let-f-x-0-tsin-tx-1-t-4-dt-with-x-gt-0-1-find-a-explicit-form-of-f-x-2-find-the-value-of-0-tsin-2t-1-t-4-dt-

Question Number 53785 by maxmathsup by imad last updated on 25/Jan/19

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

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

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

Commented by maxmathsup by imad last updated on 26/Jan/19

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

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}}

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

R e s (φ, - e^{- \frac{i π}{4}}) = \frac{- e^{- \frac{i π}{4}} e^{i x (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}{(- 2 {i e}^{- \frac{i π}{4}}) (- 2 i)} = - \frac{1}{4 i} e^{- \frac{x}{\sqrt{2}}} e^{- \frac{i x}{\sqrt{2}}} \Rightarrow

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

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

★ f (x) = \frac{π}{2} e^{- \frac{x}{\sqrt{2}}} s i n (\frac{x}{\sqrt{2}}) ★

2) t h i s i n t e g r a l i s a s p a c i a l c a s e

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