Menu Close

calculate-0-e-x-x-sinx-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 78708 by abdomathmax last updated on 20/Jan/20
calculate ∫_0 ^∞ (e^(−x) /x)(sinx)^(2 ) dx
calculate∫0∞e−xx(sinx)2dx
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)
letf(t)=∫0∞e−xtxsin2xdxwitht>0wehavef′(t)=−∫0∞e−xtsin2xdx=−∫0∞e−xt(1−cos(2x))dx=12∫0∞e−xtcos(2x)dx−12∫0∞e−xtdx∫0∞e−xtdx=[−1te−xt]x=0∞=1t∫0∞e−xtcos(2x)dx=Re(∫0∞e−xt+2ixdx)∫0∞e(−t+2i)xdx=[1−t+2ie(−t+2i)x]0∞=−1−t+2i=1t−2i=t+2it2+4⇒∫0∞e−xtcos(2x)dx=tt2+4⇒2f′(t)=tt2+4−1t⇒2f(t)=∫1tuu2+4−∫1tduu+c=12[ln(u2+4)]1t−lnt+c⇒f(t)=14{ln(t2+4)−ln(5)}−12ln(t)+c∃m>/∫0∞∣e−xtsin2xx∣dx⩽m∫0∞e−xtdx=mt→0(t→+∞)⇒limt→+∞f(t)=0⇒limt→+∞ln((4t2+1t)−14ln(5)+c=0⇒c=14ln(5)⇒f(t)=14ln(t2+4)−12ln(t)(t>0)∫0∞e−xxsin2xdx=f(1)=ln(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)
existencein0sin2(x)∼x2e−xxisintegrablein∞sin2(x)x<1⇒e−xsin2(x)x<e−xobviousletf(t)=∫0+∞e−xtxsin2(x)dx,t∈[1,∞[=IfwelldefindeC∞inIlimft→∞(t)=0proofu=xt⇒∫0+∞e−uusin2(ut)du∣sin(ut)∣⩽ut⇒∫0+∞e−uusin2(ut)du⩽∫0+∞e−uuu2t2du=1t2∫0+∞ue−udu=Γ(1)t2→0ast→∞∂f∂t=∫0+∞−e−xtsin2(x)dx=−∫0+∞e−xt(1−cos(2x)2)dx=−12∫0+∞e−xtdx+12∫0+∞e−xtcos(2x)dx=12[e−xtt]0+∞+12Re{∫0+∞ex(−t+2i)dt}=−12t+12Re[ex(−t+2i)−t+2i]0+∞=−12t+12Re[1t−2i]=−12t+12tt2+4=f′(t)f(1)=∫∞1f′(t)dt=∫∞1{−12t+t2(t2+4)}dt=limx→∞[x1−12ln(t)+14ln(t2+4)]=limx→∞[x114ln(t2+4t2)]=ln(5)4−limx→∞ln(x2+4x2)4=ln(5)4
Commented by mathmax by abdo last updated on 20/Jan/20
thank you sir.
thankyousir.
Commented by mind is power last updated on 21/Jan/20
y′re welcom
y′rewelcom
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{sin2(x)}=12L{(1−cos2x)}=12{1s−ss2+4}F(s)=L{sin2xU(x)x}=∫s∞12{1s−ss2+4}ds=12[ln(s)−12ln(s2+4)]s∞=14[ln()]s∞=−14(s2s2+4)∫0∞sin2(x)e−xx=F(1)=−14ln(15)=14ln(5)
Commented by M±th+et£s last updated on 20/Jan/20
so sorry there is a typo (1/4)[ln((s^2 /(s^2 +4)))]_s ^∞
sosorrythereisatypo14[ln(s2s2+4)]s∞

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *