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

$. . . a d v a n c e d i n t e g r a l . . .$ $p r o v e t h a t ::$ $Ω = \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{x^{2} (1 + x^{2})} d x = \frac{π}{4} (1 + \frac{π}{e^{2}})$
	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
	$Ω=∫_0 ^∞ (((1+x^2 −x^2 )sin^2 (x)dx)/(x^2 (1+x^2 ))) ∫((sin^2 (x))/x^2 )−∫_0 ^∞ ((sin^2 (x))/(1+x^2 ))dx sin^2 (x)=((1−cos(2x))/2) Ω=∫((sin^2 (x))/x_(=A) ^2 )−∫(dx/(2(1+x^2 )))+(1/2)∫_0 ^∞ ((cos(2x))/(1+x^2 ))dx A by part [−((sin^2 (x))/x)]_0 ^∞ +∫_0 ^∞ ((sin(2x))/x)dx =∫_0 ^∞ 2.sin(2x).(dx/(2x))=∫_0 ^∞ ((sin(t))/t)dt=(π/2) ∫_0 ^∞ (dx/(2(1+x^2 )))=(1/2)arctan(x)]_0 ^∞ =(π/4) Ω=(π/2)−(π/4)+(1/2)∫((cos(2x))/(1+x^2 ))dx(π/4)+(1/2)B B=(1/2)∫_(−∞) ^∞ ((cos(2x))/(1+x^2 ))dx=(1/2)Re∫_(−∞) ^∞ (e^(2ix) /(1+x^2 ))dx =(1/2).2iπRes((e^(2ix) /(1+x^2 )),x=i{ =iπ(e^(−2) /(2i))=(π/e^2 ) Ω=(π/4)+.(1/2)(π/(2e^2 ))=(π/4)(1+(1/e^2 ))$
	$Ω = \int_{0}^{\infty} \frac{(1 + x^{2} - x^{2}) {s i n}^{2} (x) d x}{x^{2} (1 + x^{2})}$ $\int \frac{{s i n}^{2} (x)}{x^{2}} - \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{1 + x^{2}} d x$ ${s i n}^{2} (x) = \frac{1 - c o s (2 x)}{2}$ $Ω = \int \frac{{s i n}^{2} (x)}{x_{= A}^{2}} - \int \frac{d x}{2 (1 + x^{2})} + \frac{1}{2} \int_{0}^{\infty} \frac{c o s (2 x)}{1 + x^{2}} d x$ $A b y p a r t$ ${[- \frac{{s i n}^{2} (x)}{x}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x$ $= \int_{0}^{\infty} 2 . s i n (2 x) . \frac{d x}{2 x} = \int_{0}^{\infty} \frac{s i n (t)}{t} d t = \frac{π}{2}$ ${\int_{0}^{\infty} \frac{d x}{2 (1 + x^{2})} = \frac{1}{2} a r c t a n (x)]}_{0}^{\infty} = \frac{π}{4}$ $Ω = \frac{π}{2} - \frac{π}{4} + \frac{1}{2} \int \frac{c o s (2 x)}{1 + x^{2}} d x \frac{π}{4} + \frac{1}{2} B$ $B = \frac{1}{2} \int_{- \infty}^{\infty} \frac{c o s (2 x)}{1 + x^{2}} d x = \frac{1}{2} R e \int_{- \infty}^{\infty} \frac{e^{2 i x}}{1 + x^{2}} d x$ $= \frac{1}{2} . 2 i π R e s (\frac{e^{2 i x}}{1 + x^{2}}, x = i {$ $= i π \frac{e^{- 2}}{2 i} = \frac{π}{e^{2}}$ $Ω = \frac{π}{4} + . \frac{1}{2} \frac{π}{2 e^{2}} = \frac{π}{4} (1 + \frac{1}{e^{2}})$
	Commented by mnjuly1970 last updated on 29/Nov/20

	$t h a n k s a l o t s i r m i n d s p o w e r$ $n i c e a s a l w a y s . . .$
	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

	$i h o p e y o u a r e s u c c e s s f u l$ $i n a l l s t a g e s o f y o u r l i f e$ $g o d k e e p y o u . .$
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