let-I-n-x-0-t-sin-t-t-2-x-2-n-dt-1-find-a-relation-between-I-n-1-and-I-n-2-calculate-I-2-x-and-I-3-x-3-calculate-0-tsin-t-2-t-2-2-dt-

Question Number 36187 by prof Abdo imad last updated on 30/May/18

l e t I_{n} (x) = \int_{0}^{\infty} \frac{t s i n (t)}{{(t^{2} + x^{2})}^{n}} d t

1) f i n d a r e l a t i o n b e t w e e n I_{n + 1} a n d I_{n}

2) c a l c u l a t e I_{2} (x) a n d I_{3} (x)

3) c a l c u l a t e \int_{0}^{\infty} \frac{t s i n (t)}{{(2 + t^{2})}^{2}} d t

Commented by maxmathsup by imad last updated on 15/Aug/18

3) l e t A = \int_{0}^{\infty} \frac{t s i n t}{{(2 + t^{2})}^{2}} d t c h a n g e m e n t t = \sqrt{2} x g i v e

A = \int_{0}^{\infty} \frac{\sqrt{2} x s i n (\sqrt{2} x)}{4 (1 + x^{2})} \sqrt{2} d x = \frac{1}{2} \int_{0}^{\infty} \frac{x s i n (x \sqrt{2})}{{(x^{2} + 1)}^{2}} d x

= \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{x s i n (x \sqrt{2})}{{(x^{2} + 1)}^{2}} d x \Rightarrow

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

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

(d o u b l e s) 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) b u t

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

= {l i m}_{z \to i} {\frac{z e^{i z \sqrt{2}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{i z \sqrt{2}} + i \sqrt{2} z e^{i z \sqrt{2}}) {(z + i)}^{2} - 2 (z + i) {z e}^{i z \sqrt{2}}}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{(z + i) (e^{i z \sqrt{2}} + i \sqrt{2} z e^{i z \sqrt{2}}) - 2 z e^{i z \sqrt{2}}}{{(z + i)}^{3}}

= \frac{(2 i) (e^{- \sqrt{2}} - \sqrt{2} e^{- \sqrt{2}}) - 2 i e^{- \sqrt{2}}}{{(2 i)}^{3}}

\frac{- 2 \sqrt{2} i e^{- \sqrt{2}}}{- 8 i} = \frac{\sqrt{2}}{4} e^{- \sqrt{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{\sqrt{2}}{4} e^{- \sqrt{2}} = \frac{i π \sqrt{2}}{2} e^{- \sqrt{2}}

4 A = I m (\int_{- \infty}^{+ \infty} φ (z) d z) = \frac{π \sqrt{2}}{2} e^{- \sqrt{2}} \Rightarrow A = \frac{π \sqrt{2}}{8} e^{- \sqrt{2}} .

Commented by maxmathsup by imad last updated on 19/Aug/18

2) w e h a v e I_{2} (x) = \int_{0}^{\infty} \frac{t s i n t}{{(t^{2} + x^{2})}^{2}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t s i n t}{{(t^{2} + x^{2})}^{2}} d t \Rightarrow

2 I_{2} (x) = I m (\int_{- \infty}^{+ \infty} \frac{t e^{i t}}{{(t^{2} + x^{2})}^{2}} d t) = I m (A (x)) c h a n g e m e n t t = x u g i v e

A (x) = \int_{- \infty}^{+ \infty} \frac{x u e^{i x u}}{x^{4} {(1 + u^{2})}^{2}} x d u = \frac{1}{x^{2}} \int_{- \infty}^{+ \infty} \frac{u e^{i x u}}{{(u^{2} + 1)}^{2}} d u (w e s u p p o s e x > 0)

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{z e^{i x z}}{{(z^{2} + 1)}^{2}}

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

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

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

= {l i m}_{z \to i} {\frac{z e^{i x z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{i x z} + i x z e^{i x z}) {(z + i)}^{2} - 2 (z + i) z e^{i x z}}{{(z + i)}^{4}}

= {l i m}_{z \to i} \frac{(1 + i x z) e^{i x z} (z + i) - 2 z e^{i x z}}{{(z + i)}^{3}}

= \frac{(1 - x) e^{- x} (2 i) - 2 i e^{- x}}{{(2 i)}^{3}} = \frac{2 i (1 - x) e^{- x} - 2 i e^{- x}}{- 8 i}

= \frac{(2 - 2 x - 2) e^{- x}}{- 8} = \frac{x e^{- x}}{4} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{x e^{- x}}{4} = \frac{i π}{2} x e^{- x} \Rightarrow

2 I_{2} (x) = \frac{π x}{2} e^{- x} \Rightarrow I_{2} (x) = \frac{π x}{4} e^{- x} .

i f x < 0 x = - α w i t h α > 0 \Rightarrow I_{2} (x) = I_{2} (- α) = \int_{0}^{\infty} \frac{t s i n t}{{(t^{2} + α^{2})}^{2}} = I_{2} (α)

= \frac{π α}{4} e^{- α} = - \frac{π x}{4} e^{x} .

Commented by maxmathsup by imad last updated on 19/Aug/18

e r r o r a t t h e f i n a l l i n e w e h a v e A (x) = \frac{1}{x^{2}} \int_{- \infty}^{+ \infty} φ (z) d z

= \frac{1}{x^{2}} \frac{i π}{2} x e^{- x} = \frac{i π}{2 x} e^{- x} \Rightarrow I_{2} (x) = \frac{1}{2} I m (A (x)) = \frac{1}{2} \frac{π}{2 x} e^{- x} \Rightarrow

I_{2} (x) = \frac{π}{4 x} e^{- x} w i t h x > 0 a n d i f x < 0

I_{2} (x) = - \frac{π}{4 x} e^{x} .