calculate-0-sin-x-x-dx-

Question Number 105545 by 4635 last updated on 29/Jul/20

c a l c u l a t e \int_{0}^{\infty} \frac{\sin x}{x} d x

Answered by Ar Brandon last updated on 29/Jul/20

Let f (a) = \int_{0}^{\infty} \frac{sinx}{x} \cdot e^{- ax} dx \Rightarrow f^{'} (a) = - \int_{0}^{\infty} sinx \cdot e^{- ax} dx

Let; p = \int_{0}^{\infty} cosx \cdot e^{- ax} dx, q = \int_{0}^{\infty} sinx \cdot e^{- ax} dx

\Rightarrow p - iq = \int_{0}^{\infty} (cosx - isinx) e^{- ax} dx = \int_{0}^{\infty} e^{- ix - ax} dx = \int_{0}^{\infty} e^{- (a + i) x} dx

= - \frac{1}{a + i} \cdot {[e^{- (a + i) x}]}_{0}^{\infty} = \frac{1}{a + i} {{\underset{x \to \infty}{\lim e}}^{- (a + i) x} = 0}

= \frac{a - i}{a^{2} + 1} = \frac{a}{a^{2} + 1} - i (\frac{1}{a^{2} + 1})

\Rightarrow I m (p - iq) = I m {\frac{a}{a^{2} + 1} - i (\frac{1}{a^{2} + 1})} \Rightarrow q = \frac{1}{a^{2} + 1}

\Rightarrow f^{'} (a) = - q = - \frac{1}{a^{2} + 1} \Rightarrow f (a) = - [Arctan (a) + C]

But f (a) = \int_{0}^{\infty} \frac{sinx}{x} \cdot e^{- ax} dx, \underset{a \to + \infty}{\lim f} (a) = 0

\Rightarrow \lim_{a \to + \infty} [Arctan (a) + C] = 0 \Rightarrow C = - \frac{π}{2}

f (0) = \int_{0}^{\infty} \frac{sinx}{x} dx = - [Arctan (0) + C] = \frac{π}{2}

Hence \int_{0}^{\infty} \frac{sinx}{x} dx = \frac{π}{2}

Answered by Aziztisffola last updated on 29/Jul/20

Using Laplace transforms .

L {\frac{\sin x}{x}} = \int_{s}^{\infty} L {\sin x} dt = \int_{s}^{\infty} \frac{1}{1 + t^{2}} dt

= {[\arctan (t)]}_{s}^{t \to \infty} = \frac{π}{2} - \arctan (s)

then \int_{0}^{\infty} e^{- s x} \frac{\sin x}{x} d x = \frac{π}{2} - \arctan (s)

let s = 0 \Rightarrow \int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2} - \arctan (0) = \frac{π}{2}

Hence \int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}

Commented by 4635 last updated on 29/Jul/20

t h a n k y o u v e r y m u c h

Commented by Aziztisffola last updated on 29/Jul/20

{you}^{'} re welcome

Answered by 1549442205PVT last updated on 30/Jul/20

Set I (k, λ) = \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx

Differentiate the integration I by λ we get

\frac{dI}{d λ} = \int_{0}^{\infty} e^{- kx} . \frac{1}{x} . {(\sin λ x)}^{'} dx = \int_{0}^{\infty} e^{- kx} \cos λ xdx

= \frac{1}{λ} \int_{0}^{\infty} e^{- kx} dsin λ x = \frac{e^{- kx} \sin λ x}{λ} ∣_{x = 0}^{\infty} + \frac{k}{λ} \int_{0}^{\infty} e^{- kx} \sin λ xdx

= \frac{e^{- kx} \sin λ x}{λ} ∣_{0}^{\infty} - \frac{k}{λ^{2}} \int_{0}^{\infty} e^{- kx} dcos λ x

= (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty} - \frac{k^{2}}{λ^{2}} \int e^{- kx} \cos λ xdx

Thus, \frac{dI}{d λ} = (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty} - \frac{k^{2}}{λ} . \frac{dI}{d λ}

\Rightarrow\Rightarrow (1 + \frac{k^{2}}{λ^{2}}) . \frac{dI}{d λ} = (\frac{e^{- kx} \sin λ x}{λ} - \frac{{ke}^{- kx} \cos λ x}{λ^{2}}) ∣_{0}^{\infty}

\frac{dI}{d λ} = \frac{e^{- kx} (λ \sin λ x - kcos λ x)}{k^{2} + λ^{2}} ∣_{0}^{\infty} = \frac{k}{λ^{2} + k^{2}}

\Rightarrow I = \tan^{- 1} (\frac{λ}{k}) (as \int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}))

Consequently, I (k, λ) = \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx = \tan^{- 1} (\frac{λ}{k})

It follows that

J = \int_{0}^{\infty} \frac{\sin λ x}{x} dx = \lim_{k \to 0} \int_{0}^{\infty} e^{- kx} \frac{\sin λ x}{x} dx

= {\underset{k \to 0}{\lim t a n}}^{- 1} (\frac{λ}{k}) = \frac{π}{2} (if λ > 0)

\boldsymbol Therefore, \boldsymbol K = \int_{0}^{\infty} \frac{\boldsymbol sinx}{\boldsymbol x} \boldsymbol dx = \frac{\boldsymbol π}{2}