let-f-a-dx-x-2-1-a-e-ix-with-a-gt-0-1-find-a-explicit-form-of-f-a-2-determine-also-g-a-dx-x-2-1-a-e-ix-2-3-let-I-Re-dx-

Question Number 67542 by mathmax by abdo last updated on 28/Aug/19

l e t f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (a + e^{i x})} w i t h a > 0

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

2) d e t e r m i n e a l s o g (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) {(a + e^{i x})}^{2}}

3) l e t I = R e (\int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (2 + e^{i x})}) a n d J = I m (\int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (2 + e^{x})})

d e t e r m i n e I a n d J a n d i t s v a l u e s .

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

l e t c h a n g e x = t a n u

f (a) = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d u}{h (u)} w i t h h (u) = a + e^{i t a n u}

g h a s o n l y o n e r o o t o n [\frac{- π}{2}; \frac{π}{2}] l e t n a m e d i t θ

W h e n u s i n g t h e R T

f (a) = 2 π i R e s (\frac{1}{h}, θ) = \frac{2 i π}{h^{'} (θ)} = \frac{2 i π}{i (1 + {t a n}^{2} θ) e^{i t a n θ}}

θ r o o t o f h \Rightarrow e^{i t a n θ} = - a a n d t a n θ = π - i l n a

S o f (a) = \frac{2 π}{- a (1 + π^{2} + {(l n a)}^{2} - 2 π i l n a)}

w e c a n a s c e r t a i n t h a t g (a) = \frac{d f}{d a}

W h e n s t a t i n g x (a) = 1 + π^{2} + {(l n a)}^{2} a n d y (a) = 2 π l n a

R e (f (a)) = - \frac{2 π}{a} \frac{x (a)}{x^{2} (a) + y^{2} (a)} a n d I m (f (a)) = \frac{- 2 π}{a} \frac{y (a)}{x^{2} (a) + y^{2} (a)}

Commented by mathmax by abdo last updated on 29/Aug/19

t h a n k y o u s i r .

Commented by mathmax by abdo last updated on 29/Aug/19

1) l e t c o n s i d e r t b e c o m p l e x f u n c t i o n W (z) = \frac{1}{(z^{2} + 1) (a + e^{i z})}

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

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

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

f (a) = \frac{π e}{a e + 1} w i t h a > 0

2) w e h a v e f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) {(a + e^{i x})}^{2}} = - g (a) \Rightarrow

g (a) = - f^{^{'}} (a) b u t f (a) = \frac{π e}{a e + 1} \Rightarrow f^{^{'}} (a) = - \frac{π e^{2}}{{(a e + 1)}^{2}} \Rightarrow

g (a) = \frac{π e^{2}}{{(a e + 1)}^{2}} .

Commented by mathmax by abdo last updated on 29/Aug/19

3) w e h a v e f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (a + c o s x + i s i n x)}

= \int_{- \infty}^{+ \infty} \frac{a + c o s x - i s i n x}{(x^{2} + 1) ({(a + c o s x)}^{2} + {s i n}^{2} x)} d x

= \int_{- \infty}^{+ \infty} \frac{a + c o s x}{(x^{2} + 1) {a^{2} + 2 a c o s x + 1)}} d x - i \int_{- \infty}^{+ \infty} \frac{s i n x}{(x^{2} + 1) {a^{2} + 2 a c o s x + 1}} d x

\Rightarrow R e (f (a)) = \int_{- \infty}^{+ \infty} \frac{a + c o s x}{(x^{2} + 1) {a^{2} + 2 a c o s x + 1}} d x a n d

I m (f (a)) = \int_{- \infty}^{+ \infty} \frac{s i n x d x}{(x^{2} + 1) {a^{2} + 2 a c o s x + 1}} = 0 b e c a u s e t h e

f u n c t i o n x \to \frac{s i n x}{(x^{2} + 1) {a^{2} + 2 a c o s x + 1}} i s o d d .