Prove ∫ 0 ∞ sin 4 ⁡ x x 4 d x = π 3

Question

Prove       ∫    0    ∞                      sin        4            ⁡      x              x      4        d  x  =      π    3

Zoe Andersen · Accepted Answer

You are likely expected to integrate by parts (twice)

\begin{array}{rcl} \int \frac{\sin^{4} (x)}{x^{4}} d x & = & - \frac{1}{3} \frac{\sin^{4} (x)}{x^{3}} + \frac{4}{3} \int \frac{\cos (x) \sin^{3} (x)}{x^{3}} d x \\ = & - \frac{1}{3} \frac{\sin^{4} (x)}{x^{3}} - \frac{2 \cos (x) \sin^{3} (x)}{3 x^{2}} + \frac{2}{3} \int \frac{3 \cos^{2} (x) \sin^{2} (x) - \sin^{4} (x)}{x^{2}} d x \\ = & - \frac{1}{3} \frac{\sin^{4} (x)}{x^{3}} - \frac{2 \cos (x) \sin^{3} (x)}{3 x^{2}} + \frac{2}{3} \int (\frac{\sin^{2} (2 x)}{x^{2}} - \frac{\sin^{2} (x)}{x^{2}}) d x \end{array}

where the last equality used

\begin{array}{rcl} 3 \cos^{2} (x) \sin^{2} (x) - \sin^{4} (x) & = & 3 \cos^{2} (x) \sin^{2} (x) - \sin^{2} (x) (1 - \cos^{2} (x)) \\ = & {(2 \sin (x) \cos (x))}^{2} - \sin^{2} (x) = \sin^{2} (2 x) - \sin^{2} (x) \end{array}

Now

\begin{array}{rcl} \int_{0}^{\infty} \frac{\sin^{4} (x)}{x^{4}} d x & = & \frac{2}{3} \int_{0}^{\infty} \frac{\sin^{2} (2 x)}{x^{2}} d x - \frac{2}{3} \int_{0}^{\infty} \frac{\sin^{2} (x)}{x^{2}} d x \\ = & \frac{4}{3} \int_{0}^{\infty} \frac{\sin^{2} (y)}{y^{2}} d y - \frac{2}{3} \int_{0}^{\infty} \frac{\sin^{2} (x)}{x^{2}} d x \\ = & \frac{2}{3} \int_{0}^{\infty} \frac{\sin^{2} (x)}{x^{2}} d x = \frac{π}{3} \end{array}

evitagimm9h · Accepted Answer

That is, the FT of   (  sin  ⁡      x        /    x      )    2   is      ∫          −      ∞              ∞        d  x                      sin        2            ⁡              x                    x      2            e          i      k      x        =      {                                                π                      (            1            −                                                            |                                k                                  |                                            2                        )                                                |                    k                      |                    ≤          2                                      0                                      |                    k                      |                    &amp;gt;          2                        Plancherel/Parseval says that      ∫          −      ∞              ∞        d  x                      sin        4            ⁡              x                    x      4        =      1          2      π            ∫          −      2              2        d  k        π    2              (      1      −                                    |                    k                      |                          2            )        2    =      π    2        4    3    =            2      π        3

Prove int_0^infty sin^4x/x^4 dx=pi/3

Answered question

Answer & Explanation

New Questions in Calculus 1