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

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e ::$ $I = \int_{0}^{\infty} (\frac{s i n (x) . s i n (2 x)}{x}) d x = ? ? ?$ $. . . m . n . 1970 . . .$
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
	$let take a try we have sinx.sin(2x)=cos((π/2)−x)cos((π/2)−2x) =(1/2){cos(π−3x)+cos(x)} =(1/2){cosx−cos(3x)} ⇒ I =(1/2)∫_0 ^∞ ((cosx−cos(3x))/x)dx I =∫_0 ^∞ ((1−cos(3x))/(2x))dx−∫_0 ^∞ ((1−cosx)/(2x))dx =(1/2)(u−v) let f(a) =∫_0 ^∞ ((1−cosx)/x)e^(−ax) dx with a>0 f^′ (a) =−∫_0 ^∞ (1−cosx)e^(−ax) dx =∫_0 ^∞ e^(−ax) cosxdx−∫_0 ^∞ e^(−ax) dx ∫_0 ^∞ e^(−ax) dx=[−(1/a)e^(−ax) ]_0 ^∞ =(1/a) ∫_0 ^∞ e^(−ax+ix) dx =∫_0 ^∞ e^((−a+i)x) dx =[(1/(−a+i)) e^((−a+i)x) ]_0 ^∞ =−(1/(−a+i))=(1/(a−i)) =((a+i)/(a^2 +1)) ⇒∫_0 ^∞ e^(−ax) cosxdx =Re(...) =(a/(1+a^2 )) ⇒ f^′ (a) =(a/(1+a^2 ))−(1/a) ⇒ f(a) =(1/2)ln(1+a^2 )−lna +c =ln(((√(1+a^2 ))/a)) +c lim_(a→+∞) f(a) =0 =c ⇒ f(a) =ln(((√(1+a^2 ))/a)) ∫_0 ^∞ ((1−cosx)/x)dx =lim_(a→0) f(a) dont exist also ∫_0 ^∞ ((1−cos(3x))/x)dxdont exist but we see ∫_0 ^∞ ((1−cos(3x))/x)dx=_(3x=t) 3∫_0 ^∞ ((1−cost)/t)(dt/3) =∫_0 ^∞ ((1−cosx)/x)dx I =0$
	$l e t t a k e a t r y w e h a v e$ $s i n x . s i n (2 x) = c o s (\frac{π}{2} - x) c o s (\frac{π}{2} - 2 x)$ $= \frac{1}{2} {c o s (π - 3 x) + c o s (x)}$ $= \frac{1}{2} {c o s x - c o s (3 x)} \Rightarrow$ $I = \frac{1}{2} \int_{0}^{\infty} \frac{c o s x - c o s (3 x)}{x} d x$ $I = \int_{0}^{\infty} \frac{1 - c o s (3 x)}{2 x} d x - \int_{0}^{\infty} \frac{1 - c o s x}{2 x} d x$ $= \frac{1}{2} (u - v)$ $l e t f (a) = \int_{0}^{\infty} \frac{1 - c o s x}{x} e^{- a x} d x w i t h a > 0$ $f^{^{'}} (a) = - \int_{0}^{\infty} (1 - c o s x) e^{- a x} d x$ $= \int_{0}^{\infty} e^{- a x} c o s x d x - \int_{0}^{\infty} e^{- a x} d x$ $\int_{0}^{\infty} e^{- a x} d x = {[- \frac{1}{a} e^{- a x}]}_{0}^{\infty}$ $= \frac{1}{a}$ $\int_{0}^{\infty} e^{- a x + i x} d x = \int_{0}^{\infty} e^{(- a + i) x} d x$ $= {[\frac{1}{- a + i} e^{(- a + i) x}]}_{0}^{\infty} = - \frac{1}{- a + i} = \frac{1}{a - i}$ $= \frac{a + i}{a^{2} + 1} \Rightarrow \int_{0}^{\infty} e^{- a x} c o s x d x$ $= R e (. . .) = \frac{a}{1 + a^{2}} \Rightarrow$ $f^{^{'}} (a) = \frac{a}{1 + a^{2}} - \frac{1}{a} \Rightarrow$ $f (a) = \frac{1}{2} l n (1 + a^{2}) - l n a + c$ $= l n (\frac{\sqrt{1 + a^{2}}}{a}) + c$ ${l i m}_{a \to + \infty} f (a) = 0 = c \Rightarrow$ $f (a) = l n (\frac{\sqrt{1 + a^{2}}}{a})$ $\int_{0}^{\infty} \frac{1 - c o s x}{x} d x = {l i m}_{a \to 0} f (a)$ $d o n t e x i s t a l s o \int_{0}^{\infty} \frac{1 - c o s (3 x)}{x} d x d o n t$ $e x i s t b u t w e s e e$ $\int_{0}^{\infty} \frac{1 - c o s (3 x)}{x} d x =_{3 x = t} 3 \int_{0}^{\infty} \frac{1 - c o s t}{t} \frac{d t}{3}$ $= \int_{0}^{\infty} \frac{1 - c o s x}{x} d x I = 0$
	Commented by Bird last updated on 05/Oct/20

	$u = v \Rightarrow I = 0$
	Commented by mnjuly1970 last updated on 05/Oct/20
	$solution: method 1 : I =L [((sin(x).sin(2x)/x)]_(s=0) = L [sin(x) sin(2x)]=(1/2) L [cos(x−2x) −cos(x+2x)] =(1/2) L [cos(x)]− (1/2) L [cos(3x)] =(1/2)∗ (s/(s^2 +1)) −(1/2)∗ (s/(s^2 +9)) I = {(1/2)∫_s ^( ∞) (u/(u^2 +1)) du − (1/2) ∫_s ^( ∞) (u/(u^2 +9))du}_(s=0) ={(1/4) [ln (((u^2 +1)/(u^2 +9)))]_( u=s) ^( u=∞) = (1/4) ln((1/9))−(1/4) ln(((s^2 +1)/(s^2 +9)) )}_(s=0) =[((−1)/2) ln(3) − (1/4)ln(((s^2 +1)/(s^2 +9)))]_( s=0) = thank you your effort and work is admirable ...$
	$s o l u t i o n :$ $m e t h o d 1 : I = L {[\frac{s i n (x) . s i n (2 x}{x}]}_{s = 0}$ $= L [s i n (x) s i n (2 x)] = \frac{1}{2} L [c o s (x - 2 x) - c o s (x + 2 x)]$ $= \frac{1}{2} L [c o s (x)] - \frac{1}{2} L [c o s (3 x)]$ $= \frac{1}{2} * \frac{s}{s^{2} + 1} - \frac{1}{2} * \frac{s}{s^{2} + 9}$ $I = {\frac{1}{2} \int_{s}^{\infty} \frac{u}{u^{2} + 1} d u - \frac{1}{2} \int_{s}^{\infty} \frac{u}{u^{2} + 9} d u}_{s = 0}$ $= {\frac{1}{4} {[l n (\frac{u^{2} + 1}{u^{2} + 9})]}_{u = s}^{u = \infty} = \frac{1}{4} l n (\frac{1}{9}) - \frac{1}{4} l n (\frac{s^{2} + 1}{s^{2} + 9})}_{s = 0}$ $= {[\frac{- 1}{2} l n (3) - \frac{1}{4} l n (\frac{s^{2} + 1}{s^{2} + 9})]}_{s = 0}$ $=$ $t h a n k y o u y o u r e f f o r t a n d$ $w o r k i s a d m i r a b l e . . .$
	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

	$t h a n k y o u$ $v e r y n i c e a s a l w a y s . . .$
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