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Question Number 123900 by mnjuly1970 last updated on 29/Nov/20

\dots a d v a n c e d i n t e g r a l \dots

p r o v e t h a t ::

Ω = \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{x^{2} (1 + x^{2})} d x = \frac{π}{4} (1 + \frac{π}{e^{2}})

Answered by mathmax by abdo last updated on 29/Nov/20

I = \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2} (1 + x^{2})} dx \Rightarrow I = \int_{0}^{\infty} (\frac{1}{x^{2}} - \frac{1}{1 + x^{2}}) \sin^{2} x dx

= \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} - \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2} + 1} dx we have by [psrts

\int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} dx = {[- \frac{\sin^{2} x}{x}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{2 sinx cosx}{x} dx

= \int_{0}^{\infty} \frac{\sin (2 x)}{x} dx =_{2 x = t} \int_{0}^{\infty} \frac{\sin (t)}{\frac{t}{2}} \frac{dt}{2} = \int_{0}^{\infty} \frac{sint}{t} dt = \frac{π}{2}

\int_{0}^{\infty} \frac{\sin^{2} x}{x^{2} + 1} dx = \int_{0}^{\infty} \frac{1 - \cos (2 x)}{2 (x^{2} + 1)} dx = \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2} + 1} - \frac{1}{2} \int_{0}^{\infty} \frac{\cos (2 x)}{x^{2} + 1} dx

= \frac{π}{4} - \frac{1}{2} \int_{0}^{\infty} \frac{\cos (2 x)}{x^{2} + 1} dx we have

\int_{0}^{\infty} \frac{\cos (2 x)}{x^{2} + 1} dx = \frac{1}{2} Re (\int_{- \infty}^{+ \infty} \frac{e^{2 ix}}{x^{2} + 1} dx)

\int_{- \infty}^{+ \infty} \frac{e^{2 ix}}{x^{2} + 1} dx = 2 i π \times \frac{e^{- 2}}{2 i} = \frac{π}{e^{2}} \Rightarrow \int_{0}^{\infty} \frac{\cos (2 x)}{x^{2} + 1} dx = \frac{π}{{2 e}^{2}} \Rightarrow

I = \frac{π}{2} - \frac{π}{4} + \frac{1}{2} . \frac{π}{{2 e}^{2}} = \frac{π}{4} + \frac{π}{{4 e}^{2}} = \frac{π}{4} (1 + \frac{1}{e^{2}})

Commented by mnjuly1970 last updated on 29/Nov/20

g r a t e f u l m y m a s t e r s i r m a x

Commented by mathmax by abdo last updated on 29/Nov/20

you are welcome sir

Answered by mindispower last updated on 29/Nov/20

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

\int \frac{{s i n}^{2} (x)}{x^{2}} - \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{1 + x^{2}} d x

{s i n}^{2} (x) = \frac{1 - c o s (2 x)}{2}

Ω = \int \frac{{s i n}^{2} (x)}{x_{= A}^{2}} - \int \frac{d x}{2 (1 + x^{2})} + \frac{1}{2} \int_{0}^{\infty} \frac{c o s (2 x)}{1 + x^{2}} d x

A b y p a r t

{[- \frac{{s i n}^{2} (x)}{x}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x

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

{\int_{0}^{\infty} \frac{d x}{2 (1 + x^{2})} = \frac{1}{2} a r c t a n (x)]}_{0}^{\infty} = \frac{π}{4}

Ω = \frac{π}{2} - \frac{π}{4} + \frac{1}{2} \int \frac{c o s (2 x)}{1 + x^{2}} d x \frac{π}{4} + \frac{1}{2} B

B = \frac{1}{2} \int_{- \infty}^{\infty} \frac{c o s (2 x)}{1 + x^{2}} d x = \frac{1}{2} R e \int_{- \infty}^{\infty} \frac{e^{2 i x}}{1 + x^{2}} d x

= \frac{1}{2} . 2 i π R e s (\frac{e^{2 i x}}{1 + x^{2}}, x = i {

= i π \frac{e^{- 2}}{2 i} = \frac{π}{e^{2}}

Ω = \frac{π}{4} + . \frac{1}{2} \frac{π}{2 e^{2}} = \frac{π}{4} (1 + \frac{1}{e^{2}})

Commented by mnjuly1970 last updated on 29/Nov/20

t h a n k s a l o t s i r m i n d s p o w e r

n i c e a s a l w a y s \dots

Commented by mindispower last updated on 29/Nov/20

w i t h e p l e a s u r

s a d n o t t i m e s t o o d o m a t h s t h e s e d a y s

Commented by mnjuly1970 last updated on 29/Nov/20

i h o p e y o u a r e s u c c e s s f u l

i n a l l s t a g e s o f y o u r l i f e

g o d k e e p y o u . .