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Question-185104

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Question Number 185104 by emmanuelson123 last updated on 17/Jan/23
Answered by aba last updated on 17/Jan/23
πln(6+4(√2))
$$\pi\mathrm{ln}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$
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))
$$\mathrm{f}\left(\mathrm{a}\right)=\int_{−\infty} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx},\mathrm{a}\geqslant\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{8}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=\mathrm{4a}^{\mathrm{3}} \int_{−\infty} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} \right)} \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} =\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{z}_{\mathrm{k}} =\mathrm{ae}^{\mathrm{i}\frac{\left(\mathrm{1}+\mathrm{2k}\right)}{\mathrm{4}}\pi} ,\mathrm{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$$\mathrm{z}_{\mathrm{0}} =\mathrm{ae}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} ,\mathrm{z}_{\mathrm{1}} =\mathrm{ae}^{\frac{\mathrm{3i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{f}'\left(\mathrm{a}\right)=\mathrm{4a}^{\mathrm{3}} .\mathrm{2i}\pi\mathrm{Res}\left(\mathrm{i},\mathrm{z}_{\mathrm{1}} ,\mathrm{z}_{\mathrm{0}} \right) \\ $$$$=\mathrm{8i}\pi\mathrm{a}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2i}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{1}\right)}+\frac{\mathrm{1}}{\left(\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4z}_{\mathrm{1}} ^{\mathrm{3}} \right)}+\frac{\mathrm{1}}{\mathrm{z}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{4z}_{\mathrm{0}} ^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{4}\pi\mathrm{a}^{\mathrm{3}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }+\frac{\mathrm{8i}\pi\mathrm{a}^{\mathrm{3}} }{\left(\mathrm{1}+\mathrm{ia}^{\mathrm{2}} \right)\left(\mathrm{4a}^{\mathrm{3}} \mathrm{e}^{\mathrm{3i}\frac{\pi}{\mathrm{4}}} \right)}.+\frac{\mathrm{8i}\pi\mathrm{a}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{ia}^{\mathrm{2}} \right)\left(\mathrm{4a}^{\mathrm{3}} \mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\frac{\pi\mathrm{4a}^{\mathrm{3}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }+\mathrm{2i}\pi\left(\frac{\mathrm{e}^{−\frac{\mathrm{3i}\pi}{\mathrm{4}}} }{\mathrm{1}+\mathrm{ia}^{\mathrm{2}} }+\frac{\mathrm{e}^{−\mathrm{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{1}−\mathrm{ia}^{\mathrm{2}} }\right) \\ $$$$=\pi.\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }+\mathrm{2i}\pi\left(\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} +\mathrm{e}^{\frac{−\mathrm{3i}\pi}{\mathrm{4}}} +\mathrm{ia}^{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} −\mathrm{e}^{−\frac{\mathrm{3i}\pi}{\mathrm{4}}} \right)}{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }\right) \\ $$$$=\pi.\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }+\mathrm{2i}\pi\left(\frac{−\mathrm{i}\sqrt{\mathrm{2}}+\mathrm{ia}^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }\right) \\ $$$$=\pi\left(\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }+\mathrm{2}\sqrt{\mathrm{2}}.\left(\frac{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }\right)\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\pi\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\pi\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }{\mathrm{1}+\mathrm{a}^{\mathrm{4}} }\mathrm{da} \\ $$$$\pi\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\pi\mathrm{coth}^{−} \left(\sqrt{\mathrm{2}}\right) \\ $$$$\pi\mathrm{ln}\left(\mathrm{2}\right)+\pi\mathrm{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right) \\ $$$$\pi\mathrm{ln}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)+\pi\mathrm{ln}\left(\mathrm{2}\right)=\pi\mathrm{ln}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$

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