We have an answer to "Show that: ∫0∞sin3(x)x3dx=3π8"

Francisca Rodden

Answered question

2022-01-03

Show that:

\int_{0}^{\infty} \frac{\sin^{3} (x)}{x^{3}} d x = \frac{3 π}{8}

Answer & Explanation

Ana Robertson

Beginner2022-01-04Added 26 answers

Let

f (y) = \int_{0}^{\infty} \frac{\sin^{3} y x}{x^{3}} d x

Then,

f^{'} (y) = 3 \int_{0}^{\infty} \frac{\sin^{2} y x \cos y x}{x^{2}} d x = {\frac{34}{\int}}_{0}^{\infty} \frac{\cos y x - \cos 3 y x}{x^{2}} d x

f^{″} (y) = \frac{3}{4} \int_{0}^{\infty} \frac{- \sin y x + 3 \sin y x}{x} d x

Therefore,

f^{″} (y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin 3 y x}{x} d x - \frac{3}{4} \int_{0}^{\infty} \frac{s u n y x}{x} d x

Now, it is quite easy to prove that

\int_{0}^{\infty} \frac{\sin a x}{x} d x = \frac{π}{2} signum a

Therefore,

f^{″} (y) = \frac{9 π}{8} signum y - \frac{3 π}{8} signum y = \frac{3 π}{4} signum y

Then,

f^{'} (y) = \frac{3 π}{4} | y | + C

Note that,

f^{'} (0) = 0

, therefore, C=0

f (y) = \frac{3 π}{8} y^{2} signum y + D

Again,

f (0) = 0

, therefore,

D = 0

Hence,

f (1) = \int_{0}^{\infty} \frac{\sin^{3} x}{x^{3}} = \frac{3 π}{8}

Janet Young

Beginner2022-01-05Added 32 answers

Now, it is quite easy to prove that

\int_{- \infty}^{\infty} d x f (x) g \cdot (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} d k F (k) G \cdot (K)

where f,g and F,G are respective Fourier transform pairs, e.g.,

F (k) = \int_{- \infty}^{\infty} d x f (x) e^{i k x}

etc. If

f (x) = \frac{\sin x}{x}

, then

F (k) = {\begin{array}{cc} π & | k | \leq 1 \\ 0 & | k | > 2 \end{array}

Further, if

g (x) = \frac{\sin^{2} x}{x^{2}}

, then

G (k) = {\begin{array}{cc} π (1 - \frac{| k |}{2}) & | k | \leq 2 \\ 0 & | k | > 2 \end{array}

Then

\int_{0}^{\infty} d x \frac{\sin^{3} x}{x^{3}} = \frac{1}{2 π} \int_{- 1}^{1} d k π^{2} (1 - \frac{| k |}{2}) = π - \frac{π}{2} \int_{0}^{1} d k k = π - \frac{π}{4}

therefore

\int_{0}^{\infty} d x \frac{\sin^{3} x}{x^{3}} = \frac{3 π}{8}

karton

Expert2022-01-11Added 613 answers

$\int_{0}^{\infty} \frac{\sin^{3} x}{x^{3}} d x = \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin x - 3 \sin x}{x^{3}} d x = \frac{1}{8} \int_{0}^{\infty} \frac{(3 \sin x - \sin 3 x)^{(2)}}{x} d x = \frac{1}{8} \int_{0}^{\infty} \frac{- 3 \sin x + 9 \sin 3 x}{x} d x = \frac{1}{8} (- \frac{3 π}{2} + \frac{9 π}{2}) = \frac{3 π}{8}$

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