advanced-calculus-evaluate-I-0-sin-x-sin-2x-x-dx-m-n-1970-

Question Number 116436 by mnjuly1970 last updated on 04/Oct/20

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

e v a l u a t e ::

I = \int_{0}^{\infty} (\frac{s i n (x) . s i n (2 x)}{x}) d x = ? ? ?

\dots m . n . 1970 \dots

Commented by Bird last updated on 04/Oct/20

i t s e e m s t h a t i n t e g r a l i s d i v e r g e n t!

Commented by mathdave last updated on 04/Oct/20

i s n o t d i v e r g e n t o o o o

Answered by mnjuly1970 last updated on 04/Oct/20

Answered by Bird last updated on 05/Oct/20

l e t t a k e a t r y w e h a v e

s i n x . s i n (2 x) = c o s (\frac{π}{2} - x) c o s (\frac{π}{2} - 2 x)

= \frac{1}{2} {c o s (π - 3 x) + c o s (x)}

= \frac{1}{2} {c o s x - c o s (3 x)} \Rightarrow

I = \frac{1}{2} \int_{0}^{\infty} \frac{c o s x - c o s (3 x)}{x} d x

I = \int_{0}^{\infty} \frac{1 - c o s (3 x)}{2 x} d x - \int_{0}^{\infty} \frac{1 - c o s x}{2 x} d x

= \frac{1}{2} (u - v)

l e t f (a) = \int_{0}^{\infty} \frac{1 - c o s x}{x} e^{- a x} d x w i t h a > 0

f^{^{'}} (a) = - \int_{0}^{\infty} (1 - c o s x) e^{- a x} d x

= \int_{0}^{\infty} e^{- a x} c o s x d x - \int_{0}^{\infty} e^{- a x} d x

\int_{0}^{\infty} e^{- a x} d x = {[- \frac{1}{a} e^{- a x}]}_{0}^{\infty}

= \frac{1}{a}

\int_{0}^{\infty} e^{- a x + i x} d x = \int_{0}^{\infty} e^{(- a + i) x} d x

= {[\frac{1}{- a + i} e^{(- a + i) x}]}_{0}^{\infty} = - \frac{1}{- a + i} = \frac{1}{a - i}

= \frac{a + i}{a^{2} + 1} \Rightarrow \int_{0}^{\infty} e^{- a x} c o s x d x

= R e (\dots) = \frac{a}{1 + a^{2}} \Rightarrow

f^{^{'}} (a) = \frac{a}{1 + a^{2}} - \frac{1}{a} \Rightarrow

f (a) = \frac{1}{2} l n (1 + a^{2}) - l n a + c

= l n (\frac{\sqrt{1 + a^{2}}}{a}) + c

{l i m}_{a \to + \infty} f (a) = 0 = c \Rightarrow

f (a) = l n (\frac{\sqrt{1 + a^{2}}}{a})

\int_{0}^{\infty} \frac{1 - c o s x}{x} d x = {l i m}_{a \to 0} f (a)

d o n t e x i s t a l s o \int_{0}^{\infty} \frac{1 - c o s (3 x)}{x} d x d o n t

e x i s t b u t w e s e e

\int_{0}^{\infty} \frac{1 - c o s (3 x)}{x} d x =_{3 x = t} 3 \int_{0}^{\infty} \frac{1 - c o s t}{t} \frac{d t}{3}

= \int_{0}^{\infty} \frac{1 - c o s x}{x} d x I = 0

Commented by Bird last updated on 05/Oct/20

u = v \Rightarrow I = 0

Commented by mnjuly1970 last updated on 05/Oct/20

s o l u t i o n :

m e t h o d 1 : I = L {[\frac{s i n (x) . s i n (2 x}{x}]}_{s = 0}

= L [s i n (x) s i n (2 x)] = \frac{1}{2} L [c o s (x - 2 x) - c o s (x + 2 x)]

= \frac{1}{2} L [c o s (x)] - \frac{1}{2} L [c o s (3 x)]

= \frac{1}{2} * \frac{s}{s^{2} + 1} - \frac{1}{2} * \frac{s}{s^{2} + 9}

I = {\frac{1}{2} \int_{s}^{\infty} \frac{u}{u^{2} + 1} d u - \frac{1}{2} \int_{s}^{\infty} \frac{u}{u^{2} + 9} d u}_{s = 0}

= {\frac{1}{4} {[l n (\frac{u^{2} + 1}{u^{2} + 9})]}_{u = s}^{u = \infty} = \frac{1}{4} l n (\frac{1}{9}) - \frac{1}{4} l n (\frac{s^{2} + 1}{s^{2} + 9})}_{s = 0}

= {[\frac{- 1}{2} l n (3) - \frac{1}{4} l n (\frac{s^{2} + 1}{s^{2} + 9})]}_{s = 0}

=

t h a n k y o u y o u r e f f o r t a n d

w o r k i s a d m i r a b l e \dots

Answered by mathdave last updated on 05/Oct/20

s o l u t i o n

w e k n o w \sin x \sin 2 x = \frac{1}{2} (\cos x - \cos 3 x)

l e t I = \int_{0}^{\infty} \frac{\sin x \sin 2 x}{x} d x = \frac{1}{2} \int_{0}^{\infty} (\frac{\cos x - \cos 3 x}{x}) d x

f r o m m a z i d e n t i t y

\int_{0}^{\infty} f (t) G (t) d t = \int^{\infty} f (s) g (s) d s b u t n o t e

L (f (t)) = f (s) a n d L^{- 1} (G (t)) = g (s)

I = \frac{1}{2} \int_{0}^{\infty} L (\cos x - \cos 3 x) L^{- 1} {(\frac{1}{x})}_{s = x} d s b u t l^{- 1} (\frac{1}{x}) = 1

I = \frac{1}{2} \int_{0}^{\infty} (\frac{s}{s^{2} + 1} - \frac{s}{s^{2} + 9}) d s \frac{1}{2} \int_{0}^{\infty} \frac{2 s}{2 (s^{2} + 1)} d s - \frac{1}{2} \int_{0}^{\infty} \frac{2 s}{2 (s^{2} + 9)} d s = \frac{1}{4} \ln {(\frac{s^{2} + 1}{s^{2} + 9})}_{0}^{\infty}

I = \frac{1}{4} (0 - \ln (\frac{1}{9})) = \frac{\ln (3)}{2}

∵ \int_{0}^{\infty} \frac{\sin x \sin 2 x}{x} d x = \frac{\ln (3)}{2}

b y m a t h d a v e (05 / 10 / 2020)

Commented by mnjuly1970 last updated on 06/Oct/20

t h a n k y o u

v e r y n i c e a s a l w a y s \dots