calculate-cos-cosx-sinx-x-2-4-dx-

Question Number 92993 by mathmax by abdo last updated on 10/May/20

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{c o s (c o s x - s i n x)}{x^{2} + 4} d x

Commented by mathmax by abdo last updated on 10/May/20

I = \int_{- \infty}^{+ \infty} \frac{c o s (c o s x - s i n x)}{x^{2} + 4} d x \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{c o s (\sqrt{2} c o s (x + \frac{π}{4}))}{x^{2} + 4} d x

=_{x + \frac{π}{4} = t} \int_{- \infty}^{+ \infty} \frac{c o s (\sqrt{2} t)}{{(t - \frac{π}{4})}^{2} + 4} d t = R e (\int_{- \infty}^{+ \infty} \frac{e^{i \sqrt{2} t}}{{(t - \frac{π}{4})}^{2} + 4})

l e t φ (z) = \frac{e^{i z \sqrt{2}}}{{(z - \frac{π}{4})}^{2} + 4} p o l e s o f φ ?

φ (z) = \frac{e^{i z \sqrt{2}}}{{(z - \frac{π}{4})}^{2} - {(2 i)}^{2}} = \frac{e^{i z \sqrt{2}}}{(z - \frac{π}{4} - 2 i) (z - \frac{π}{4} + 2 i)}

s o t h e p o l e s o f φ a r e \frac{π}{4} + 2 i a n d \frac{π}{4} - 2 i

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{π}{4} + 2 i) = 2 i π \frac{e^{i \sqrt{2} (\frac{π}{4} + 2 i)}}{4 i}

= \frac{π}{2} e^{\frac{i π \sqrt{2}}{4}} e^{- 2 \sqrt{2}} = \frac{π}{2} e^{- 2 \sqrt{2}} {c o s (\frac{π \sqrt{2}}{4}) + i s i n (\frac{π \sqrt{2}}{4})} \Rightarrow

I = \frac{π}{2} e^{- 2 \sqrt{2}} c o s (\frac{π \sqrt{2}}{4})