let-f-x-0-sin-t-2-x-2-t-2-2-dt-with-x-gt-0-1-determine-a-explicit-form-for-f-x-2-find-also-g-x-0-sin-t-2-x-2-t-2-3-dt-3-give-f-n-x-at-form-of-int

Question Number 67744 by mathmax by abdo last updated on 31/Aug/19

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

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

2) f i n d a l s o g (x) = \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{3}} d t

3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l a n d c a l c u l a t e f^{(n)} (1) .

4) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{2}} d t a n d \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{3}} d t

Commented by mathmax by abdo last updated on 31/Aug/19

1) w e h a v e f (x) = \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{2}} d t c h a 7 g e m e n t t = x u g i v e

f (x) = \int_{0}^{\infty} \frac{s i n (x^{2} u^{2})}{x^{4} {(1 + u^{2})}^{2}} x d u = \frac{1}{x^{3}} \int_{0}^{\infty} \frac{s i n (x^{2} u^{2})}{{(u^{2} + 1)}^{2}} d u

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

W (z) = \frac{e^{{i x}^{2} z^{2}}}{{(z^{2} + 1)}^{2}} \Rightarrow W (z) = \frac{e^{{i x}^{2} z^{2}}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f W 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} W (z) d z = 2 i π R e s (W, i) a n d

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

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

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

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

= \frac{(- 4 {i x}^{2} - 2) e^{- {i x}^{2}}}{- 8 i} = \frac{(2 {i x}^{2} - 1) e^{- {i x}^{2}}}{4 i} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times \frac{(2 {i x}^{2} - 1) e^{- {i x}^{2}}}{4 i} = \frac{π}{2} (2 {i x}^{2} - 1) e^{- {i x}^{2}}

= \frac{π}{2} (2 {i x}^{2} - 1) (c o s (x^{2}) - i s i n (x^{2}))

= \frac{π}{2} {2 {i x}^{2} c o s (x^{2}) + 2 x^{2} s i n (x^{2}) - c o s (x^{2}) + i s i n (x^{2})}

= \frac{π}{2} {2 x^{2} s i n (x^{2}) - c o s (x^{2}) + i (2 x^{2} c o s (x^{2}) + s i n (x^{2})} \Rightarrow

f (x) = \frac{2 x^{2} c o s (x^{2}) + s i n (x^{2})}{2 x^{3}} .

Commented by mathmax by abdo last updated on 31/Aug/19

s o r r y f (x) = \frac{π}{4 x^{3}} {2 x^{2} c o s (x^{2}) + s i n (x^{2})}

Commented by mathmax by abdo last updated on 31/Aug/19

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{\infty} \frac{- 2 (2 x) (x^{2} + t^{2})}{{(x^{2} + t^{2})}^{4}} s i n (t^{2}) d t = 4 x \int_{0}^{\infty} \frac{s i n (t^{2})}{{(x^{2} + t^{2})}^{3}}

= 4 x g (x) \Rightarrow g (x) = \frac{1}{4 x} f^{^{'}} (x) w e h a v e

f (x) = \frac{π}{4 x^{3}} {2 x^{2} c o s (x^{2}) + s i n (x^{2})} \Rightarrow

f^{^{'}} (x) = \frac{π}{4} (\frac{- 3 x^{2}}{x^{6}}) {2 x^{2} c o s (x^{2}) + s i n (x^{2})}

+ \frac{π}{4 x^{3}} {4 x c o s (x^{2}) - 4 x^{3} s i n (x^{2}) + 2 x c o s (x^{2})} = \dots

Commented by mathmax by abdo last updated on 31/Aug/19

4) \int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{2}} d t = f (1) = \frac{π}{4} {2 c o s (1) + s i n (1)}

= \frac{π}{2} c o s (1) + \frac{π}{4} s i n (1) .

\int_{0}^{\infty} \frac{s i n (t^{2})}{{(1 + t^{2})}^{3}} d t = g (1) r e s t t o c a l c u l s t e g (1) \dots