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Question Number 224510 by Rogerkeyter last updated on 15/Sep/25
Answered by maths2 last updated on 16/Sep/25
∫_0 ^1 ((ln(1−x+x^2 ))/x)+((ln(1−x+x^2 ))/(1−x))dx =2∫_0 ^1 ((ln(1−x+x^2 ))/x)dx =2∫_0 ^1 ((ln(1−x+x^2 )+ln(1+x)−ln(1+x))/x)dx =2∫_0 ^1 ((ln(1+x^3 ))/x)−((ln(1+x))/x)dx =2∫_0 ^1 ((ln(1+x^3 ))/x^3 ).x^2 dx−2∫_0 ^1 ((ln(1+x))/x)dx =2.∫_0 ^1 ((ln(1+x))/x).(dx/3)−2∫_0 ^1 ((ln(1+x))/x)dx =−(4/3)∫_0 ^1 ((ln(1−(−x)))/(−x)).d(−x) =(4/3)[Li_2 (−x)]_0 ^1 =(4/3)li_2 (−1)=(4/3)((−π^2 )/(12))=((−π^2 )/9)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}+\frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)+{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}−\frac{\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }.{x}^{\mathrm{2}} {dx}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=\mathrm{2}.\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}.\frac{{dx}}{\mathrm{3}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$=−\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{x}\right)\right)}{−{x}}.{d}\left(−{x}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left[{Li}_{\mathrm{2}} \left(−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}{li}_{\mathrm{2}} \left(−\mathrm{1}\right)=\frac{\mathrm{4}}{\mathrm{3}}\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}=\frac{−\pi^{\mathrm{2}} }{\mathrm{9}} \\ $$
Commented by Tawa11 last updated on 16/Sep/25
Commented by Tawa11 last updated on 16/Sep/25
Wolfram alpha did not put minus sign. and your work is correct sir. Or where does the minus went to in wolfram alpha?.
$$\mathrm{Wolfram}\:\mathrm{alpha}\:\mathrm{did}\:\mathrm{not}\:\mathrm{put}\:\mathrm{minus}\:\mathrm{sign}. \\ $$$$\mathrm{and}\:\mathrm{your}\:\mathrm{work}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{sir}. \\ $$$$\mathrm{Or}\:\mathrm{where}\:\mathrm{does}\:\mathrm{the}\:\mathrm{minus}\:\mathrm{went}\:\mathrm{to}\:\mathrm{in}\:\mathrm{wolfram}\:\mathrm{alpha}?. \\ $$
Commented by Tawa11 last updated on 16/Sep/25
I have seen your mistake sir. Wolfram alpha is correct.
$$\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{your}\:\mathrm{mistake}\:\mathrm{sir}. \\ $$$$\mathrm{Wolfram}\:\mathrm{alpha}\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by Tawa11 last updated on 16/Sep/25
∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(x^2 − x)) dx = ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(− x + x^2 )) dx = ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(− (x − x^2 ))) dx = − ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/((x − x^2 ))) dx = − ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(x (1 − x))) dx = − [∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/x) dx + ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(1 − x)) dx] = − ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/x) dx − ∫_( 0) ^( 1) ((ln(1 − x + x^2 ))/(1 − x)) dx That is where your mistake started sir.
$$\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \:\:−\:\:\mathrm{x}}\:\mathrm{dx} \\ $$$$\:=\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{−\:\:\mathrm{x}\:\:\:+\:\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\:=\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{−\:\:\left(\mathrm{x}\:\:\:−\:\:\:\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx} \\ $$$$\:=\:\:\:−\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{x}\:\:\:−\:\:\:\mathrm{x}^{\mathrm{2}} \right)}\:\mathrm{dx} \\ $$$$\:=\:\:\:−\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}\:\left(\mathrm{1}\:\:\:−\:\:\:\mathrm{x}\right)}\:\mathrm{dx} \\ $$$$\:=\:\:\:−\:\:\:\left[\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\:\mathrm{dx}\:\:\:+\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}\:\:\:−\:\:\:\mathrm{x}\:\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}\:\:\:−\:\:\:\mathrm{x}}\:\mathrm{dx}\right] \\ $$$$\:=\:\:\:−\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}\:\:−\:\:\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}}\:\mathrm{dx}\:\:\:−\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}\:\:\:−\:\:\:\mathrm{x}\:\:\:+\:\:\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}\:\:\:−\:\:\:\mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{where}\:\mathrm{your}\:\mathrm{mistake}\:\mathrm{started}\:\mathrm{sir}. \\ $$
Answered by Nicholas666 last updated on 17/Sep/25
I = ∫_0 ^1 ((ln (1−x+x^2 ))/(x^2 −x)) dx = (1/(x^2 −x)) = (1/(x−1)) −(1/x) I = −2∫_0 ^1 ((ln(1−x+x^2 ))/x) dx = { ((1−x+x^2 =(1−e^(iπ/3) ∙x)(1−e^(−iπ/3) ∙x))),((ln(1−x+x^2 )=ln(1−e^(iπ/3) ∙x)+(1−e^(−iπ/3) ∙x))) :} Using dilogarithm identities ; ∫_0 ^1 ((ln(1−ax))/x) dx = −Li_2 (a) I = 2 (Li_2 (e^(iπ/3) )+ Li_2 (e^(iπ/3) )=4R(Li_2 (e^(iπ/3) )) R(Li_2 (e^(iθ) ))=(π^2 /6)−((πθ)/2)+(θ^2 /4) R(Li_2 (e^(iπ/3) ))= (π^2 /(36)) I = 4 ∙ (π^2 /(36)) = (π^2 /9) ≈1.0966....
$$\:\:\:\:\mathcal{I}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} −{x}}\:{dx}\:=\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}}\:=\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\mathcal{I}\:=\:−\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}\:{dx}\:=\:\begin{cases}{\mathrm{1}−{x}+{x}^{\mathrm{2}} \:=\left(\mathrm{1}−{e}^{{i}\pi/\mathrm{3}} \centerdot{x}\right)\left(\mathrm{1}−{e}^{−{i}\pi/\mathrm{3}} \centerdot{x}\right)}\\{\mathrm{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)=\mathrm{ln}\left(\mathrm{1}−{e}^{{i}\pi/\mathrm{3}} \centerdot{x}\right)+\left(\mathrm{1}−{e}^{−{i}\pi/\mathrm{3}} \centerdot{x}\right)}\end{cases}\: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Using}}\:\boldsymbol{\mathrm{dilogarithm}}\:\boldsymbol{\mathrm{identities}}\:;\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{{ax}}\right)}{\boldsymbol{{x}}}\:\boldsymbol{{dx}}\:=\:−\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\boldsymbol{{a}}\right) \\ $$$$\:\:\:\:\:\:\mathcal{I}\:=\:\mathrm{2}\:\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\pi/\mathrm{3}} \right)+\:\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\pi/\mathrm{3}} \right)=\mathrm{4}\mathfrak{R}\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\pi/\mathrm{3}} \right)\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathfrak{R}\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\theta} \right)\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi\theta}{\mathrm{2}}+\frac{\theta^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathfrak{R}\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{{i}\pi/\mathrm{3}} \right)\right)=\:\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$\:\:\:\:\:\:\:\mathcal{I}\:=\:\mathrm{4}\:\centerdot\:\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{9}}\:\approx\mathrm{1}.\mathrm{0966}…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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