sérgio

Para calcular esta integral, note que

$\dfrac{1}{s^{2}}=\displaystyle\int_{0}^{\infty }xe^{-sx}dx$

Assim,

$I:=\displaystyle\int_{0}^{\infty }\dfrac{\cos \left( s\right) -\cos \left( 3s\right) }{s^{2}}ds$

$=\displaystyle\int_{0}^{\infty }\int_{0}^{\infty }xe^{-sx}\left[ \cos \left( s\right) -\cos \left( 3s\right) \right] dxds$

Invertendo a ordem de integração e usando a definição de transformada de Laplace, temos:

$I=\displaystyle\int_{0}^{\infty }x\int_{0}^{\infty }\left[ \mathcal{L}\left\{ \cos\left( s\right) \right\} -\mathcal{L}\left\{ \cos \left( 3s\right) \right\}\right] dx$

$=\displaystyle\int_{0}^{\infty }\left( \dfrac{x^{2}}{1+x^{2}}-\dfrac{x^{2}}{9+x^{2}}\right) dx$

ou seja,

$I=\displaystyle\int_{0}^{\infty }\dfrac{9}{9+x^{2}}dx-\displaystyle\int_{0}^{\infty }\dfrac{dx}{1+x^{2}}=\left[ 3\arctan \left( \dfrac{x}{3}\right) -\arctan \left( x\right) \right] _{0}^{\infty }$

$=\dfrac{3\pi }{2}-\dfrac{\pi }{2}=\dfrac{2\pi }{2}=\pi$.

Outra resolução: fatima

* * *

Problem

Prove or disprove: $\pi =\displaystyle\int_{0}^{\infty }\dfrac{\cos x-\cos 3x}{x^{2}}dx$.

Solution by Prof. Paulo Sérgio

To evaluate this integral note that

$\dfrac{1}{s^{2}}=\displaystyle\int_{0}^{\infty }xe^{-sx}dx$

Hence,

$I:=\displaystyle\int_{0}^{\infty }\dfrac{\cos \left( s\right) -\cos \left( 3s\right) }{s^{2}}ds$

$=\displaystyle\int_{0}^{\infty }\int_{0}^{\infty }xe^{-sx}\left[ \cos \left( s\right) -\cos \left( 3s\right) \right] dxds$

By reversing the order of integration and using the definition of the Laplace transform we get:

$I=\displaystyle\int_{0}^{\infty }x\int_{0}^{\infty }\left[ \mathcal{L}\left\{ \cos\left( s\right) \right\} -\mathcal{L}\left\{ \cos \left( 3s\right) \right\}\right] dx$

$=\displaystyle\int_{0}^{\infty }\left( \dfrac{x^{2}}{1+x^{2}}-\dfrac{x^{2}}{9+x^{2}}\right) dx$

therefore,

$I=\displaystyle\int_{0}^{\infty }\dfrac{9}{9+x^{2}}dx-\displaystyle\int_{0}^{\infty }\dfrac{dx}{1+x^{2}}=\left[ 3\arctan \left( \dfrac{x}{3}\right) -\arctan \left( x\right) \right] _{0}^{\infty }$

$=\dfrac{3\pi }{2}-\dfrac{\pi }{2}=\dfrac{2\pi }{2}=\pi$.

Other Solver: fatima

## Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Calculus, Cálculo, Integrais, Integrais impróprios, Matemática, Math, Problem Of The Month, Problema do mês, Transformadas de Laplace com as etiquetas , , , , , . ligação permanente.

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