Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({x}\right){dx}\:+\:\int_{\mathrm{1}} ^{\:\mathrm{2}} {ln}\left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2}} {ln}\left({x}\right){dx}=\left[{xln}\left({x}\right)−{x}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}\:={ln}\left(\mathrm{4}\right)−\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com