x = x 2 x = x^2 ?

Calculus Level 3

True or False?

sin x x d x = sin 2 x x 2 d x \int_{-\infty}^\infty \frac{ \sin x } { x} \, dx= \int_{-\infty}^\infty \frac{ \sin^2 x } { x^2} \, dx

True False

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1 solution

Leonel Castillo
Jan 8, 2018

First, let's do integration by parts for the integral on the right. Let u = sin 2 x , d u = 2 sin x cos x = sin 2 x , d v = x 2 , v = x 1 u = \sin ^2 x, du = 2 \sin x \cos x = \sin 2x, dv = x^{-2}, v = - x^{-1} . Then the integral to the right is equal to sin 2 x x + sin 2 x x d x - \frac{\sin^2 x}{x} \big|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \frac{ \sin 2x}{x} dx . Because sin 2 ( x ) = o ( x ) \sin^2(x) = o(x) the term outside the integral vanishes so we are left with just sin 2 x x d x \int_{-\infty}^{\infty} \frac{ \sin 2x}{x} dx . Now, let u = 2 x , d u = 2 d x u = 2x, du = 2dx . Then the integral is equal to sin u u 2 d u 2 = 2 2 sin u u d u = sin u u d u \int_{-\infty}^{\infty} \frac{ \sin u}{\frac{u}{2}} \frac{du}{2} = \frac{2}{2} \int_{-\infty}^{\infty} \frac{ \sin u}{u} du = \int_{-\infty}^{\infty} \frac{ \sin u}{u} du .

Sin(x) /x isn't continuous on R

Pranav Iyer - 1 year, 3 months ago

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