calculale-A-n-cos-x-n-1-x-2-dx-with-n-integr-natural-

Question Number 42680 by prof Abdo imad last updated on 31/Aug/18

c a l c u l a l e A_{n} (α) = \int_{- \infty}^{+ \infty} \frac{c o s (α x^{n})}{1 + x^{2}} d x w i t h

n i n t e g r n a t u r a l .

Commented by prof Abdo imad last updated on 31/Aug/18

α r e a l .

Commented by maxmathsup by imad last updated on 01/Sep/18

w e h a v e A_{n} (α) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i α x^{n}}}{1 + x^{2}} d x) l e t c o n s i d e r t h e c o m l e x f u n c t i o n

φ (z) = \frac{e^{i α z^{n}}}{1 + z^{2}} \Rightarrow φ (z) = \frac{e^{i α z^{n}}}{(z - i) (z + i)} t h e p o l e s o f φ a r e \bar{+} i

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t R e s (φ, i) = \frac{e^{i α i^{n}}}{2 i} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{i α i^{n}}}{2 i} = π e^{i α {(- 1)}^{\frac{n}{2}}} = π {c o s ({(- 1)}^{\frac{n}{2}} α) + i s i n ({(- 1)}^{\frac{n}{2}} α)} \Rightarrow

A_{n} = π c o s {{(- 1)}^{\frac{n}{2}} α} \Rightarrow A_{2 n} = π c o s {{(- 1)}^{n} α} a n d

A_{2 n + 1} = π c o s {i {(- 1)}^{n} α} b u t c o s z = \frac{e^{i z} + e^{- i z}}{2} \Rightarrow c o s (i x) = \frac{e^{- x} + e^{x}}{2} = c h (x)

(x \in R) \Rightarrow A_{2 n + 1} = π c h {{(- 1)}^{n} α} .