calculate ∫_0 ^∞ (e^(−x) /x)(sinx)^(2 ) dx


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Question Number 78708 by abdomathmax last updated on 20/Jan/20

$c a l c u l a t e \int_{0}^{\infty} \frac{e^{- x}}{x} {(s i n x)}^{2} d x$
Commented by mathmax by abdo last updated on 21/Jan/20
$let f(t) =∫_0 ^∞ (e^(−xt) /x)sin^2 x dx with t >0 we have f^′ (t) =−∫_0 ^∞ e^(−xt) sin^2 xdx =−∫_0 ^∞ e^(−xt) (1−cos(2x))dx =(1/2)∫_0 ^∞ e^(−xt) cos(2x)dx−(1/2)∫_0 ^∞ e^(−xt) dx ∫_0 ^∞ e^(−xt) dx =[−(1/t)e^(−xt) ]_(x=0) ^∞ =(1/t) ∫_0 ^∞ e^(−xt) cos(2x)dx =Re(∫_0 ^∞ e^(−xt+2ix) dx) ∫_0 ^∞ e^((−t+2i)x) dx =[(1/(−t+2i)) e^((−t+2i)x) ]_0 ^∞ =−(1/(−t+2i)) =(1/(t−2i)) =((t+2i)/(t^2 +4)) ⇒ ∫_0 ^∞ e^(−xt) cos(2x)dx =(t/(t^2 +4)) ⇒2f^′ (t)=(t/(t^2 +4))−(1/t) ⇒ 2f(t) =∫_1 ^t (u/(u^2 +4))−∫_1 ^t (du/u) +c =(1/2)[ln(u^2 +4)]_1 ^t −lnt +c ⇒ f(t)=(1/4){ln(t^2 +4)−ln(5)} −(1/2)ln(t) +c ∃m> /∫_0 ^∞ ∣((e^(−xt) sin^2 x)/x)∣dx≤m∫_0 ^∞ e^(−xt) dx =(m/t)→0(t→+∞) ⇒ lim_(t→+∞) f(t)=0 ⇒lim_(t→+∞) ln((((^4 (√(t^2 +1)))/(√t))) −(1/4)ln(5)+c =0 ⇒ c=(1/4)ln(5) ⇒f(t)=(1/4)ln(t^2 +4)−(1/2)ln(t) (t>0) ∫_0 ^∞ (e^(−x) /x)sin^2 x dx =f(1) =((ln(5))/4)$
$l e t f (t) = \int_{0}^{\infty} \frac{e^{- x t}}{x} {s i n}^{2} x d x w i t h t > 0 w e h a v e$ $f^{^{'}} (t) = - \int_{0}^{\infty} e^{- x t} {s i n}^{2} x d x = - \int_{0}^{\infty} e^{- x t} (1 - c o s (2 x)) d x$ $= \frac{1}{2} \int_{0}^{\infty} e^{- x t} c o s (2 x) d x - \frac{1}{2} \int_{0}^{\infty} e^{- x t} d x$ $\int_{0}^{\infty} e^{- x t} d x = {[- \frac{1}{t} e^{- x t}]}_{x = 0}^{\infty} = \frac{1}{t}$ $\int_{0}^{\infty} e^{- x t} c o s (2 x) d x = R e (\int_{0}^{\infty} e^{- x t + 2 i x} d x)$ $\int_{0}^{\infty} e^{(- t + 2 i) x} d x = {[\frac{1}{- t + 2 i} e^{(- t + 2 i) x}]}_{0}^{\infty} = - \frac{1}{- t + 2 i} = \frac{1}{t - 2 i} = \frac{t + 2 i}{t^{2} + 4} \Rightarrow$ $\int_{0}^{\infty} e^{- x t} c o s (2 x) d x = \frac{t}{t^{2} + 4} \Rightarrow 2 f^{^{'}} (t) = \frac{t}{t^{2} + 4} - \frac{1}{t} \Rightarrow$ $2 f (t) = \int_{1}^{t} \frac{u}{u^{2} + 4} - \int_{1}^{t} \frac{d u}{u} + c = \frac{1}{2} {[l n (u^{2} + 4)]}_{1}^{t} - l n t + c \Rightarrow$ $f (t) = \frac{1}{4} {l n (t^{2} + 4) - l n (5)} - \frac{1}{2} l n (t) + c$ $\exists m > / \int_{0}^{\infty} ∣ \frac{e^{- x t} {s i n}^{2} x}{x} ∣ d x ⩽ m \int_{0}^{\infty} e^{- x t} d x = \frac{m}{t} \to 0 (t \to + \infty) \Rightarrow$ ${l i m}_{t \to + \infty} f (t) = 0 \Rightarrow {l i m}_{t \to + \infty} l n (\frac{(^{4} \sqrt{t^{2} + 1}}{\sqrt{t}}) - \frac{1}{4} l n (5) + c = 0 \Rightarrow$ $c = \frac{1}{4} l n (5) \Rightarrow f (t) = \frac{1}{4} l n (t^{2} + 4) - \frac{1}{2} l n (t) (t > 0)$ $\int_{0}^{\infty} \frac{e^{- x}}{x} {s i n}^{2} x d x = f (1) = \frac{l n (5)}{4}$
	Answered by mind is power last updated on 20/Jan/20
	$existence in 0 sin^2 (x)∼x^2 e^(−x) x is integrable in ∞ ((sin^2 (x))/x)<1⇒e^(−x) ((sin^2 (x))/x)<e^(−x) obvious let f(t)=∫_0 ^(+∞) (e^(−xt) /x)sin^2 (x)dx,t∈[1,∞[=I f well definde C_∞ in I lim_(t→∞) f(t)=0 proof u=xt⇒∫_0 ^(+∞) (e^(−u) /u)sin^2 ((u/t))du ∣sin((u/t))∣≤(u/t)⇒∫_0 ^(+∞) (e^(−u) /u)sin^2 ((u/t))du≤∫_0 ^(+∞) (e^(−u) /u)(u^2 /t^2 )du=(1/t^2 )∫_0 ^(+∞) ue^(−u) du=((Γ(1))/t^2 ) →0 as t→∞ (∂f/∂t)=∫_0 ^(+∞) −e^(−xt) sin^2 (x)dx=−∫_0 ^(+∞) e^(−xt) (((1−cos(2x))/2))dx =−(1/2)∫_0 ^(+∞) e^(−xt) dx+(1/2)∫_0 ^(+∞) e^(−xt) cos(2x)dx =(1/2)[(e^(−xt) /t)]_0 ^(+∞) +(1/2)Re{∫_0 ^(+∞) e^(x(−t+2i)) dt} =−(1/(2t))+(1/2)Re[(e^(x(−t+2i)) /(−t+2i))]_0 ^(+∞) =−(1/(2t))+(1/2)Re[(1/(t−2i))] =−(1/(2t))+(1/2)(t/(t^2 +4))=f′(t) f(1)=∫_∞ ^1 f′(t)dt=∫_∞ ^1 {−(1/(2t))+(t/(2(t^2 +4)))}dt=lim_(x→∞) [_x ^1 −(1/2)ln(t)+(1/4)ln(t^2 +4)] =lim_(x→∞) [_x ^1 (1/4)ln(((t^2 +4)/t^2 ))]=((ln(5))/4)−lim_(x→∞) ((ln(((x^2 +4)/x^2 )))/4)=((ln(5))/4)$
	$existence in 0 \sin^{2} (x) \sim x^{2}$ $e^{- x} x is integrable$ $in \infty \frac{\sin^{2} (x)}{x} < 1 \Rightarrow e^{- x} \frac{\sin^{2} (x)}{x} < e^{- x} obvious$ $let f (t) = \int_{0}^{+ \infty} \frac{e^{- xt}}{x} \sin^{2} (x) dx, t \in [1, \infty [= I$ $f well definde C_{\infty} in I$ $\underset{t \to \infty}{\lim f} (t) = 0 proof$ $u = xt \Rightarrow \int_{0}^{+ \infty} \frac{e^{- u}}{u} \sin^{2} (\frac{u}{t}) du$ $∣ \sin (\frac{u}{t}) ∣⩽ \frac{u}{t} \Rightarrow \int_{0}^{+ \infty} \frac{e^{- u}}{u} \sin^{2} (\frac{u}{t}) du ⩽ \int_{0}^{+ \infty} \frac{e^{- u}}{u} \frac{u^{2}}{t^{2}} du = \frac{1}{t^{2}} \int_{0}^{+ \infty} {ue}^{- u} du = \frac{Γ (1)}{t^{2}}$ $\to 0 as t \to \infty$ $\frac{\partial f}{\partial t} = \int_{0}^{+ \infty} - e^{- xt} \sin^{2} (x) dx = - \int_{0}^{+ \infty} e^{- xt} (\frac{1 - \cos (2 x)}{2}) dx$ $= - \frac{1}{2} \int_{0}^{+ \infty} e^{- xt} dx + \frac{1}{2} \int_{0}^{+ \infty} e^{- xt} \cos (2 x) dx$ $= \frac{1}{2} {[\frac{e^{- xt}}{t}]}_{0}^{+ \infty} + \frac{1}{2} Re {\int_{0}^{+ \infty} e^{x (- t + 2 i)} dt}$ $= - \frac{1}{2 t} + \frac{1}{2} Re {[\frac{e^{x (- t + 2 i)}}{- t + 2 i}]}_{0}^{+ \infty}$ $= - \frac{1}{2 t} + \frac{1}{2} Re [\frac{1}{t - 2 i}]$ $= - \frac{1}{2 t} + \frac{1}{2} \frac{t}{t^{2} + 4} = f^{'} (t)$ $f (1) = \int_{\infty}^{1} f^{'} (t) dt = \int_{\infty}^{1} {- \frac{1}{2 t} + \frac{t}{2 (t^{2} + 4)}} dt = \lim_{x \to \infty} [_{x}^{1} - \frac{1}{2} \ln (t) + \frac{1}{4} \ln (t^{2} + 4)]$ $= \lim_{x \to \infty} [_{x}^{1} \frac{1}{4} \ln (\frac{t^{2} + 4}{t^{2}})] = \frac{\ln (5)}{4} - \lim_{x \to \infty} \frac{\ln (\frac{x^{2} + 4}{x^{2}})}{4} = \frac{\ln (5)}{4}$
	Commented by mathmax by abdo last updated on 20/Jan/20

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 21/Jan/20

	$y^{'} re welcom$
	Answered by M±th+et£s last updated on 20/Jan/20
	$L{sin^2 (x)}=(1/2)L{(1−cos2x)}=(1/2){(1/s)−(s/(s^2 +4))} F(s)=L{((sin^2 x U(x))/x)}=∫_s ^∞ (1/2){(1/s) −(s/(s^2 +4))}ds =(1/2)[ln(s)−(1/2)ln(s^2 +4)]_s ^∞ =(1/4)[ln()]_s ^∞ =((−1)/4) ((s^2 /(s^2 +4))) ∫_0 ^∞ ((sin^2 (x) e^(−x) )/x) = F(1)=−(1/4)ln((1/5))=(1/4) ln(5)$
	$L {{s i n}^{2} (x)} = \frac{1}{2} L {(1 - c o s 2 x)} = \frac{1}{2} {\frac{1}{s} - \frac{s}{s^{2} + 4}}$ $F (s) = L {\frac{{s i n}^{2} x U (x)}{x}} = \int_{s}^{\infty} \frac{1}{2} {\frac{1}{s} - \frac{s}{s^{2} + 4}} d s$ $= \frac{1}{2} {[l n (s) - \frac{1}{2} l n (s^{2} + 4)]}_{s}^{\infty} = \frac{1}{4} {[l n ()]}_{s}^{\infty}$ $= \frac{- 1}{4} (\frac{s^{2}}{s^{2} + 4})$ $\int_{0}^{\infty} \frac{{s i n}^{2} (x) e^{- x}}{x} = F (1) = - \frac{1}{4} l n (\frac{1}{5}) = \frac{1}{4} l n (5)$
	Commented by M±th+et£s last updated on 20/Jan/20

	$s o s o r r y t h e r e i s a t y p o$ $\frac{1}{4} {[l n (\frac{s^{2}}{s^{2} + 4})]}_{s}^{\infty}$
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