let-f-n-a-sin-x-n-x-2-a-2-dx-with-a-positif-real-not-0-and-n-from-N-1-find-a-explicit-form-of-f-a-2-calculate-g-n-a-sin-x-n-x-2-a-2-2-dx

Question Number 56699 by maxmathsup by imad last updated on 21/Mar/19

l e t f_{n} (a) = \int_{- \infty}^{\infty} \frac{s i n (x^{n})}{x^{2} + a^{2}} d x w i t h a p o s i t i f r e a l n o t 0 a n d n f r o m N

1) f i n d a e x p l i c i t f o r m o f f (a)

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

3) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{x^{2} + 4} d x a n d \int_{- \infty}^{+ \infty} \frac{s i n (x^{2})}{x^{2} + 9} d x

4) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{{(x^{2} + 4)}^{2}} d x .

Commented by maxmathsup by imad last updated on 23/Mar/19

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

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

r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i a)

R e s (φ, i a) = {l i m}_{z \to i a} (z - i a) φ (z) = \frac{e^{i {(i a)}^{n}}}{2 i a} = \frac{e^{i^{n + 1} a^{n}}}{2 i a} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{i^{n + 1} a^{n}}}{2 i a} = \frac{π}{a} e^{i^{n + 1} a^{n}}

i f n i s e v e n n = 2 p \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{a} e^{i^{2 p + 1} a^{n}} = \frac{π}{a} e^{i {(- 1)}^{p} a^{2 p}}

= \frac{π}{a} {c o s ({(- 1)}^{p} a^{2 p} + i s i n ({(- 1)}^{p} a^{2 p}} \Rightarrow f_{n} (a) = \frac{π}{a} s i n ({(- 1)}^{p} a^{2 p}) (n = 2 p)

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

f_{n} (a) = 0

2) w e h a v e b y d e r i v a t i o n f_{n}^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{\partial}{\partial a} (\frac{s i n (x^{n})}{x^{2} + a^{2}}) d x

= - \int_{- \infty}^{+ \infty} \frac{2 a s i n (x^{n})}{{(x^{2} + a^{2})}^{2}} d x = - 2 a g_{n} (a) \Rightarrow g_{n} (a) = - \frac{1}{2 a} f_{n}^{^{'}} (a)

g_{2 p} (a) = - \frac{1}{2 a} f_{2 p}^{^{'}} (a) = - \frac{1}{2 a} {- \frac{π}{a^{2}} s i n {{(- 1)}^{p} a^{2 p}} + \frac{π}{a} (2 p) {(- 1)}^{p} a^{2 p - 1} c o s {{(- 1)}^{p} a^{2 p}}

= \frac{π}{2 a^{3}} s i n {{(- 1)}^{p} a^{2 p}} + \frac{p π}{a^{2}} {(- 1)}^{p + 1} a^{2 p - 1} c o s {{(- 1)}^{p} a^{2 p}}

g_{2 p + 1} (a) = 0 d u e t o f_{2 p + 1} (a) = 0 .

3) \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{x^{2} + 4} d x = f_{3} (2) = 0 b e c a u s e 3 i s o d d .

\int_{- \infty}^{+ \infty} \frac{s i n (x^{2})}{x^{2} + 9} = f_{2} (3) = \frac{π}{3} s i n (- 9) = - \frac{π}{3} s i n (9) .

Commented by maxmathsup by imad last updated on 23/Mar/19

4) \int_{- \infty}^{+ \infty} \frac{s i n (x^{3})}{{(x^{2} + 4)}^{2}} d x = g_{3} (2) = 0 b e c a u s e 3 i s o d d .