calculate-A-0-cos-sinx-sin-cosx-x-2-2-dx-from-R-

Question Number 48255 by Abdo msup. last updated on 21/Nov/18

c a l c u l a t e A_{λ} = \int_{0}^{\infty} \frac{c o s (λ s i n x) - s i n (λ c o s x)}{x^{2} + λ^{2}} d x

λ f r o m R .

Commented by Abdo msup. last updated on 23/Nov/18

w e h a v e A_{λ} = \int_{0}^{\infty} \frac{c o s (λ s i n x)}{x^{2} + λ^{2}} d x - \int_{0}^{\infty} \frac{s i n (λ c o s x)}{x^{2} + λ^{2}} d x

= H - K c a s e 1 λ > 0

2 H = \int_{- \infty}^{+ \infty} \frac{c o s (λ s i n x)}{x^{2} + λ^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{e^{i λ s i n x}}{x^{2} + λ^{2}} d x) l e t

φ (z) = \frac{e^{i λ s i n z}}{z^{2} + λ^{2}} \Rightarrow φ (z) = \frac{e^{i λ s i n z}}{(z - i λ) (z + i λ)}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i λ)

= 2 i π \frac{e^{i λ s i n (i λ)}}{2 i λ} = \frac{π}{λ} e^{i λ s i n (i λ)} b u t s i n (i λ) = \frac{e^{i (i λ)} - e^{- i (i λ)}}{2 i}

= \frac{e^{λ} - e^{- λ}}{2} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{λ} e^{i λ \frac{e^{λ} - e^{- λ}}{2}}

= \frac{π}{λ} {c o s (\frac{λ}{2} (e^{λ} - e^{- λ})) + i s i n (\frac{λ}{2} (e^{λ} - e^{- λ})) \Rightarrow

H = \frac{π}{2 λ} c o s (\frac{λ}{2} (e^{λ} - e^{- λ})) a l s o

2 K = \int_{- \infty}^{+ \infty} \frac{s i n (λ c o s x)}{x^{2} + λ^{2}} d x = I m (\int_{- \infty}^{+ \infty} \frac{e^{i λ c o s x}}{x^{2} + λ^{2}} d x) b u t

\int_{- \infty}^{+ \infty} \frac{e^{i λ c o s x}}{x^{2} + λ^{2}} d x ? = 2 i π \frac{e^{i λ c o s (i λ)}}{2 i λ} = \frac{π}{λ} e^{i λ c o s (i λ)} b u t

c o s (i λ) = \frac{e^{i (i λ)} + e^{- i (i λ)}}{2} = \frac{e^{λ} + e^{- λ}}{2} \Rightarrow

\int_{- \infty}^{+ \infty} \frac{e^{i λ c o s x}}{x^{2} + λ^{2}} d x = \frac{π}{λ} e^{i λ \frac{e^{λ} + e^{- λ}}{2}}

= \frac{π}{λ} {c o s (\frac{λ}{2} (e^{λ} + e^{- λ})) + i s i n (\frac{λ}{2} (e^{λ} + e^{- λ}))} \Rightarrow

K = \frac{π}{2 λ} s i n (\frac{λ}{2} (e^{λ} + e^{- λ})) \Rightarrow

A_{λ} = \frac{π}{2 λ} {c o s (\frac{λ}{2} (e^{λ} - e^{- λ}) - s i n (\frac{λ}{2} (e^{λ} + e^{- λ}))}

i f λ < 0 w e f o l l o w t h e s a m e m e t h o d .