0-sin-4-x-x-4-dx-pi-3-

Question Number 143312 by Ar Brandon last updated on 12/Jun/21

\int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} dx = \frac{π}{3}

Answered by Olaf_Thorendsen last updated on 12/Jun/21

Let f (x) = 1 (constant function unity)

As f is a continuous function satisfying

the π - periodic assumption, we can

apply the extended {Lobachevsky}^{'} s

Dirichlet integral formula :

\int_{0}^{\infty} \frac{\sin^{4}}{x^{4}} f (x) d x = \int_{0}^{\frac{π}{2}} f (x) d x - \frac{2}{3} \int_{0}^{\frac{π}{2}} \sin^{2} x . f (x) d x

\Rightarrow \int_{0}^{\infty} \frac{\sin^{4}}{x^{4}} d x = \int_{0}^{\frac{π}{2}} d x - \frac{2}{3} \int_{0}^{\frac{π}{2}} \sin^{2} x d x

\int_{0}^{\infty} \frac{\sin^{4}}{x^{4}} d x = \frac{π}{2} - \frac{2}{3} \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x

\int_{0}^{\infty} \frac{\sin^{4}}{x^{4}} d x = \frac{π}{2} - \frac{1}{3} {[x - \frac{1}{2} \sin 2 x]}_{0}^{\frac{π}{2}}

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

\int_{0}^{\infty} \frac{\sin^{4}}{x^{4}} d x = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}

Commented by Ar Brandon last updated on 12/Jun/21

Thanks Sir

Commented by Ar Brandon last updated on 12/Jun/21

I tried this;

Φ = \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} dx,

\sin^{4} x = \frac{(1 - 2 \cos 2 x + \cos^{2} 2 x)}{4} = \frac{1}{4} - \frac{\cos 2 x}{2} + \frac{(1 + \cos 4 x)}{8}

Φ = \frac{3}{8} \int_{0}^{\infty} \frac{1}{x^{4}} dx - \frac{1}{2} \int_{0}^{\infty} \frac{\cos 2 x}{x^{4}} dx + \frac{1}{8} \int_{0}^{\infty} \frac{\cos 4 x}{x^{4}} dx

= - {[\frac{1}{{8 x}^{3}}]}_{0}^{\infty} - \frac{π 2^{3}}{4 Γ (4) \cos (\frac{π}{2} \times 4)} + \frac{π 4^{3}}{16 Γ (4) \cos (\frac{π}{2} \times 4)}

= - {[\frac{1}{{8 x}^{3}}]}_{0}^{\infty} - \frac{8 π}{24} + \frac{64 π}{48} = \dots

Why did it not go ? Do you know why ?