dx-1-x-4-

Question Number 171371 by pticantor last updated on 13/Jun/22

\int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{4}} = ?

Answered by aleks041103 last updated on 13/Jun/22

s o l v e u s i n g c o m p l e x a n a l y s i s

w e i n r e g r a t e t h e c o m p l e x f u n c t i o n

\frac{1}{1 + z^{4}} o v e r t h e c o n t o u r Γ = \underset{r \to \infty}{l i m} Γ_{1} \cup Γ_{2} .

Γ_{1} := [- r, r] a n d Γ_{2} := {{r e}^{i t} ∣ t \in (0, π)}

\Rightarrow I = \oint_{Γ} \frac{d z}{1 + z^{4}} = \int_{Γ_{1}} \frac{d z}{1 + z^{4}} + \int_{Γ_{2}} \frac{d z}{1 + z^{4}}

\int_{Γ_{1}} \frac{d z}{1 + z^{4}} = \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{4}}

\int_{Γ_{2}} \frac{d z}{1 + z^{4}} = \underset{r \to \infty}{l i m} \int_{0}^{π} \frac{{i r e}^{i t} d t}{1 + r^{4} e^{4 i t}} =

= \underset{r \to \infty}{l i m} \frac{i r}{r^{4}} \int_{0}^{π} e^{- 3 i t} d t = 0

\Rightarrow A n s . = \oint_{Γ} \frac{d z}{1 + z^{4}} = 2 π i Σ R e s (z_{i})

1 + z^{4} = 0

z^{4} = e^{i π} = e^{i (π + 2 k π)} \Rightarrow z = e^{i (\frac{π}{4} + \frac{k π}{2})}

t h e r e f o r e w e h a v e

z_{1} = e^{i π / 4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}, I m (z_{1}) > 0

z_{2} = e^{3 i π / 4} = - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}, I m (z_{2}) > 0

z_{3} = e^{5 i π / 4} = - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}, I m (z_{3}) < 0

z_{4} = e^{7 i π / 4} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}, I m (z_{4}) < 0

1 + z^{4} = (z - z_{1}) (z - z_{2}) (z - z_{3}) (z - z_{4})

\Rightarrow w e h a v e p o l e s o f \frac{1}{1 + z^{4}} i n b o u n d s

o f Γ a t z_{1} a n d z_{2} a n d t h o s e p o l e s a r e

s i m p l e .

\Rightarrow R e s (z_{1}) = \underset{z \to z_{1}}{l i m} \frac{z - z_{1}}{1 + z^{4}} = \frac{1}{4 z_{1}^{3}} = \frac{1}{4} \frac{z_{1}}{z_{1}^{4}} = - \frac{1}{4} e^{i π / 4}

R e s (z_{2}) = \underset{z \to z_{2}}{l i m} \frac{z - z_{2}}{1 + z^{4}} = \frac{1}{4 z_{2}^{3}} = \frac{1}{4} \frac{z_{2}}{z_{2}^{4}} = - \frac{1}{4} {i e}^{i π / 2}

Σ R e s (z_{i}) = - \frac{1}{4} (1 + i) e^{i π / 4} =

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

= - \frac{\sqrt{2}}{4} e^{i π / 2} = - \frac{i}{2 \sqrt{2}}

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

Answered by Mathspace last updated on 14/Jun/22

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

= \frac{1}{2 i} \int_{- \infty}^{+ \infty} (\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}) d x

= \frac{1}{2 i} {\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} - c o n j (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i})}

= i m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i}) l e t f (z) = \frac{1}{z^{2} - i}

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

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

= 2 i π \times \frac{1}{2 e^{\frac{i π}{4}}} = i π e^{- \frac{i π}{4}}

= i π {\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i} = \frac{i π}{\sqrt{2}} + \frac{π}{\sqrt{2}} \Rightarrow

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

Answered by floor(10²Eta[1]) last updated on 14/Jun/22

\int_{- \infty}^{\infty} \frac{dx}{{(x^{2} + 1)}^{2} - {2 x}^{2}} = \int_{- \infty}^{\infty} \frac{dx}{(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)}

\frac{1 + i}{4} \int_{- \infty}^{\infty} \frac{dx}{x^{2} + \sqrt{2} x + 1} + \frac{1 - i}{4} \int_{- \infty}^{\infty} \frac{dx}{x^{2} - \sqrt{2} x + 1}

\frac{1 + i}{4} \int_{- \infty}^{\infty} \frac{dx}{{(x + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} + \frac{1 - i}{4} \int_{- \infty}^{\infty} \frac{dx}{{(x - \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}}

\frac{1 + i}{4} {(\sqrt{2} arctg (x \sqrt{2} + 1))}_{- \infty}^{\infty} + \frac{1 - i}{4} {(\sqrt{2} arctg (x \sqrt{2} - 1))}_{- \infty}^{\infty}

\frac{1 + i}{4} \sqrt{2} (\frac{π}{2} - (- \frac{π}{2})) + \frac{1 - i}{4} \sqrt{2} (\frac{π}{2} - (- \frac{π}{2}))

= \frac{π \sqrt{2} + i π \sqrt{2}}{4} + \frac{π \sqrt{2} - i π \sqrt{2}}{4} = \frac{π \sqrt{2}}{2} = \frac{π}{\sqrt{2}}