calculate-0-xsin-x-1-x-4-dx-with-real-

Question Number 70237 by mathmax by abdo last updated on 02/Oct/19

c a l c u l a t e \int_{0}^{\infty} \frac{x s i n (α x)}{1 + x^{4}} d x w i t h α r e a l

Commented by mind is power last updated on 02/Oct/19

n i c e i d e l e a t m y p o s t m i s t a c k [

\frac{z s i n (z)}{1 + z^{4}} w h e n ∣ z ∣\to \infty i s n o t d e f i n d

z = {r e}^{i a} a \in [0, π [

s i n (z) = \frac{e^{{i r e}^{i a}} - e^{{i r e}^{i a}}}{2 i}

t h e f a c t o r e^{- {i r e}^{i a}} = e^{- i r v o s (a)} . e^{r s i n (a)} ∣ e^{- {i r e}^{i a}} ∣= e^{r s i n (a)} w i t c h i s g e t b i g a s w e w a n t

\Rightarrow l i m z f (z) ⇏ 0 j o r d a n l e m m a d o n t n o t a p p l i e

\int_{- \infty}^{+ \infty} f (z) d z \neq 2 i π r e s i d (\dots . .)

i h a v e d o n e i t q u i c k l y w i t h o u t s e e i n g w h a t g o i n g o n S o r r y s i r

Commented by mathmax by abdo last updated on 02/Oct/19

l e t A = \int_{0}^{\infty} \frac{x s i n (α x)}{x^{4} + 1} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{x s i n (α x)}{x^{4} + 1} d x

= I m (\int_{- \infty}^{+ \infty} \frac{x e^{i α x}}{x^{4} + 1} d x) l e t φ (z) = \frac{z e^{i α z}}{z^{4} + 1} p o l e s o f φ ?

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

= \frac{z e^{i α z}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} s o t h e p o l e s o f φ a r e

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

R e s (φ, e^{\frac{i π}{4}}) = {l i m}_{z \to e^{\frac{i π}{4}}} (z - e^{\frac{i π}{4}}) φ (z)

= \frac{e^{\frac{i π}{4}} e^{i α e^{\frac{i π}{4}}}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{e^{i α (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}{4 i} = \frac{1}{4 i} e^{\frac{i α}{\sqrt{2}}} e^{- \frac{α}{\sqrt{2}}}

R e s (φ, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{- \frac{i π}{4}}) φ (z)

= \frac{- e^{- \frac{i π}{4}} e^{i α (- e^{- \frac{i π}{4}})}}{(- 2 i) (- 2 e^{- \frac{i π}{4}})} = \frac{1}{4 i} e^{- i α (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})} = - \frac{1}{4 i} e^{- \frac{i α}{\sqrt{2}}} e^{- \frac{α}{\sqrt{2}}}

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{4 i} e^{- \frac{α}{\sqrt{2}}} e^{\frac{i α}{\sqrt{2}}} - \frac{1}{4 i} e^{- \frac{α}{\sqrt{2}}} e^{- \frac{i α}{\sqrt{2}}}}

= \frac{π}{2} e^{- \frac{α}{\sqrt{2}}} {2 i s i n (\frac{α}{\sqrt{2}})} = i π e^{- \frac{α}{\sqrt{2}}} s i n (\frac{α}{\sqrt{2}}) \Rightarrow

2 A = π e^{- \frac{α}{\sqrt{2}}} s i n (\frac{α}{\sqrt{2}}) \Rightarrow A = \frac{π}{2} e^{- \frac{α}{\sqrt{2}}} s i n (\frac{α}{\sqrt{2}}) .

Commented by mathmax by abdo last updated on 03/Oct/19

n e v e r m i n d s i r .