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Question Number 49003 by mondodotto@gmail.com last updated on 01/Dec/18

∫((logx)/(√(1−x^x )))dx please help this

$$\int\frac{\boldsymbol{\mathrm{log}{x}}}{\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\boldsymbol{{x}}} }}\boldsymbol{\mathrm{d}{x}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{this}} \\ $$

Commented by maxmathsup by imad last updated on 01/Dec/18

I =∫ ((ln(x))/(√(1−e^(xln(x)) )))dx changement xln(x)=t give (ln(x)+1)dx =dt I =∫ ((ln(x)+1−1)/(√(1−x^x )))dx =∫ ((ln(x)+1)/(√(1−e^(xlnx) )))dx −∫ (dx/(√(1−x^x ))) =∫ (dt/(√(1−t))) −∫ (dx/(√(1−x^x ))) =−2(√(1−t)) −∫ (dx/(√(1−x^x ))) =−2(√(1−x^x )) −∫ (dx/(√(1−x^x ))) let find ∫ (dx/(√(1−x^x ))) at form of serie due to 0<x^x <1 we have (1+u)^α =1+((αu)/(1!)) +((α(α−1))/(2!)) u^2 +....+((α(α−1)...(α−n+1))/(n!)) u^n +... ⇒ (1−u)^α =1−αu +((α(α−1))/(2!)) u^2 +((α(α−1)...(α−n+1))/(n!))(−1)^n u^n +... ⇒ (1−x^x )^(−(1/2)) =1+(1/2) x^x +(((−(1/2))(−(3/2)))/(2!)) x^(2x) +...(((−(1/2))(−(3/2))...(−(1/2)−n+1))/(n!)) (−1)^n x^(nx) +... ⇒∫ (dx/(√(1−x^x ))) =x +(1/2) ∫ x^x dx +∫ (3/(4 2!)) x^(2x) dx +... +(((−(1/2))(−(3/2))....(−(1/2)−n+1))/(n!)) (−1)^(n ) ∫ x^(nx) dx+....

$${I}\:=\int\:\:\:\frac{{ln}\left({x}\right)}{\sqrt{\mathrm{1}−{e}^{{xln}\left({x}\right)} }}{dx}\:\:{changement}\:{xln}\left({x}\right)={t}\:{give}\:\left({ln}\left({x}\right)+\mathrm{1}\right){dx}\:={dt} \\ $$$${I}\:=\int\:\frac{{ln}\left({x}\right)+\mathrm{1}−\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{{x}} }}{dx}\:=\int\:\:\frac{{ln}\left({x}\right)+\mathrm{1}}{\sqrt{\mathrm{1}−{e}^{{xlnx}} }}{dx}\:−\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }} \\ $$$$=\int\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}}}\:\:−\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }}\:=−\mathrm{2}\sqrt{\mathrm{1}−{t}}\:−\int\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }}\:=−\mathrm{2}\sqrt{\mathrm{1}−{x}^{{x}} }\:−\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }} \\ $$$${let}\:{find}\:\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }}\:\:{at}\:{form}\:{of}\:{serie}\:{due}\:{to}\:\:\mathrm{0}<{x}^{{x}} <\mathrm{1} \\ $$$${we}\:{have}\:\left(\mathrm{1}+{u}\right)^{\alpha} \:=\mathrm{1}+\frac{\alpha{u}}{\mathrm{1}!}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}\:{u}^{\mathrm{2}} \:+....+\frac{\alpha\left(\alpha−\mathrm{1}\right)...\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\:{u}^{{n}} \:+...\:\Rightarrow \\ $$$$\left(\mathrm{1}−{u}\right)^{\alpha} \:=\mathrm{1}−\alpha{u}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}\:{u}^{\mathrm{2}} \:\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)...\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:+...\:\Rightarrow \\ $$$$\left(\mathrm{1}−{x}^{{x}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:{x}^{{x}} \:\:+\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}!}\:{x}^{\mathrm{2}{x}} \:+...\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)...\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}\:\left(−\mathrm{1}\right)^{{n}} \:{x}^{{nx}} +... \\ $$$$\Rightarrow\int\:\:\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{{x}} }}\:={x}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{x}^{{x}} {dx}\:+\int\:\:\frac{\mathrm{3}}{\mathrm{4}\:\mathrm{2}!}\:{x}^{\mathrm{2}{x}} {dx}\:+... \\ $$$$+\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)....\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)}{{n}!}\:\left(−\mathrm{1}\right)^{{n}\:} \:\int\:\:\:{x}^{{nx}} {dx}+.... \\ $$

Commented by MJS last updated on 01/Dec/18

where did you find this?

$$\mathrm{where}\:\mathrm{did}\:\mathrm{you}\:\mathrm{find}\:\mathrm{this}? \\ $$

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