find-0-e-ax-sin-bx-dx-a-gt-0-and-b-gt-0-

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 = \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}}}