advanced-calculus-prove-that-sin-4-x-cos-4-x-x-2-dx-pi-16-m-n-

Question Number 140956 by mnjuly1970 last updated on 14/May/21

\dots . . a d v a n c e d \dots \dots c a l c u l u s \dots . .

p r o v e t h a t :

ϕ := \int_{- \infty}^{\infty} \frac{{s i n}^{4} (x) . {c o s}^{4} (x)}{x^{2}} d x = \frac{π}{16}

m . n

Answered by mathmax by abdo last updated on 14/May/21

Φ = 2 \int_{0}^{\infty} \frac{{(sinx . cosx)}^{4}}{x^{2}} dx = \frac{2}{2^{4}} \int_{0}^{\infty} \frac{\sin^{4} (2 x)}{x^{2}} dx

= \frac{1}{8} \int_{0}^{\infty} \frac{{(\frac{1 - \cos (4 x)}{2})}^{2}}{x^{2}} dx = \frac{1}{32} \int_{0}^{\infty} \frac{1 - 2 \cos (4 x) + \frac{1 + \cos (8 x)}{2}}{x^{2}} dx

= \frac{1}{64} \int_{0}^{\infty} \frac{2 - 4 \cos (4 x) + 1 + \cos (8 x)}{x^{2}} dx

= \frac{1}{64} \int_{0}^{\infty} \frac{4 - 4 \cos (4 x) - (1 - \cos (8 x))}{x^{2}} dx

= \frac{1}{16} \int_{0}^{\infty} \frac{1 - \cos (4 x)}{x^{2}} dx - \frac{1}{64} \int_{0}^{\infty} \frac{1 - \cos (8 x)}{x^{2}} dx

we have \int_{0}^{\infty} \frac{1 - \cos (4 x)}{x^{2}} dx =_{2 x = t} 4 \int_{0}^{\infty} \frac{1 - \cos 2 t}{t^{2}} \frac{dt}{2}

= 2 \int_{0}^{\infty} \frac{{2 \sin}^{2} (t)}{t^{2}} dt = 4 \int_{0}^{\infty} \frac{\sin^{2} (t)}{t^{2}}

= 4 {{[- \frac{1}{t} \sin^{2} t]}_{0}^{\infty} + \int_{0}^{\infty} \frac{2 sint . cost}{t} dt}

= 4 \int_{0}^{\infty} \frac{\sin (2 t)}{t} dt =_{2 t = z} 4 \int_{0}^{\infty} \frac{sinz}{\frac{z}{2}} \frac{dz}{2} = 4 . \frac{π}{2} = 2 π

\int_{0}^{\infty} \frac{1 - \cos (8 x)}{x^{2}} dx =_{4 x = t} 16 \int_{0}^{\infty} \frac{1 - \cos (2 t)}{t^{2}} \frac{dt}{4}

= 4 \int_{0}^{\infty} \frac{{2 \sin}^{2} t}{t^{2}} dt = 8 \int_{0}^{\infty} \frac{\sin^{2} t}{t^{2}} = 8 . \frac{π}{2} = 4 π \Rightarrow

Φ = \frac{1}{16} (2 π) - \frac{1}{64} (4 π) = \frac{π}{8} - \frac{π}{16} = \frac{π}{16} \Rightarrow Φ = \frac{π}{16} ★

Commented by mnjuly1970 last updated on 14/May/21

g r a t e f u l s i r m a x \dots t h a n k i n g