let-f-x-0-x-sinx-a-2-x-4-dx-with-a-gt-0-1-find-a-explicit-form-of-f-a-2-find-g-a-0-xsinx-a-2-x-4-2-dx-3-find-the-value-of-0-x-sinx-x-4-1-d

Question Number 44466 by maxmathsup by imad last updated on 29/Sep/18

l e t f (x) = \int_{0}^{\infty} \frac{x s i n x}{a^{2} + x^{4}} d x w i t h a > 0

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

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

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

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

Commented by abdo.msup.com last updated on 29/Sep/18

f (a) = \int_{0}^{\infty} \frac{x s i n x}{a^{2} + x^{4}} d x w i t h a > 0 .

Commented by maxmathsup by imad last updated on 03/Oct/18

1) w e h a v e f (a) = \int_{0}^{\infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x \Rightarrow 2 f (a) = \int_{- \infty}^{+ \infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x

= I m (\int_{- \infty}^{+ \infty} \frac{x e^{i x}}{a^{2} + x^{4}} d x) c h a n g e m e n t x = \sqrt{a} t g i v e

I = \int_{- \infty}^{+ \infty} \frac{x e^{i x}}{a^{2} + x^{4}} d x = \int_{- \infty}^{+ \infty} \frac{\sqrt{a} t e^{i \sqrt{a} t}}{a^{2} + a^{2} t^{4}} \sqrt{a} d t = \frac{1}{a} \int_{- \infty}^{+ \infty} \frac{t e^{i \sqrt{a} t}}{t^{4} + 1} d t \Rightarrow

a I = \int_{- \infty}^{+ \infty} \frac{t e^{i \sqrt{a} t}}{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 \sqrt{a} z}}{z^{4} + 1} \Rightarrow φ (z) = \frac{z e^{i \sqrt{a} z}}{(z^{2} - i) (z^{2} + i)} = \frac{z e^{i \sqrt{a} 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}}

\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}})} b u t

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

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

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

= \frac{π}{2} e^{i \frac{\sqrt{a}}{\sqrt{2}}} {e^{\frac{\sqrt{a}}{\sqrt{2}}} + e^{- \frac{\sqrt{a}}{\sqrt{2}}}} = π c h (\frac{\sqrt{a}}{\sqrt{2}}) {c o s (\frac{\sqrt{a}}{\sqrt{2}}) + i s i n (\frac{\sqrt{a}}{\sqrt{2}})} \Rightarrow

I = \frac{π}{a} c h (\frac{\sqrt{a}}{\sqrt{2}}) {c o s (\frac{\sqrt{a}}{\sqrt{2}}) + i s i n (\frac{\sqrt{a}}{\sqrt{2}})} \Rightarrow 2 f (a) = \frac{π}{a} c h (\frac{\sqrt{a}}{\sqrt{2}}) s i n (\frac{\sqrt{a}}{\sqrt{2}}) \Rightarrow

f (a) = \frac{π}{2 a} c h (\frac{\sqrt{a}}{\sqrt{2}}) s i n (\frac{\sqrt{a}}{\sqrt{2}}) w i t h a > 0 .

Commented by maxmathsup by imad last updated on 04/Oct/18

2) w e h a v e f (a) = \int_{0}^{\infty} \frac{x s i n (x)}{a^{2} + x^{4}} d x \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{2 a x s i n x}{{(a^{2} + x^{4})}^{2}} d x \Rightarrow

\int_{0}^{\infty} \frac{x s i n x}{{(a^{2} + x^{4})}^{2}} = - \frac{1}{2 a} f^{^{'}} (a) a n d f (a) i s k n o w n .

Commented by maxmathsup by imad last updated on 04/Oct/18

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