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Question Number 217235 by Tawa11 last updated on 06/Mar/25
∫_( 0) ^( 1) ((x ln^2 (x))/(1 + x^2 )) dx
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:+\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Answered by MrGaster last updated on 06/Mar/25
=∫_0 ^1 ln^2 (x)((x/(1+x^2 )))dx =∫_0 ^1 ln^2 (x)(x−x^3 +x^5 −x^7 +…)dx =∫_0 ^1 (xln^2 (x)−x^3 ln^2 (x)+x^5 ln^2 (x)−x^7 ln^2 (x)+…)dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n+1) ln^2 (x)dx =Σ_(n=0) ^∞ (−1)^n ((1/((2n+2)^3 ))) =(1/8)Σ_(n=0) ^∞ (((−1)^n )/((n+1)^3 )) =(1/8)((7/8)ζ(3)) = determinant ((((3ζ(3))/(16))))
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left({x}\right)\left(\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left({x}\right)\left({x}−{x}^{\mathrm{3}} +{x}^{\mathrm{5}} −{x}^{\mathrm{7}} +\ldots\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}\mathrm{ln}^{\mathrm{2}} \left({x}\right)−{x}^{\mathrm{3}} \mathrm{ln}^{\mathrm{2}} \left({x}\right)+{x}^{\mathrm{5}} \mathrm{ln}^{\mathrm{2}} \left({x}\right)−{x}^{\mathrm{7}} \mathrm{ln}^{\mathrm{2}} \left({x}\right)+\ldots\right){dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right) \\ $$$$=\begin{array}{|c|}{\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{16}}}\\\hline\end{array} \\ $$
Commented by Tawa11 last updated on 07/Mar/25
Thanks sir. I appreciate
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$
Commented by mr W last updated on 08/Mar/25
how (1/8)((7/8)ζ(3))=^? (3/(16))ζ(3)
$${how} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\right)\overset{?} {=}\frac{\mathrm{3}}{\mathrm{16}}\zeta\left(\mathrm{3}\right) \\ $$
Answered by MathematicalUser2357 last updated on 07/Mar/25
Answered by mr W last updated on 08/Mar/25
∫_0 ^1 ((x ln^2 x)/(1+x^2 )) dx =∫_0 ^1 ln^2 x(x−x^3 +x^5 −x^7 +...) dx =∫_0 ^1 Σ_(n=0) ^∞ (−1)^n x^(2n+1) ln^2 x dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n+1) ln^2 x dx I_n =∫_0 ^1 x^(2n+1) ln^2 x dx =(1/(2(n+1)))∫_0 ^1 ln^2 x d(x^(2n+2) ) =(1/(2(n+1)))[(x^(2(n+1)) ln^2 x)_0 ^1 −2∫_0 ^1 x^(2n+1) ln x dx] =−(1/(2(n+1)^2 ))∫_0 ^1 ln x d(x^(2n+2) ) =−(1/(2(n+1)^2 ))[(x^(2(n+1)) ln x)_0 ^1 −∫_0 ^1 x^(2n+1) dx] =(1/(2(n+1)^2 ))∫_0 ^1 x^(2n+1) dx =(1/(4(n+1)^3 ))(x^(2n+2) )_0 ^1 =(1/(4(n+1)^3 )) ∫_0 ^1 ((x ln^2 x)/(1+x^2 )) dx=Σ_(n=0) ^∞ (((−1)^n )/(4(n+1)^3 ))=(1/4)Σ_(n=1) ^∞ (((−1)^(n+1) )/n^3 ) S=Σ_(n=1) ^∞ (((−1)^(n+1) )/n^3 )=(1/1^3 )−(1/2^3 )+(1/3^3 )−(1/4^3 )+... =((1/1^3 )+(1/3^3 )+(1/5^3 )+...)−((1/2^3 )+(1/4^3 )+(1/6^3 )+...) =((1/1^3 )+(1/2^3 )+(1/3^3 )+(1/4^3 )+(1/5^3 )+(1/6^3 )+...)−2×((1/2^3 )+(1/4^3 )+(1/6^3 )+...) =((1/1^3 )+(1/2^3 )+(1/3^3 )+(1/4^3 )+(1/5^3 )+(1/6^3 )+...)−(1/4)((1/1^3 )+(1/2^3 )+(1/3^3 )+...) =(3/4)((1/1^3 )+(1/2^3 )+(1/3^3 )+(1/4^3 )+(1/5^3 )+(1/6^3 )+...) =((3ζ(3))/4) ⇒∫_0 ^1 ((x ln^2 x)/(1+x^2 )) dx=(1/4)×((3ζ(3))/4)=((3ζ(3))/(16)) ✓
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{ln}^{\mathrm{2}} \:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \:{x}\left({x}−{x}^{\mathrm{3}} +{x}^{\mathrm{5}} −{x}^{\mathrm{7}} +…\right)\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$ \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \:{x}\:{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \:{x}\:{d}\left({x}^{\mathrm{2}{n}+\mathrm{2}} \right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left[\left({x}^{\mathrm{2}\left({n}+\mathrm{1}\right)} \mathrm{ln}^{\mathrm{2}} \:{x}\right)_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \mathrm{ln}\:{x}\:{dx}\right] \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\:{x}\:{d}\left({x}^{\mathrm{2}{n}+\mathrm{2}} \right) \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\left[\left({x}^{\mathrm{2}\left({n}+\mathrm{1}\right)} \mathrm{ln}\:{x}\right)_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {dx}\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\left({x}^{\mathrm{2}{n}+\mathrm{2}} \right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{ln}^{\mathrm{2}} \:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} } \\ $$$$ \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+… \\ $$$$\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+…\right)−\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }+…\right) \\ $$$$\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }+…\right)−\mathrm{2}×\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }+…\right) \\ $$$$\:\:\:=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }+…\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+…\right) \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{3}} }+…\right) \\ $$$$\:\:\:=\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{4}} \\ $$$$ \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{ln}^{\mathrm{2}} \:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{4}}=\frac{\mathrm{3}\zeta\left(\mathrm{3}\right)}{\mathrm{16}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 08/Mar/25
Wow, Thanks sir. I really appreciate.
$$\mathrm{Wow},\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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