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Question Number 128954 by mathmax by abdo last updated on 11/Jan/21

calculate f(λ) =∫_0 ^1 ln(x^4 +λ^4 )dx with λ>0 then find the value of ∫_0 ^1 ln(1+x^4 )dx

$$\mathrm{calculate}\:\:\mathrm{f}\left(\lambda\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}^{\mathrm{4}} \:+\lambda^{\mathrm{4}} \right)\mathrm{dx}\:\:\:\mathrm{with}\:\lambda>\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{dx} \\ $$

Answered by Lordose last updated on 16/Jan/21

f(λ) = ∫_0 ^( 1) ln(x^4 +λ^4 )dx = ∫_0 ^( 1) ln(λ^4 (((x/λ))^4 +1))dx f(λ) = 4∫_0 ^( 1) ln(λ)dx + ∫_0 ^( 1) ln(1+(x^4 /λ^4 ))dx f(λ) = 4ln(λ) + Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^( 1) ((x/λ))^(4n) dx f(λ) = 4ln(λ) + Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(4n+1)λ^(4n) )) Ω = ∫_0 ^( 1) ln(1+x^4 )dx = 4ln(1) + Σ_(n=1) ^∞ (((−1))/(n(4n+1))) Ω = Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(4n+1))) = (π/( (√2))) − 4 + ln(2) + (√2)coth^(−1) ((√2))

$$ \\ $$ $$ \\ $$ $$\mathrm{f}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\mathrm{x}^{\mathrm{4}} +\lambda^{\mathrm{4}} \right)\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\lambda^{\mathrm{4}} \left(\left(\frac{\mathrm{x}}{\lambda}\right)^{\mathrm{4}} +\mathrm{1}\right)\right)\mathrm{dx} \\ $$ $$\mathrm{f}\left(\lambda\right)\:=\:\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\lambda\right)\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{4}} }{\lambda^{\mathrm{4}} }\right)\mathrm{dx} \\ $$ $$\mathrm{f}\left(\lambda\right)\:=\:\mathrm{4ln}\left(\lambda\right)\:+\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{x}}{\lambda}\right)^{\mathrm{4n}} \mathrm{dx}\: \\ $$ $$\mathrm{f}\left(\lambda\right)\:=\:\mathrm{4ln}\left(\lambda\right)\:+\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}\left(\mathrm{4n}+\mathrm{1}\right)\lambda^{\mathrm{4n}} } \\ $$ $$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{dx}\:=\:\mathrm{4ln}\left(\mathrm{1}\right)\:+\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)}{\mathrm{n}\left(\mathrm{4n}+\mathrm{1}\right)} \\ $$ $$\Omega\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}\left(\mathrm{4n}+\mathrm{1}\right)}\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:−\:\mathrm{4}\:+\:\mathrm{ln}\left(\mathrm{2}\right)\:+\:\sqrt{\mathrm{2}}\mathrm{coth}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right) \\ $$

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