∫_(−∞) ^(+∞) (dx/(1+x^4 ))=?


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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
	$solve using complex analysis we inregrate the complex function (1/(1+z^4 )) over the contour Γ=lim_(r→∞) Γ_1 ∪Γ_2 . Γ_1 :=[−r,r] and Γ_2 :={re^(it) ∣t∈(0,π)} ⇒I=∮_Γ (dz/(1+z^4 ))=∫_Γ_1 (dz/(1+z^4 ))+∫_Γ_2 (dz/(1+z^4 )) ∫_Γ_1 (dz/(1+z^4 ))=∫_(−∞) ^(+∞) (dx/(1+x^4 )) ∫_Γ_2 (dz/(1+z^4 ))=lim_(r→∞) ∫_0 ^π ((ire^(it) dt)/(1+r^4 e^(4it) ))= =lim_(r→∞) ((ir)/r^4 )∫_0 ^π e^(−3it) dt=0 ⇒Ans.=∮_Γ (dz/(1+z^4 ))=2πiΣRes(z_i ) 1+z^4 =0 z^4 =e^(iπ) =e^(i(π+2kπ)) ⇒z=e^(i((π/4)+((kπ)/2))) therefore we have z_1 =e^(iπ/4) =(1/( (√2)))+(i/( (√2))),Im(z_1 )>0 z_2 =e^(3iπ/4) =−(1/( (√2)))+(i/( (√2))),Im(z_2 )>0 z_3 =e^(5iπ/4) =−(1/( (√2)))−(i/( (√2))),Im(z_3 )<0 z_4 =e^(7iπ/4) =(1/( (√2)))−(i/( (√2))),Im(z_4 )<0 1+z^4 =(z−z_1 )(z−z_2 )(z−z_3 )(z−z_4 ) ⇒we have poles of (1/(1+z^4 )) in bounds of Γ at z_1 and z_2 and those poles are simple. ⇒Res(z_1 )=lim_(z→z_1 ) ((z−z_1 )/(1+z^4 ))=(1/(4z_1 ^3 ))=(1/4) (z_1 /z_1 ^4 )=−(1/4)e^(iπ/4) Res(z_2 )=lim_(z→z_2 ) ((z−z_2 )/(1+z^4 ))=(1/(4z_2 ^3 ))=(1/4) (z_2 /z_2 ^4 )=−(1/4)ie^(iπ/2) ΣRes(z_i )=−(1/4)(1+i)e^(iπ/4) = =−((√2)/4)((1/( (√2)))+(i/( (√2))))e^(iπ/4) = =−((√2)/4)e^(iπ/2) =−(i/(2(√2))) ⇒∫_(−∞) ^(+∞) (dx/(1+x^4 ))=2πi(−(i/(2(√2))))=(π/( (√2)))$
	$s o l v e u s i n g c o m p l e x a n a l y s i s$ $w e i n r e g r a t e t h e c o m p l e x f u n c t i o n$ $\frac{1}{1 + z^{4}} o v e r t h e c o n t o u r Γ = \underset{r \to \infty}{l i m} Γ_{1} \cup Γ_{2} .$ $Γ_{1} := [- r, r] a n d Γ_{2} := {{r e}^{i t} ∣ t \in (0, π)}$ $\Rightarrow I = \oint_{Γ} \frac{d z}{1 + z^{4}} = \int_{Γ_{1}} \frac{d z}{1 + z^{4}} + \int_{Γ_{2}} \frac{d z}{1 + z^{4}}$ $\int_{Γ_{1}} \frac{d z}{1 + z^{4}} = \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{4}}$ $\int_{Γ_{2}} \frac{d z}{1 + z^{4}} = \underset{r \to \infty}{l i m} \int_{0}^{π} \frac{{i r e}^{i t} d t}{1 + r^{4} e^{4 i t}} =$ $= \underset{r \to \infty}{l i m} \frac{i r}{r^{4}} \int_{0}^{π} e^{- 3 i t} d t = 0$ $\Rightarrow A n s . = \oint_{Γ} \frac{d z}{1 + z^{4}} = 2 π i Σ R e s (z_{i})$ $1 + z^{4} = 0$ $z^{4} = e^{i π} = e^{i (π + 2 k π)} \Rightarrow z = e^{i (\frac{π}{4} + \frac{k π}{2})}$ $t h e r e f o r e w e h a v e$ $z_{1} = e^{i π / 4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}, I m (z_{1}) > 0$ $z_{2} = e^{3 i π / 4} = - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}, I m (z_{2}) > 0$ $z_{3} = e^{5 i π / 4} = - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}, I m (z_{3}) < 0$ $z_{4} = e^{7 i π / 4} = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}, I m (z_{4}) < 0$ $1 + z^{4} = (z - z_{1}) (z - z_{2}) (z - z_{3}) (z - z_{4})$ $\Rightarrow w e h a v e p o l e s o f \frac{1}{1 + z^{4}} i n b o u n d s$ $o f Γ a t z_{1} a n d z_{2} a n d t h o s e p o l e s a r e$ $s i m p l e .$ $\Rightarrow R e s (z_{1}) = \underset{z \to z_{1}}{l i m} \frac{z - z_{1}}{1 + z^{4}} = \frac{1}{4 z_{1}^{3}} = \frac{1}{4} \frac{z_{1}}{z_{1}^{4}} = - \frac{1}{4} e^{i π / 4}$ $R e s (z_{2}) = \underset{z \to z_{2}}{l i m} \frac{z - z_{2}}{1 + z^{4}} = \frac{1}{4 z_{2}^{3}} = \frac{1}{4} \frac{z_{2}}{z_{2}^{4}} = - \frac{1}{4} {i e}^{i π / 2}$ $Σ R e s (z_{i}) = - \frac{1}{4} (1 + i) e^{i π / 4} =$ $= - \frac{\sqrt{2}}{4} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) e^{i π / 4} =$ $= - \frac{\sqrt{2}}{4} e^{i π / 2} = - \frac{i}{2 \sqrt{2}}$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{4}} = 2 π i (- \frac{i}{2 \sqrt{2}}) = \frac{π}{\sqrt{2}}$
	Answered by Mathspace last updated on 14/Jun/22
	$∫_(−∞) ^(+∞) (dx/(x^4 +1))=∫_(−∞) ^(+∞) (dx/((x^2 −i)(x^2 +i))) =(1/(2i))∫_(−∞) ^(+∞) ((1/(x^2 −i))−(1/(x^2 +i)))dx =(1/(2i)){∫_(−∞) ^(+∞) (dx/(x^2 −i))−conj(∫_(−∞) ^(+∞) (dx/(x^2 −i)))} =im(∫_(−∞) ^(+∞) (dx/(x^2 −i))) let f(z)=(1/(z^2 −i)) ⇒f(z)=(1/((z−e^((iπ)/4) )(z+e^((iπ)/4) ))) ∫_(−∞) ^(+∞) f(z)dz=2iπRe(f,e^((iπ)/4) ) =2iπ×(1/(2e^((iπ)/4) ))=iπe^(−((iπ)/4)) =iπ{(1/( (√2)))−(1/( (√2)))i}=((iπ)/( (√2)))+(π/( (√2))) ⇒ ∫_(−∞) ^(+∞) (dx/(1+x^4 ))=(π/( (√2)))$
	$\int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + 1} = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - i) (x^{2} + i)}$ $= \frac{1}{2 i} \int_{- \infty}^{+ \infty} (\frac{1}{x^{2} - i} - \frac{1}{x^{2} + i}) d x$ $= \frac{1}{2 i} {\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i} - c o n j (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i})}$ $= i m (\int_{- \infty}^{+ \infty} \frac{d x}{x^{2} - i}) l e t f (z) = \frac{1}{z^{2} - i}$ $\Rightarrow f (z) = \frac{1}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} f (z) d z = 2 i π R e (f, e^{\frac{i π}{4}})$ $= 2 i π \times \frac{1}{2 e^{\frac{i π}{4}}} = i π e^{- \frac{i π}{4}}$ $= i π {\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i} = \frac{i π}{\sqrt{2}} + \frac{π}{\sqrt{2}} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{4}} = \frac{π}{\sqrt{2}}$
	Answered by floor(10²Eta[1]) last updated on 14/Jun/22

	$\int_{- \infty}^{\infty} \frac{dx}{{(x^{2} + 1)}^{2} - {2 x}^{2}} = \int_{- \infty}^{\infty} \frac{dx}{(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)}$ $\frac{1 + i}{4} \int_{- \infty}^{\infty} \frac{dx}{x^{2} + \sqrt{2} x + 1} + \frac{1 - i}{4} \int_{- \infty}^{\infty} \frac{dx}{x^{2} - \sqrt{2} x + 1}$ $\frac{1 + i}{4} \int_{- \infty}^{\infty} \frac{dx}{{(x + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} + \frac{1 - i}{4} \int_{- \infty}^{\infty} \frac{dx}{{(x - \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}}$ $\frac{1 + i}{4} {(\sqrt{2} arctg (x \sqrt{2} + 1))}_{- \infty}^{\infty} + \frac{1 - i}{4} {(\sqrt{2} arctg (x \sqrt{2} - 1))}_{- \infty}^{\infty}$ $\frac{1 + i}{4} \sqrt{2} (\frac{π}{2} - (- \frac{π}{2})) + \frac{1 - i}{4} \sqrt{2} (\frac{π}{2} - (- \frac{π}{2}))$ $= \frac{π \sqrt{2} + i π \sqrt{2}}{4} + \frac{π \sqrt{2} - i π \sqrt{2}}{4} = \frac{π \sqrt{2}}{2} = \frac{π}{\sqrt{2}}$
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