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Question Number 143546 by mathmax by abdo last updated on 15/Jun/21

calculate ∫_0 ^∞ ((log^2 x)/((8+x^4 )^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{8}+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Answered by mathmax by abdo last updated on 16/Jun/21

f(a)=∫_0 ^∞ ((log^2 x)/(a^4 +x^4 ))dx ⇒f^′ (a)=−∫_0 ^∞ ((4a^3 log^2 x)/((a^4 +x^4 )^2 )) dx ⇒ f^′ (2^(3/4) )=−4 (2)^(3/4) ∫_0 ^∞ ((log^2 x)/((8+x^4 )^2 ))dx =−2^(2+(3/4)) ∫_0 ^∞ ((log^2 x)/((8+x^4 )^2 ))dx ⇒∫_0 ^∞ ((log^2 x)/((8+x^4 )^2 ))dx=−(1/2^((11)/4) )f^′ (2^(3/4) ) f(a)=_(x=at) ∫_0 ^∞ ((log^2 (at))/(a^4 (1+t^4 )))adt =(1/a^3 )∫_0 ^∞ ((log^2 a +2logalogt +log^2 t)/(1+t^4 ))dt =((log^2 a)/a^3 )∫_0 ^∞ (dt/(1+t^4 )) +((2loga)/a^3 )∫_0 ^∞ ((logt)/(1+t^4 ))dt +(1/a^3 )∫_0 ^∞ ((log^2 (t))/(1+t^4 ))dt ∫_0 ^∞ (dt/(1+t^4 ))=_(t^4 =y) (1/4) ∫_0 ^∞ (y^((1/4)−1) /(1+y))dy =(1/4)×(π/(sin((π/4))))=(π/(4×((√2)/2))) =(π/(2(√2))) ∫_0 ^∞ ((logt)/(1+t^4 ))dt =_(t^4 =y) (1/4) ∫_0 ^∞ ((logy)/(1+y))×(1/4)y^((1/4)−1) dy =(1/(16))∫_0 ^∞ ((y^((1/4)−1) logy)/(1+y))dy =(1/(16))w^′ ((1/4)) with w(λ)=∫_0 ^∞ (y^(λ−1) /(1+y))dy ⇒w^′ (λ)=∫_0 ^∞ ((y^(λ−1) logy)/(1+y))dy w(λ)=(π/(sin(πλ))) ⇒w^′ (λ)=−((π^2 cos(πλ))/(sin^2 (πλ))) ⇒ w^′ ((1/4))=−π^2 ×((1/( (√2)))/(((1/( (√2))))^2 ))=−2π^2 .(1/( (√2)))=−π^2 (√2) ⇒ ∫_0 ^∞ ((logt)/(1+t^4 ))dt =−((√2)/(16))π^2 ∫_0 ^∞ ((log^2 t)/(1+t^4 ))dt =_(t^4 =y) (1/(16))∫_0 ^∞ ((log^2 y)/(1+y))×(1/4)y^((1/4)−1) dt =(1/(64))∫_0 ^∞ ((y^((1/4)−1) log^2 y)/(1+y))dy=(1/(64))w^((2)) ((1/4)) w^((2)) (λ)=−π^2 ×((−πsin(πλ)sin^2 (πλ)−2πsin(πλ)cos(πλ))/(sin^4 (πλ))) =−π^3 ×((sin^2 (πλ)−2cos(πλ))/(sin^3 (πλ))) ⇒ w^((2)) ((1/4))=−π^3 ×(((1/2)−2×(1/( (√2))))/(((1/( (√2))))^3 ))=−π^3 ×(((1/2)−(√2))/(1/(2(√2)))) =−2(√2)π^3 (((1−2(√2))/2))=(√2)π^3 (2(√2)−1) ⇒ f(a)=((log^2 a)/a^3 )×(π/(2(√2))) +((2loga)/a^3 )(−((√2)/(16)))π^2 +(1/a^3 )×(1/(64))(√2)π^3 (2(√2)−1) Φ=−(1/2^((11)/4) )f^′ (2^(3/4) )....

$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\mathrm{a}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4a}^{\mathrm{3}} \mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{a}^{\mathrm{4}} \:+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \right)=−\mathrm{4}\:\left(\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{8}+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx}\:=−\mathrm{2}^{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{8}+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{8}+\mathrm{x}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{4}}} }\mathrm{f}^{'} \left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=_{\mathrm{x}=\mathrm{at}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{at}\right)}{\mathrm{a}^{\mathrm{4}} \left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)}\mathrm{adt}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{a}\:+\mathrm{2logalogt}\:+\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{log}^{\mathrm{2}} \mathrm{a}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:+\frac{\mathrm{2loga}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }=_{\mathrm{t}^{\mathrm{4}} =\mathrm{y}} \frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+\mathrm{y}}\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\pi}{\mathrm{4}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=_{\mathrm{t}^{\mathrm{4}} \:=\mathrm{y}} \:\frac{\mathrm{1}}{\mathrm{4}}\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{logy}}{\mathrm{1}+\mathrm{y}}×\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{logy}}{\mathrm{1}+\mathrm{y}}\mathrm{dy}\:\:=\frac{\mathrm{1}}{\mathrm{16}}\mathrm{w}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\mathrm{with} \\ $$$$\mathrm{w}\left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{y}^{\lambda−\mathrm{1}} }{\mathrm{1}+\mathrm{y}}\mathrm{dy}\:\Rightarrow\mathrm{w}^{'} \left(\lambda\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\lambda−\mathrm{1}} \:\mathrm{logy}}{\mathrm{1}+\mathrm{y}}\mathrm{dy} \\ $$$$\mathrm{w}\left(\lambda\right)=\frac{\pi}{\mathrm{sin}\left(\pi\lambda\right)}\:\Rightarrow\mathrm{w}^{'} \left(\lambda\right)=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\pi\lambda\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\lambda\right)}\:\Rightarrow \\ $$$$\mathrm{w}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{2}} \:×\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=−\mathrm{2}\pi^{\mathrm{2}} .\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=−\pi^{\mathrm{2}} \sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=−\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\pi^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=_{\mathrm{t}^{\mathrm{4}} =\mathrm{y}} \:\:\:\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{y}}{\mathrm{1}+\mathrm{y}}×\frac{\mathrm{1}}{\mathrm{4}}\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \mathrm{log}^{\mathrm{2}} \mathrm{y}}{\mathrm{1}+\mathrm{y}}\mathrm{dy}=\frac{\mathrm{1}}{\mathrm{64}}\mathrm{w}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\mathrm{w}^{\left(\mathrm{2}\right)} \left(\lambda\right)=−\pi^{\mathrm{2}} ×\frac{−\pi\mathrm{sin}\left(\pi\lambda\right)\mathrm{sin}^{\mathrm{2}} \left(\pi\lambda\right)−\mathrm{2}\pi\mathrm{sin}\left(\pi\lambda\right)\mathrm{cos}\left(\pi\lambda\right)}{\mathrm{sin}^{\mathrm{4}} \left(\pi\lambda\right)} \\ $$$$=−\pi^{\mathrm{3}} ×\frac{\mathrm{sin}^{\mathrm{2}} \left(\pi\lambda\right)−\mathrm{2cos}\left(\pi\lambda\right)}{\mathrm{sin}^{\mathrm{3}} \left(\pi\lambda\right)}\:\Rightarrow \\ $$$$\mathrm{w}^{\left(\mathrm{2}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{3}} ×\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{3}} }=−\pi^{\mathrm{3}} ×\frac{\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$=−\mathrm{2}\sqrt{\mathrm{2}}\pi^{\mathrm{3}} \left(\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\sqrt{\mathrm{2}}\pi^{\mathrm{3}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{log}^{\mathrm{2}} \mathrm{a}}{\mathrm{a}^{\mathrm{3}} }×\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:+\frac{\mathrm{2loga}}{\mathrm{a}^{\mathrm{3}} }\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\right)\pi^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{64}}\sqrt{\mathrm{2}}\pi^{\mathrm{3}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\Phi=−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{11}}{\mathrm{4}}} }\mathrm{f}^{'} \left(\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{4}}} \right).... \\ $$

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