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Calculus Level 5

Given that I = ∫ 0 ∞ ( 1 − cos ⁡ ( 42 3 x ) x 2 ) 2 d x = A π . I=\displaystyle\int \limits^{\infty }_{0}\left( \dfrac{1-\cos ( \sqrt[3]{42} x) }{x^{2}} \right) ^{2}\, dx= A\pi. Find A A .

Bonus Find the closed form of ∫ 0 ∞ ( 1 − cos ⁡ ( α x ) x 2 ) 2 d x \displaystyle\int \limits^{\infty }_{0}\left( \dfrac{1-\cos ( \alpha x) }{x^{2}} \right) ^{2}\, dx


The answer is 7.

Refaat M. Sayed by Refaat M. Sayed , 24, Egypt

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3 solutions

Mark Hennings
Apr 30, 2016

The function f ( x ) = { π 2 ( α − ∣ x ∣ ) ∣ x ∣ < α 0 ∣ x ∣ > α f(x) \; = \; \left\{ \begin{array}{lll} \sqrt{\tfrac{\pi}{2}}\big(\alpha - |x|\big) & \qquad & |x| < \alpha \\ 0 & & |x| > \alpha \end{array} \right. has Fourier transform ( F f ) ( y ) = 1 2 π ∫ − ∞ ∞ f ( x ) e − i x y d x = 1 − cos ⁡ α y y 2 . (\mathcal{F}f)(y) \; = \; \tfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x) e^{-ixy}\,dx \; = \; \frac{1 - \cos \alpha y}{y^2} \;. Hence ∫ 0 ∞ ( 1 − cos ⁡ α y y 2 ) 2 d y = 1 2 ∫ − ∞ ∞ ∣ ( F f ) ( y ) ∣ 2 d y = 1 2 ∫ − ∞ ∞ ∣ f ( x ) ∣ 2 d x = 1 4 π ∫ − α α ( α − ∣ x ∣ ) 2 d x = 1 2 π [ − 1 3 ( α − x ) 3 ] 0 α = 1 6 π α 3 \begin{array}{rcl} \displaystyle \int_0^\infty \left(\frac{1-\cos\alpha y}{y^2}\right)^2\,dy & = & \displaystyle \tfrac12 \int_{-\infty}^\infty \big|(\mathcal{F}f)(y)\big|^2\,dy \; = \; \tfrac12\int_{-\infty}^\infty \big|f(x)\big|^2\,dx \\ & = & \displaystyle \tfrac14\pi\int_{-\alpha}^\alpha (\alpha-|x|)^2\,dx \; = \; \tfrac12\pi \Big[ -\tfrac13(\alpha-x)^3\Big]_0^\alpha \\ & = & \displaystyle \tfrac16\pi \alpha^3 \end{array} and so A = 1 6 α 3 A = \tfrac16\alpha^3 . For this particular problem, we deduce that A = 1 6 × 42 = 7 A = \tfrac16\times42 = \boxed{7} .

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Nice Solution!!

I think another (much more computational) way to do this is to set the bounds of the integral from 0 to N (for N large),and then differentiate under the integral with respect to alpha four times. Then compute this integral w.r.t. x. Then integrate four times to get the integral in question. (You will have to use that the sinc function integrated twice is cosine plus the integral of the sinc function.) Finally send N to infinity to get the answer (using that the integral of sinc from 0 to Infinity is Pi/2).

Christopher Criscitiello - 3 years, 9 months ago

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This would work, but I would not want to perform all those calculations, since there would be a shed-load of terms coming from the fact that the integral is from 0 0 to N N , not ∞ \infty . In addition, you would have to pay careful attention with the potential for singularities at y = 0 y=0 .

Mark Hennings - 3 years, 9 months ago

I think we can also solve this with contour integration. Here is my solution:

Sources: Stein et al. Complex Analysis

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Sahil Silare
Mar 15, 2017

Hint:The answer is written in his profile.

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