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Question Number 185104 by emmanuelson123 last updated on 17/Jan/23

	Answered by aba last updated on 17/Jan/23

	$π \ln (6 + 4 \sqrt{2})$
	Commented by emmanuelson123 last updated on 17/Jan/23
	Can you show your work, thank you!
	Answered by witcher3 last updated on 24/Jan/23
	$f(a)=∫_(−∞) ^∞ ((ln(x^4 +a^4 ))/((1+x^2 )))dx,a≥0 f(0)=8∫_0 ^∞ ((ln(x))/(1+x^2 ))dx=0 f′(a)=4a^3 ∫_(−∞) ^∞ (dx/((1+x^2 )(x^4 +a^4 ))) x^4 +a^4 =0⇒x=z_k =ae^(i(((1+2k))/4)π) ,k∈{0,1,2,3} z_0 =ae^(i(π/4)) ,z_1 =ae^((3iπ)/4) f′(a)=4a^3 .2iπRes(i,z_1 ,z_0 ) =8iπa^3 ((1/(2i(a^4 +1)))+(1/((z_1 ^2 +1)(4z_1 ^3 )))+(1/(z_0 ^2 +1)).(1/(4z_0 ^3 ))) =((4πa^3 )/(1+a^4 ))+((8iπa^3 )/((1+ia^2 )(4a^3 e^(3i(π/4)) ))).+((8iπa^3 )/((1−ia^2 )(4a^3 e^((iπ)/4) ))) =((π4a^3 )/(1+a^4 ))+2iπ((e^(−((3iπ)/4)) /(1+ia^2 ))+(e^(−i(π/4)) /(1−ia^2 ))) =π.((4a^3 )/(1+a^4 ))+2iπ(((e^(−((iπ)/4)) +e^((−3iπ)/4) +ia^2 (e^(−((iπ)/4)) −e^(−((3iπ)/4)) ))/(1+a^4 ))) =π.((4a^3 )/(1+a^4 ))+2iπ(((−i(√2)+ia^2 (√2))/(1+a^4 ))) =π(((4a^3 )/(1+a^4 ))+2(√2).(((1−a^2 )/(1+a^4 )))) f(a)=πln(2)+2π(√2)∫_0 ^1 ((1−a^2 )/(1+a^4 ))da πln(2)+2πcoth^− ((√2)) πln(2)+πln(((1+(√2))/( (√2)−1))) πln(3+2(√2))+πln(2)=πln(6+4(√2))$
	$f (a) = \int_{- \infty}^{\infty} \frac{\ln (x^{4} + a^{4})}{(1 + x^{2})} dx, a ⩾ 0$ $f (0) = 8 \int_{0}^{\infty} \frac{\ln (x)}{1 + x^{2}} dx = 0$ $f^{'} (a) = {4 a}^{3} \int_{- \infty}^{\infty} \frac{dx}{(1 + x^{2}) (x^{4} + a^{4})}$ $x^{4} + a^{4} = 0 \Rightarrow x = z_{k} = {ae}^{i \frac{(1 + 2 k)}{4} π}, k \in {0, 1, 2, 3}$ $z_{0} = {ae}^{i \frac{π}{4}}, z_{1} = {ae}^{\frac{3 i π}{4}}$ $f^{'} (a) = {4 a}^{3} . 2 i π Res (i, z_{1}, z_{0})$ $= 8 i π a^{3} (\frac{1}{2 i (a^{4} + 1)} + \frac{1}{(z_{1}^{2} + 1) ({4 z}_{1}^{3})} + \frac{1}{z_{0}^{2} + 1} . \frac{1}{{4 z}_{0}^{3}})$ $= \frac{4 π a^{3}}{1 + a^{4}} + \frac{8 i π a^{3}}{(1 + {ia}^{2}) ({4 a}^{3} e^{3 i \frac{π}{4}})} . + \frac{8 i π a^{3}}{(1 - {ia}^{2}) ({4 a}^{3} e^{\frac{i π}{4}})}$ $= \frac{π {4 a}^{3}}{1 + a^{4}} + 2 i π (\frac{e^{- \frac{3 i π}{4}}}{1 + {ia}^{2}} + \frac{e^{- i \frac{π}{4}}}{1 - {ia}^{2}})$ $= π . \frac{{4 a}^{3}}{1 + a^{4}} + 2 i π (\frac{e^{- \frac{i π}{4}} + e^{\frac{- 3 i π}{4}} + {ia}^{2} (e^{- \frac{i π}{4}} - e^{- \frac{3 i π}{4}})}{1 + a^{4}})$ $= π . \frac{{4 a}^{3}}{1 + a^{4}} + 2 i π (\frac{- i \sqrt{2} + {ia}^{2} \sqrt{2}}{1 + a^{4}})$ $= π (\frac{{4 a}^{3}}{1 + a^{4}} + 2 \sqrt{2} . (\frac{1 - a^{2}}{1 + a^{4}}))$ $f (a) = π \ln (2) + 2 π \sqrt{2} \int_{0}^{1} \frac{1 - a^{2}}{1 + a^{4}} da$ $π \ln (2) + 2 π \coth^{-} (\sqrt{2})$ $π \ln (2) + π \ln (\frac{1 + \sqrt{2}}{\sqrt{2} - 1})$ $π \ln (3 + 2 \sqrt{2}) + π \ln (2) = π \ln (6 + 4 \sqrt{2})$
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