let-f-x-0-sin-xt-2-1-t-4-1-dt-1-find-a-explicit-form-of-f-x-2-let-g-x-0-t-2-cos-xt-2-1-t-4-1-dt-find-a-explicit-form-of-g-x-3-calculate-0-sin-

Question Number 57237 by maxmathsup by imad last updated on 31/Mar/19

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

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

2) l e t g (x) = \int_{0}^{\infty} \frac{t^{2} c o s ({x t}^{2} - 1)}{t^{4} + 1} d t

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

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

Commented by maxmathsup by imad last updated on 05/Apr/19

1) w e h a v e 2 f (x) = \int_{- \infty}^{+ \infty} \frac{s i n ({x t}^{2} - 1)}{t^{4} + 1} d t = I m (\int_{- \infty}^{+ \infty} \frac{e^{i ({x t}^{2} - 1)}}{t^{4} + 1} d t)

l e t φ (z) = \frac{e^{i ({x z}^{2} - 1)}}{z^{4} + 1} \Rightarrow φ (z) = \frac{e^{i ({x z}^{2} - 1)}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{i ({x z}^{2} - 1)}}{(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}} 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, 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^{i (x (i) - 1)}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{e^{- x} e^{- i}}{4 i e^{\frac{i π}{4}}} = \frac{e^{- x}}{4 i} e^{- (1 + \frac{π}{4}) i}

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^{i (- x i - 1)}}{- 2 e^{- \frac{i π}{4}} (- 2 i)}

= \frac{e^{x}}{4 i} e^{- i} e^{\frac{i π}{4}} = \frac{e^{x}}{4 i} e^{(- 1 + \frac{π}{4}) i} \Rightarrow

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

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

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

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

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

= \frac{π}{\sqrt{2}} {c o s (1) c h (x) + i c o s (1) s h (x) - i s i n (1) c h (x) + s i n (1) s h (x)} \Rightarrow

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

2) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2} c o s ({x t}^{2} - 1)}{t^{4} + 1} d t = g (x) \Rightarrow

g (x) = \frac{π}{2 \sqrt{2}} {c o s (1) c h (x) - s i n (1) s h (x)} .

Commented by maxmathsup by imad last updated on 05/Apr/19

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

\int_{0}^{\infty} \frac{t^{2} s i n (3 t^{2} - 1)}{t^{4} + 1} d t = g (3) = \frac{π}{2 \sqrt{2}} {c o s (1) c h (3) - s i n (1) s h (3)} .