calculate ∫_0 ^∞ ((sin^2 (x))/(x^2 (1+x^2 )))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 63033 by mathmax by abdo last updated on 28/Jun/19

$c a l c u l a t e \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{x^{2} (1 + x^{2})} d x$
Commented by mathmax by abdo last updated on 28/Jun/19
$let A =∫_0 ^∞ ((sin^2 x)/(x^2 (1+x^2 ))) dx ⇒ A =∫_0 ^∞ sin^2 x{(1/x^2 ) −(1/(1+x^2 ))}dx =∫_0 ^∞ ((sin^2 x)/x^2 )dx −∫_0 ^∞ ((sin^2 x)/(1+x^2 ))dx =H−K by parts H =[−(1/x)sin^2 x]_0 ^(+∞) −∫_0 ^∞ −(1/x)2sinx cosxdx =∫_0 ^∞ ((sin(2x))/x)dx =_(2x=t) ∫_0 ^∞ ((sin(t))/(t/2)) (dt/2) =∫_0 ^∞ ((sint)/t)dt =(π/2) (result proved) K =∫_0 ^∞ ((1−cos(2x))/(2(1+x^2 )))dx =(1/2)∫_0 ^∞ (dx/(1+x^2 )) −(1/2)∫_0 ^∞ ((cos(2x))/(1+x^2 ))dx =(π/4) −(1/4)∫_(−∞) ^(+∞) ((cos(2x))/(1+x^2 ))dx ∫_(−∞) ^(+∞) ((cos(2x))/(1+x^2 ))dx =Re(∫_(−∞) ^(+∞) (e^(i2x) /(x^2 +1))dx) let W(z) =(e^(i2z) /(z^2 +1)) the poles of W are +^− i residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) =2iπ (e^(−2) /(2i)) =(π/e^2 ) ⇒ ∫_(−∞) ^(+∞) ((cos(2x))/(1+x^2 ))dx =(π/e^2 ) ⇒ K =(π/4)−(π/(4e^2 )) ⇒ A =(π/2) −(π/4) +(π/(4e^2 )) =(π/4) +(π/(4e^2 )) ⇒ A =(π/4)(1+(1/e^2 )).$
$l e t A = \int_{0}^{\infty} \frac{{s i n}^{2} x}{x^{2} (1 + x^{2})} d x \Rightarrow A = \int_{0}^{\infty} {s i n}^{2} x {\frac{1}{x^{2}} - \frac{1}{1 + x^{2}}} d x$ $= \int_{0}^{\infty} \frac{{s i n}^{2} x}{x^{2}} d x - \int_{0}^{\infty} \frac{{s i n}^{2} x}{1 + x^{2}} d x = H - K$ $b y p a r t s H = {[- \frac{1}{x} {s i n}^{2} x]}_{0}^{+ \infty} - \int_{0}^{\infty} - \frac{1}{x} 2 s i n x c o s x d x$ $= \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x =_{2 x = t} \int_{0}^{\infty} \frac{s i n (t)}{\frac{t}{2}} \frac{d t}{2} = \int_{0}^{\infty} \frac{s i n t}{t} d t = \frac{π}{2} (r e s u l t p r o v e d)$ $K = \int_{0}^{\infty} \frac{1 - c o s (2 x)}{2 (1 + x^{2})} d x = \frac{1}{2} \int_{0}^{\infty} \frac{d x}{1 + x^{2}} - \frac{1}{2} \int_{0}^{\infty} \frac{c o s (2 x)}{1 + x^{2}} d x$ $= \frac{π}{4} - \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + x^{2}} d x$ $\int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + x^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i 2 x}}{x^{2} + 1} d x) l e t W (z) = \frac{e^{i 2 z}}{z^{2} + 1} t h e p o l e s o f W a r e \bar{+} i$ $r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i) = 2 i π \frac{e^{- 2}}{2 i} = \frac{π}{e^{2}} \Rightarrow \int_{- \infty}^{+ \infty} \frac{c o s (2 x)}{1 + x^{2}} d x = \frac{π}{e^{2}} \Rightarrow$ $K = \frac{π}{4} - \frac{π}{4 e^{2}} \Rightarrow A = \frac{π}{2} - \frac{π}{4} + \frac{π}{4 e^{2}} = \frac{π}{4} + \frac{π}{4 e^{2}} \Rightarrow$ $A = \frac{π}{4} (1 + \frac{1}{e^{2}}) .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum