Show-that-0-sin-x-x-pi-2-

Question Number 58791 by Tawa1 last updated on 30/Apr/19

Show that : \int_{0}^{\infty} \frac{\sin (x)}{x} = \frac{π}{2}

Answered by tanmay last updated on 30/Apr/19

i t h a s s o m e t r i c k t o s o l v e \dots

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

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

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

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

n o w l e t

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

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

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

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

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

p = \frac{a}{a^{2} + 1} a n d i q = i (\frac{1}{a^{2} + 1})

q = \frac{1}{a^{2} + 1}

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

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

\int d F (a) = (- 1) \int \frac{d a}{a^{2} + 1}

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

a s a \to \infty F (a) \to 0

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

0 = - \frac{π}{2} + c

c = \frac{π}{2}

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

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

n o w i f y o u p u t a = 0

w e g e t

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

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

Commented by Tawa1 last updated on 30/Apr/19

God bless you sir

Commented by tanmay last updated on 30/Apr/19

t h a n y o u \dots b l e s s i n g s h o w e r t o a l l

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