Prove-that-0-sin-x-x-dx-pi-2-

Question Number 86830 by Ar Brandon last updated on 31/Mar/20

P r o v e t h a t \int_{0}^{\infty} \frac{s i n x}{x} d x = \frac{π}{2}

Answered by mind is power last updated on 31/Mar/20

l e t f (z) = \frac{e^{i z}}{z}

\forall ϵ, R > 0 ϵ < R w e h a v e

\int_{C_{ϵ}} f (z) d z = 2 i π R e s (f, 0)

C_{ϵ} =] - R, - ϵ [\cup [ϵ e^{i θ}, θ \in [- π, 0]]] ϵ, + R [\cup {R e}^{i θ} = 2 i π

\int_{ϵ}^{R} (- f (- z) + f (z)) d z = 2 i π - \int_{- π}^{0} f (ϵ e^{i θ}) . i ϵ e^{i θ} d θ (\int_{0}^{π} {i e}^{i R c o s (θ) - R s i n (θ)}

\int_{ϵ}^{R} 2 i \frac{s i n (x)}{x} d x = 2 i π - \int_{0}^{π} i (\frac{e^{i ϵ (c o s (θ) + i s i n (θ))}}{ϵ e^{i θ}}) . ϵ e^{i θ} d θ - \int_{0}^{π} {i e}^{i R c o s (θ) - R s i n (θ)} d θ

2 i \int_{0}^{R} \frac{s i n (x)}{x} d x = 2 i π - \lim_{ϵ \to 0} \int_{0}^{π} {i e}^{i ϵ e^{i θ}} d θ - i \int_{0}^{π} e^{i R c o s (θ) - R s i n (θ)} d θ = i π

\Rightarrow 2 i \int_{0}^{\infty} \frac{s i n (z)}{z} d z = i π \Rightarrow \int_{0}^{+ \infty} \frac{s i n (z)}{z} = \frac{π}{2}

Commented by Ar Brandon last updated on 01/Apr/20

T h a n k s S i r, b u t i t h i n k y o u a p p l i e d a t h e o r y w h i c h i {d o n}^{'} t y e t k n o w .

C a n y o u p l e a s e e m p h a s i z e m o r e o n t h i s ?

Commented by mathmax by abdo last updated on 01/Apr/20

s i r a r b r o n d o n t r y t o s t u d y c o m p l e x a n s l y s i s s t e p b y s t e p a n d

t h a t t h e c o c e p t b e v o m e e a z y f o r y o u \dots

Commented by Ar Brandon last updated on 02/Apr/20

O K t h a n k s s o m u c h .

Answered by TANMAY PANACEA. last updated on 31/Mar/20

f (a) = \int_{0}^{\infty} \frac{s i n x}{x} \times e^{- a x} d x

\frac{d f (a)}{d a} = \int_{0}^{\infty} \frac{s i n x}{x} \times e^{- a x} \times - x d x

(- 1) \frac{d f}{d a} = \int_{0}^{\infty} e^{- a x} \times s i n x d x

q = \int_{0}^{\infty} e^{- a x} \times s i n x d x

p = \int_{0}^{\infty} e^{- a x} \times c o s x d x

p + i q = \int_{0}^{\infty} e^{- a x} \times e^{i x} d x

= \int_{0}^{\infty} e^{- x (a - i)} d x

=∣ \frac{e^{- x (a - i)}}{a - i} ∣_{0}^{\infty} = \frac{1}{a - i} = \frac{a + i}{a^{2} + 1}

s o q = \int_{0}^{\infty} e^{- a x} s i n x d x = \frac{1}{a^{2} + 1}

n o w \frac{- d f (a)}{d a} = \frac{1}{a^{2} + 1}

\int d f = \int \frac{- d a}{a^{2} + 1}

f (a) = - {t a n}^{- 1} (a) + c

n o w l o o k f (a) = \int_{0}^{\infty} \frac{s i n x}{x} \times e^{- a x} d x

w h e n a \to \infty t h e n e^{- a x} \to 0 s o f (a) = 0

h e n c e . .

0 = - {t a n}^{- 1} (\infty) + c

c = \frac{π}{2}

f (a) = - {t a n}^{- 1} (a) + \frac{π}{2}

f (a) = \int_{0}^{\infty} \frac{s i n x}{x} \times e^{- a x} d x = - {t a n}^{- 1} (a) + \frac{π}{2}

n o w l o o k

f (0) = \int_{0}^{\infty} \frac{s i n x}{x} \times e^{- 0} d x = \frac{π}{2}

\int_{0}^{\infty} \frac{s i n x}{x} = \frac{π}{2}

Commented by Ar Brandon last updated on 01/Apr/20

g o t i t, t h a n k s