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Question Number 35620 by abdo mathsup 649 cc last updated on 21/May/18
let f(x) =e^(−x) sinx odd 2π periodic developp f at fourier serie .
letf(x)=e−xsinxodd2πperiodicdeveloppfatfourierserie.
Commented by abdo mathsup 649 cc last updated on 25/May/18
we have f(x)=Σ_(n=1) ^∞ a_n sin(nx) with a_n = (2/T) ∫_([T]) f(x)sin(nx)dx =(2/(2π)) ∫_(−π) ^π e^(−x) sinx sin(nx)dx = (2/π) ∫_0 ^π e^(−x) sin(nx)sin(x)dx cos(a+b) =cosa cosb −sinasinb cos(a−b)=cosa cosb +sina sinb ⇒I sina sinb =(1/2){cos(a−b)−cos(a+b)} π a_n = ∫_0 ^π ( e^(−x) cos(n−1)x −e^(−x) cos(n+1)x) but I(λ) = ∫_0 ^π e^(−x) cos(λx)dx =Re( ∫_0 ^π e^(−x) e^(iλx) dx) =Re( ∫_0 ^π e^((−1+iλ)x) dx) ∫_0 ^π e^((−1+iλ)x) dx = [(1/(−1+iλ)) e^((−1+iλ)x) ]_0 ^π =((−1)/(1−iλ)){ e^((−1+iλ)π) −1} =((−1−iλ)/(1+λ^2 )) { e^(−π) (cos(λπ) +isin(λπ) −1} =−(((1+iλ){ e^(−π) cos(λπ)−1 +i e^(−π) sin(λπ)})/(1+λ^2 )) =−(1/(1+λ^2 )){ e^(−π) cos(λπ) −1 +i e^(−π) sin(λπ) +iλ e^(−π) cos(λπ) −iλ −λ e^(−π) sin(λπ)}−⇒ I(λ) =((−1)/(1+λ^2 )){ e^(−π) cos(λπ) −λ e^(−π) sin(λπ) −1} I(n−1)= ((−1)/(1+(n−1)^2 )){ e^(−π) cos(n−1)π −(n−1) e^(−π) sin(n−1)π −1} =((−1)/(1+(n−1)^2 )){ (−1)^(n−1) e^(−π) −1} = (((−1)^n e^(−π) +1)/(1+(n−1)^2 )) I(n) = ((−1)/(1+n^2 )){ (−1)^n e^(−π) −1}=((1−(−1)^n e^(−π) )/(1+n^2 )) π a_n = I(n−1)−I(n)= ((1+(−1)^n e^(−π) )/(1+(n−1)^2 )) −((1 −(−1)^n e^(−π) )/(1+n^2 )) ⇒ e^(−x) sinx =Σ_(n=1) ^∞ (1/π){ ((1+(−1)^n e^(−π) )/(1+(n−1)^2 )) +(((−1)^n e^(−π) −1)/(1+n^2 ))}sin(nx)
wehavef(x)=∑n=1∞ansin(nx)withan=2T∫[T]f(x)sin(nx)dx=22π∫−ππe−xsinxsin(nx)dx=2π∫0πe−xsin(nx)sin(x)dxcos(a+b)=cosacosb−sinasinbcos(a−b)=cosacosb+sinasinb⇒Isinasinb=12{cos(a−b)−cos(a+b)}πan=∫0π(e−xcos(n−1)x−e−xcos(n+1)x)butI(λ)=∫0πe−xcos(λx)dx=Re(∫0πe−xeiλxdx)=Re(∫0πe(−1+iλ)xdx)∫0πe(−1+iλ)xdx=[1−1+iλe(−1+iλ)x]0π=−11−iλ{e(−1+iλ)π−1}=−1−iλ1+λ2{e−π(cos(λπ)+isin(λπ)−1}=−(1+iλ){e−πcos(λπ)−1+ie−πsin(λπ)}1+λ2=−11+λ2{e−πcos(λπ)−1+ie−πsin(λπ)+iλe−πcos(λπ)−iλ−λe−πsin(λπ)}−⇒I(λ)=−11+λ2{e−πcos(λπ)−λe−πsin(λπ)−1}I(n−1)=−11+(n−1)2{e−πcos(n−1)π−(n−1)e−πsin(n−1)π−1}=−11+(n−1)2{(−1)n−1e−π−1}=(−1)ne−π+11+(n−1)2I(n)=−11+n2{(−1)ne−π−1}=1−(−1)ne−π1+n2πan=I(n−1)−I(n)=1+(−1)ne−π1+(n−1)2−1−(−1)ne−π1+n2⇒e−xsinx=∑n=1∞1π{1+(−1)ne−π1+(n−1)2+(−1)ne−π−11+n2}sin(nx)

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