calculate-0-sin-2-x-x-2-1-x-2-dx-

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

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}}) .