Question

مطلوب است محاسبه: $ \int _0^1 ln( a^{x} - b^{-x})dx=I(a,b) $ به طوری که$ - \frac{ \pi ^{2} }{6ln(ab)}+ \frac{1}{ln(ab)}Li_2( \frac{1}{ab} )+ln \sqrt{a}=I(a,b) $

Answer 1

قبل از هر راهکاری توجه داریم که اگر $(|x|<1)$ :

$Li_2(x)=- \int _0^x \frac{Ln(1-t)}{t} dt=- \int _0^1 \frac{Ln(1-tx)}{t} dt= \frac{ \pi ^2}{6} - \int _1^x \frac{Ln(1-t)}{t} dt$

$ \Rightarrow I(a,b)=\int _0^1Ln(a^x-b^{-x})dx=\int _0^1Lna^x(1-(ab)^{-x})dx$

$=\int _0^1Lna^xdx+\int _0^1Ln(1-(ab)^{-x})dx$

$= \frac{1}{2} Lna+\int _0^1Ln(1-(ab)^{-x})dx$

حالا قرار دهید:

$t:=(ab)^{-x} \Rightarrow t=e^{-Ln(ab)x} \Rightarrow dt=-Ln(ab)e^{-Ln(ab)x}dx=-Ln(ab).(ab)^{-x}dx$

$ \Rightarrow dx= -\frac{dt}{Ln(ab).t} $

$ \Rightarrow \int _0^1Ln(1-(ab)^{-x})dx=- \frac{1}{Ln(ab)} \int _1^{(ab)^{-1}} \frac{Ln(1-t)}{t}dt=\frac{1}{Ln(ab)}( Li_2((ab)^{-1}-\frac{ \pi ^2}{6})$

$I(a,b)= \frac{1}{2} Lna+\frac{1}{Ln(ab)}( Li_2((ab)^{-1}-\frac{ \pi ^2}{6})=-\frac{ \pi ^2}{6Ln(ab)}+ \frac{1}{Ln(ab)}Li_2( \frac{1}{ab} )+Ln \sqrt{a} $

$ \Box $

Answer 2

$$ \int _0^1Ln( \frac{ (ab)^{x}-1 }{ b^{x} } )dx= \int _0^1Ln( (ab)^{x}-1)-xLnb dx= \int _0^1Ln( ((ab)^{x}(1- (ab)^{-x}-Lnb \int _0^1xdx= \int _0^1Ln( (ab)^{x}dx+ \int _0^1Ln(1- (ab)^{-x})dx-Lnb \int _0^1xdx=(Ln(ab)-Lnb) \int _0^1xdx- \sum _ {n=1} ^ \infty \frac{1}{n} \int _0^1 (ab)^{-xn}dx= \frac{1}{2} Ln( \frac{ab}{b})- \sum _{n=1}^ \infty \int _0^1 ( a^{n} b^{n}) ^{-x} dx= \frac{1}{2} Lna- \sum _ {n=1}^ \infty \frac{1}{n}( \frac{1}{Ln (ab)^{n} } (1- \frac{1}{ a^{n} b^{n} } )=Ln \sqrt{a} - \frac{1}{Lnab} \sum _ {n=1}^ \infty \frac{1}{ n^{2} }(1- \frac{1}{ a^{n} b^{n} } )=Ln \sqrt{a} - \frac{1}{Lnab} \sum _{n=1} ^ \infty \frac{1}{ n^{2} } + \frac{1}{Lnab} \sum _ {n=1} ^ \infty \frac{ (\frac{1}{ab} )^{n} }{ n^{2} } =Ln \sqrt{a}- \frac{ \pi ^{2} }{6Lnab} + \frac{1}{Lnab} Li_2( \frac{1}{ab}) $$