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calculer-une-primitive-de-x-ln-1-x-2-puis-jusrifier-la-convergence-de-0-1-ln-1-x-2-dx-

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Question Number 163316 by SANOGO last updated on 06/Jan/22
calculer une primitive de x:→ln(1−x^2 ) puis jusrifier la convergence de ∫_0 ^1 ln(1−x^2 )dx
$${calculer}\:{une}\:{primitive}\:{de}\:{x}:\rightarrow{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$${puis}\:{jusrifier}\:{la}\:{convergence}\:{de}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$
Answered by amin96 last updated on 06/Jan/22
x^2 =t (dt/dx)=2x=2(√t) ∫_0 ^1 ((ln(1−t))/(2(√t)))dt=(1/2)Σ_(n=1) ^∞ (1/n)∫_0 ^1 t^(n−(1/2)) dt= =(1/2)Σ_(n=1) ^∞ (1/(n(n+(1/2))))=(1/2)Σ_(n=0) ^∞ (1/((n+1)(n+(3/2))))= =(1/2)(((𝛙(1)−𝛙((3/2)))/(1−(1/2))))=𝛙(1)−𝛙((3/2))= =𝛙(1)−𝛙((1/2))−2=−𝛄+2ln(2)+𝛄−2= =2ln(2)−2
$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} =\boldsymbol{\mathrm{t}}\:\:\:\:\frac{\boldsymbol{\mathrm{dt}}}{\boldsymbol{\mathrm{dx}}}=\mathrm{2}\boldsymbol{\mathrm{x}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{t}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{t}}\right)}{\mathrm{2}\sqrt{\boldsymbol{{t}}}}\boldsymbol{{dt}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{t}}^{\boldsymbol{{n}}−\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dt}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}\left(\boldsymbol{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\boldsymbol{\psi}\left(\mathrm{1}\right)−\boldsymbol{\psi}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)=\boldsymbol{\psi}\left(\mathrm{1}\right)−\boldsymbol{\psi}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)= \\ $$$$=\boldsymbol{\psi}\left(\mathrm{1}\right)−\boldsymbol{\psi}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{2}=−\boldsymbol{\gamma}+\mathrm{2}\boldsymbol{{ln}}\left(\mathrm{2}\right)+\boldsymbol{\gamma}−\mathrm{2}= \\ $$$$=\mathrm{2}\boldsymbol{{ln}}\left(\mathrm{2}\right)−\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 06/Jan/22
2ln(2)−2
$$\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2} \\ $$
Commented by amin96 last updated on 06/Jan/22
Oh yes .I wrote 1−(1/2)=−(1/2) )))
$$\left.{O}\left.{h}\left.\:{yes}\:.{I}\:{wrote}\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\right)\right)\right) \\ $$
Answered by smallEinstein last updated on 06/Jan/22
Answered by mnjuly1970 last updated on 06/Jan/22
Ω= ∫_0 ^( 1) ln(x)dx + ∫_1 ^( 2) ln(x)dx = ∫_0 ^( 2) ln(x)dx=[xln(x)−x]_0 ^2 2ln(2)−2 =ln(4)−2
$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){dx}\:+\:\int_{\mathrm{1}} ^{\:\mathrm{2}} {ln}\left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} {ln}\left({x}\right){dx}=\left[{xln}\left({x}\right)−{x}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}\:={ln}\left(\mathrm{4}\right)−\mathrm{2} \\ $$

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