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Question Number 205151 by mathzup last updated on 10/Mar/24
find ∫_0 ^∞ ((ln^2 x)/(1+x^4 ))dx
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$
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
$$\Omega=\int_{−\infty} ^{\mathrm{0}} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}+\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)+\left({ln}\left({x}\right)+{i}\pi\right)^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }{dx}−\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }+\mathrm{2}{i}\pi\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\left({Re}\left(\Omega\right)+\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{4}} }=\int_{\mathrm{0}} ^{\infty} \frac{{y}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{4}\left(\mathrm{1}+{y}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${C}=\left\{{z}\in\mathbb{C}\mid{Re}\left({z}\right)>\mathrm{0}\right\} \\ $$$$\int_{{C}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{4}} }{dx}=\mathrm{2}{i}\pi{Res}\left({f},{e}^{{i}\frac{\pi}{\mathrm{4}}} ,{e}^{\mathrm{3}{i}\frac{\pi}{\mathrm{4}}} \right)=\Omega \\ $$$$=\mathrm{2}{i}\pi.\left(\frac{\left(\frac{{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{4}{e}^{\mathrm{3}\frac{{i}\pi}{\mathrm{4}}} }+\frac{\left(\frac{\mathrm{3}{i}\pi}{\mathrm{4}}\right)^{\mathrm{2}} }{\mathrm{4}{e}^{\mathrm{9}{i}\frac{\pi}{\mathrm{4}}} }\right)={i}\frac{\pi}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{16}}{e}^{{i}\frac{\pi}{\mathrm{4}}} −\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{16}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{32}}\left(−\mathrm{4}{i}\sqrt{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{2}}\right);\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{x}^{\mathrm{4}} }=−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{64}}\pi^{\mathrm{3}} +\frac{\pi^{\mathrm{3}} }{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{64}}\pi^{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$

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