find ∫_0 ^∞ ((ln^2 x)/(1+x^4 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 205151 by mathzup last updated on 10/Mar/24

$f i n d \int_{0}^{\infty} \frac{{l n}^{2} x}{1 + x^{4}} d x$
	Answered by Berbere last updated on 12/Mar/24
	$Ω=∫_(−∞) ^0 ((ln^2 (x))/(1+x^4 ))dx+∫_0 ^∞ ((ln^2 (x))/(1+x^4 ))dx=∫_0 ^∞ ((ln^2 (x)+(ln(x)+iπ)^2 )/(1+x^4 )) =2∫_0 ^∞ ((ln^2 (x))/(1+x^4 ))dx−π^2 ∫_0 ^∞ (dx/(1+x^4 ))+2iπ∫_0 ^∞ ((ln(x))/(1+x^4 )) ∫_0 ^∞ ((ln^2 (x))/(1+x^4 ))=(1/2)(Re(Ω)+π^2 ∫_0 ^∞ (dx/(1+x^4 ))) ∫_0 ^∞ (dx/(1+x^4 ))=∫_0 ^∞ (y^(−(3/4)) /(4(1+y)))=(1/4)β((1/4),(3/4))=(π/(4sin((π/4))))=(π/(2(√2))) C={z∈C∣Re(z)>0} ∫_C ((ln^2 (z))/(1+z^4 ))dx=2iπRes(f,e^(i(π/4)) ,e^(3i(π/4)) )=Ω =2iπ.((((((iπ)/4))^2 )/(4e^(3((iπ)/4)) ))+(((((3iπ)/4))^2 )/(4e^(9i(π/4)) )))=i(π/2)((π^2 /(16))e^(i(π/4)) −((9π^2 )/(16))e^(−((iπ)/4)) ) =(π^3 /(32))(−4i(√2)−5(√2));∫_0 ^∞ ((ln^2 (x))/(1+x^4 ))=−((5(√2))/(64))π^3 +(π^3 /(4(√2))) =((3(√2))/(64))π^3$
	$Ω = \int_{- \infty}^{0} \frac{{l n}^{2} (x)}{1 + x^{4}} d x + \int_{0}^{\infty} \frac{{l n}^{2} (x)}{1 + x^{4}} d x = \int_{0}^{\infty} \frac{{l n}^{2} (x) + {(l n (x) + i π)}^{2}}{1 + x^{4}}$ $= 2 \int_{0}^{\infty} \frac{{l n}^{2} (x)}{1 + x^{4}} d x - π^{2} \int_{0}^{\infty} \frac{d x}{1 + x^{4}} + 2 i π \int_{0}^{\infty} \frac{l n (x)}{1 + x^{4}}$ $\int_{0}^{\infty} \frac{{l n}^{2} (x)}{1 + x^{4}} = \frac{1}{2} (R e (Ω) + π^{2} \int_{0}^{\infty} \frac{d x}{1 + x^{4}})$ $\int_{0}^{\infty} \frac{d x}{1 + x^{4}} = \int_{0}^{\infty} \frac{y^{- \frac{3}{4}}}{4 (1 + y)} = \frac{1}{4} β (\frac{1}{4}, \frac{3}{4}) = \frac{π}{4 s i n (\frac{π}{4})} = \frac{π}{2 \sqrt{2}}$ $C = {z \in C ∣ R e (z) > 0}$ $\int_{C} \frac{{l n}^{2} (z)}{1 + z^{4}} d x = 2 i π R e s (f, e^{i \frac{π}{4}}, e^{3 i \frac{π}{4}}) = Ω$ $= 2 i π . (\frac{{(\frac{i π}{4})}^{2}}{4 e^{3 \frac{i π}{4}}} + \frac{{(\frac{3 i π}{4})}^{2}}{4 e^{9 i \frac{π}{4}}}) = i \frac{π}{2} (\frac{π^{2}}{16} e^{i \frac{π}{4}} - \frac{9 π^{2}}{16} e^{- \frac{i π}{4}})$ $= \frac{π^{3}}{32} (- 4 i \sqrt{2} - 5 \sqrt{2}); \int_{0}^{\infty} \frac{{l n}^{2} (x)}{1 + x^{4}} = - \frac{5 \sqrt{2}}{64} π^{3} + \frac{π^{3}}{4 \sqrt{2}}$ $= \frac{3 \sqrt{2}}{64} π^{3}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum