find ∫_0 ^∞ e^(−ax) ∣sin(bx)∣ dx a>0 and b>0


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 136943 by Mathspace last updated on 28/Mar/21

$f i n d \int_{0}^{\infty} e^{- a x} ∣ s i n (b x) ∣ d x$ $a > 0 a n d b > 0$
	Answered by mathmax by abdo last updated on 30/Mar/21
	$Φ=∫_0 ^∞ e^(−ax) ∣sin(bx)∣dx ⇒Φ=_(bx=t) ∫_0 ^∞ e^(−(a/b)t) ∣sint∣(dt/b) =(1/b) Σ_(n=0) ^∞ ∫_(nπ) ^((n+1)π) e^(−((at)/b)) ∣sint∣ dt =_(t=nπ +y) (1/b)Σ_(n=0) ^∞ ∫^π _0 e^(−(a/b)(nπ+y)) ∣sin(nπ+y)∣ dy =(1/b)Σ_(n=0) ^∞ e^(−((naπ)/b)) ∫_0 ^π e^(−((ay)/b)) siny dy let (a/b)=λ ⇒ Φ=(1/b) Σ_(n=0) ^∞ e^(−nλπ) .∫_0 ^π e^(−λy) siny dy we have ∫_0 ^π e^(−λy) siny dy =Im(∫_0 ^π e^(−λy+iy) dy) and ∫_0 ^π e^((−λ+i)y) dy =[(1/(−λ+i))e^((−λ+i)y) ]_0 ^π =((−1)/(λ−i)){e^((−λ+i)π) −1} =((−(λ+i))/(1+λ^2 )){− e^(−λπ) −1} =(((λ+i)(e^(−λπ) +1))/(1+λ^2 )) ⇒∫_0 ^π e^(−λy) siny dy =((1+e^(−λπ) )/(1+λ^2 )) ⇒ Φ =((1+e^(−λπ) )/(b(1+λ^2 )))Σ_(n=0) ^∞ (e^(−λπ) )^n =((1+e^(−λπ) )/(b(1+λ^2 )))×(1/(1−e^(−λπ) )) ⇒ Φ =((1+e^(−((aπ)/b)) )/(b(1−e^(−((aπ)/b)) )(1+(a^2 /b^2 )))) ⇒Φ=(b/(a^2 +b^2 ))×((1+e^(−((aπ)/b)) )/(1−e^(−((aπ)/b)) ))$
	$Φ = \int_{0}^{\infty} e^{- ax} ∣ \sin (bx) ∣ dx \Rightarrow Φ =_{bx = t} \int_{0}^{\infty} e^{- \frac{a}{b} t} ∣ sint ∣ \frac{dt}{b}$ $= \frac{1}{b} \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- \frac{at}{b}} ∣ sint ∣ dt$ $=_{t = n π + y} \frac{1}{b} \sum_{n = 0}^{\infty} {\int_{0}}^{π} e^{- \frac{a}{b} (n π + y)} ∣ \sin (n π + y) ∣ dy$ $= \frac{1}{b} \sum_{n = 0}^{\infty} e^{- \frac{na π}{b}} \int_{0}^{π} e^{- \frac{ay}{b}} siny dy let \frac{a}{b} = λ \Rightarrow$ $Φ = \frac{1}{b} \sum_{n = 0}^{\infty} e^{- n λ π} . \int_{0}^{π} e^{- λ y} siny dy we have$ $\int_{0}^{π} e^{- λ y} siny dy = Im (\int_{0}^{π} e^{- λ y + iy} dy) and$ $\int_{0}^{π} e^{(- λ + i) y} dy = {[\frac{1}{- λ + i} e^{(- λ + i) y}]}_{0}^{π} = \frac{- 1}{λ - i} {e^{(- λ + i) π} - 1}$ $= \frac{- (λ + i)}{1 + λ^{2}} {- e^{- λ π} - 1} = \frac{(λ + i) (e^{- λ π} + 1)}{1 + λ^{2}}$ $\Rightarrow \int_{0}^{π} e^{- λ y} siny dy = \frac{1 + e^{- λ π}}{1 + λ^{2}} \Rightarrow$ $Φ = \frac{1 + e^{- λ π}}{b (1 + λ^{2})} \sum_{n = 0}^{\infty} {(e^{- λ π})}^{n} = \frac{1 + e^{- λ π}}{b (1 + λ^{2})} \times \frac{1}{1 - e^{- λ π}} \Rightarrow$ $Φ = \frac{1 + e^{- \frac{a π}{b}}}{b (1 - e^{- \frac{a π}{b}}) (1 + \frac{a^{2}}{b^{2}})} \Rightarrow Φ = \frac{b}{a^{2} + b^{2}} \times \frac{1 + e^{- \frac{a π}{b}}}{1 - e^{- \frac{a π}{b}}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum