calculate-cos-tx-1-x-4-dx-with-t-0-2-calculate-0-dx-1-x-4-

Question Number 34229 by abdo imad last updated on 03/May/18

c a l c u l a t e \int_{- \infty}^{\infty} \frac{c o s (t x)}{1 + x^{4}} d x w i t h t ⩾ 0

2) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{1 + x^{4}} .

Commented by math khazana by abdo last updated on 04/May/18

l e t p u t f (t) = \int_{- \infty}^{+ \infty} \frac{c o s (t x)}{1 + x^{4}} d x

f (t) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i t x}}{1 + x^{4}} d x) 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{e^{i t z}}{1 + z^{4}} w e h a v e

φ (z) = \frac{e^{i t z}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{i t z}}{(z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})}

= \frac{e^{i t z}}{(z - e^{i \frac{π}{4}}) (z + e^{i \frac{π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- i \frac{π}{4}})}

t h e p o l e s o f φ a r e e^{i \frac{π}{4}}, - e^{i \frac{π}{4}}, e^{- i \frac{π}{4}}, - e^{- i \frac{π}{4}}

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

l e t p (z) a l l p o l e s o f φ a r e r o o t s o f p a n d

p^{^{'}} (z) = 4 z^{3} \Rightarrow p^{^{'}} (z_{k}) = 4 z_{k}^{3} = \frac{4 z_{k}^{4}}{z_{k}} = \frac{- 4}{z_{k}}

R e s (φ, e^{i \frac{π}{4}}) = \frac{e^{i t e^{i \frac{π}{4}}}}{- 4} e^{i \frac{π}{4}} = - \frac{1}{4} e^{i \frac{π}{4}} e^{i t e^{i \frac{π}{4}}}

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

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

R e s (φ, - e^{- i \frac{π}{4}}) = \frac{1}{4} e^{- i \frac{π}{4}} e^{- {i t e}^{- i \frac{π}{4}}}

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

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

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

= \frac{i π}{2} e^{- \frac{t}{\sqrt{2}}} (- 2 i s i n (\frac{π}{4} + \frac{t}{2})) = π e^{- \frac{t}{\sqrt{2}}} s i n (\frac{π}{4} + \frac{t}{2})

f (t) = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow

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

Commented by math khazana by abdo last updated on 04/May/18

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

= \frac{π}{\sqrt{2}} = \frac{π \sqrt{2}}{2} \Rightarrow \int_{0}^{+ \infty} \frac{d x}{1 + x^{4}} = \frac{π \sqrt{2}}{4} .

Commented by math khazana by abdo last updated on 04/May/18

e r r o r a t t h e f i n a l l i n e s

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

= π e^{- \frac{t}{\sqrt{2}}} s i n (\frac{π}{4} + \frac{t}{\sqrt{2}}) \Rightarrow

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

Answered by Joel578 last updated on 03/May/18

\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{\sin (π a)}, 0 < a < 1

I = \int_{0}^{\infty} \frac{d x}{1 + x^{4}} (u = x^{4} \to d u = 4 x^{3} d x)

= \frac{1}{4} \int_{0}^{\infty} \frac{x^{- 3}}{1 + u} d u

= \frac{1}{4} \int_{0}^{\infty} \frac{u^{- \frac{3}{4}}}{1 + u} d u = \frac{1}{4} \int_{0}^{\infty} \frac{u^{\frac{1}{4} - 1}}{1 + u} d u

= \frac{1}{4} (\frac{π}{\sin (π / 4)})

= \frac{π \sqrt{2}}{4}

Answered by tanmay.chaudhury50@gmail.com last updated on 03/May/18

\underset{0}{\int^{\infty}} \frac{d x}{1 + x^{4}} = \underset{0}{\int^{\infty}} \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x

Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

\frac{1}{2} \underset{0}{\int^{\infty}} \frac{(1 + \frac{1}{x^{2}}) - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} d x

\frac{1}{2} \underset{0}{\int^{\infty}} \frac{(1 + \frac{1}{x^{2}})}{{(x - \frac{1}{x})}^{2} + 2} d x - \frac{1}{2} \underset{0}{\int^{\infty}} \frac{(1 - \frac{1}{x^{2}})}{{(x + \frac{1}{x})}^{2} - 2}

\frac{1}{2} \underset{0}{\int^{\infty}} \frac{{d t}_{1}}{t_{1}^{2} + {(\sqrt{2})}^{2}} - \frac{1}{2} \underset{0}{\int^{\infty}} \frac{{d t}_{2}}{t_{2}^{2} - {(\sqrt{2})}^{2}}

= \frac{1}{2 \sqrt{2}} ∣ {t a n}^{- 1} (\frac{t_{1}}{\sqrt{2}}) \underset{0}{\overset{\infty}{∣}} - \frac{1}{2} . \frac{1}{2 \sqrt{2}} ∣ l n ∣ \frac{\sqrt{2} - t_{2}}{\sqrt{2} + t_{2}} ∣

Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18

= \frac{1}{2 \sqrt{2}} . \frac{Π}{2} - \frac{1}{4 \sqrt{2}} . l n ∣ \frac{\frac{\sqrt{2}}{t_{2}} - 1}{\frac{\sqrt{2}}{t_{2}} + 1} ∣\to i t s v a l u e i s z e r o