Menu Close

How-to-calculate-this-integral-pi-2-0-ln-1-sint-sint-dt-

Tinku Tara 0 Comments
Pin




Question Number 197272 by Erico last updated on 12/Sep/23
How to calculate this integral ∫^( (π/2)) _( 0) ((ln(1+sint))/(sint))dt
$$\mathrm{How}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{this}\:\mathrm{integral} \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{ln}\left(\mathrm{1}+{sint}\right)}{{sint}}{dt} \\ $$
Answered by Mathspace last updated on 12/Sep/23
tan((t/2))=x ⇒ I=∫_0 ^1 ((ln(1+((2x)/(1+x^2 ))))/((2x)/(1+x^2 )))((2dx)/(1+x^2 )) =∫_0 ^1 ((ln(((1+x^2 +2x)/(1+x^2 ))))/x)dx =∫_0 ^1 ((ln(x+1)^2 −ln(1+x^2 ))/x)dx =2∫_0 ^1 ((ln(1+x))/x)dx−∫_0 ^1 ((ln(1+x^2 ))/x)dx (dx/dx)ln(1+x)=(1/(1+x))=Σ_(n=0) ^∞ (−1)^n x^n ⇒ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) =Σ_(n=1) ^∞ (−1)^(n−1) (x^n /n) ⇒ ∫_0 ^1 ((ln(1+x))/x)dx=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 x^(n−1) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=η(2) =(1−2^(1−2) )ξ(2)=(1/2).(π^2 /6)=(π^2 /(12)) ln(1+x^2 )=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^(2n) ⇒ ∫_0 ^1 ((ln(1+x^2 ))/x)dx=Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 x^(2n−1) dx =(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) =(1/2)(π^2 /(12))=(π^2 /(24)) ⇒ ∫_0 ^(π/2) ((ln(1+sinx))/(sinx))dx =2.(π^2 /(12))−(π^2 /(24))=((3π^2 )/(24))=(π^2 /8)
$${tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:\Rightarrow \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\frac{{dx}}{{dx}}{ln}\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=\eta\left(\mathrm{2}\right) \\ $$$$=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{2}} \right)\xi\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}{dx}=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{sinx}\right)}{{sinx}}{dx} \\ $$$$=\mathrm{2}.\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Answered by universe last updated on 12/Sep/23
I = ∫_0 ^(𝛑/2) ((log (1+𝛂sin x) dx)/(sin x)) (dI/d𝛂) = ∫_0 ^(𝛑/2) (1/(1+𝛂sin x))dx (dI/d𝛂) = ∫_0 ^(𝛑/2 ) ((sec^2 x/2)/(1+tan^2 x/2+2𝛂tan x/2))dx (dI/d𝛂) = ∫_0 ^1 ((2dy)/(y^2 +2𝛂y+1)) = ∫^1 _0 ((2dy)/((y+𝛂)^2 +(1−𝛂^2 ))) (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))tan^(−1) ((y+𝛂)/( (√(1−𝛂^2 )))) ∣_0 ^1 (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))[tan^(−1) ((1−𝛂)/( (√(1−𝛂^2 )))) − tan^(−1) (𝛂/( (√(1−𝛂^2 ))))] (dI/d𝛂) = (2/( (√(1−𝛂^2 ))))tan^(−1) (((1/(√(1−𝛂^2 )))/(1+(1+𝛂)𝛂/1−𝛂^2 ))) ∫(dI/d𝛂) = ∫(2/( (√(1−𝛂^2 ))))tan^(−1) (√((1−𝛂)/(1+𝛂))) d𝛂 let 𝛂 = cos𝛃 ⇒ d𝛂 = −sin 𝛃 d𝛃 I = −2∫tan^(−1) ((√((1−cos 𝛃)/(1+cos 𝛃))) )d𝛃 I = −2∫(𝛃/2) d𝛃 I = −(𝛃^2 /2) +c I = −(((cos^(−1) 𝛂)^2 )/2) + c let 𝛂 = 0 then I = 0 c = (𝛑^2 /8) I = (𝛑^2 /8) − (((cos^(−1) 𝛂)^2 )/2) now put 𝛂 = 1 I = (𝛑^2 /8)
$$\:\boldsymbol{{I}}\:\:\:\:=\:\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\boldsymbol{\mathrm{log}}\:\left(\mathrm{1}+\boldsymbol{\alpha\mathrm{sin}}\:\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}}{\boldsymbol{\mathrm{sin}}\:\boldsymbol{{x}}} \\ $$$$\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\alpha\mathrm{sin}}\:\boldsymbol{{x}}}\boldsymbol{{dx}}\:\: \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}\:\:} \frac{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \:\boldsymbol{{x}}/\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{{x}}/\mathrm{2}+\mathrm{2}\boldsymbol{\alpha\mathrm{tan}}\:\boldsymbol{{x}}/\mathrm{2}}\boldsymbol{{dx}} \\ $$$$\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\boldsymbol{{dy}}}{\boldsymbol{{y}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\alpha{y}}+\mathrm{1}}\:=\:\underset{\mathrm{0}} {\int}^{\mathrm{1}} \frac{\mathrm{2}\boldsymbol{{dy}}}{\left(\boldsymbol{{y}}+\boldsymbol{\alpha}\right)^{\mathrm{2}} +\left(\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} \right)}\: \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\boldsymbol{{y}}+\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\:\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\left[\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\mathrm{1}−\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\:−\:\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \frac{\boldsymbol{\alpha}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\right] \\ $$$$\:\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:\:=\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\frac{\mathrm{1}/\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}{\mathrm{1}+\left(\mathrm{1}+\boldsymbol{\alpha}\right)\boldsymbol{\alpha}/\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\int\frac{\boldsymbol{{dI}}}{\boldsymbol{{d}\alpha}}\:=\:\int\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\boldsymbol{\alpha}^{\mathrm{2}} }}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−\boldsymbol{\alpha}}{\mathrm{1}+\boldsymbol{\alpha}}}\:\boldsymbol{{d}\alpha} \\ $$$$\:\:\boldsymbol{{let}}\:\:\boldsymbol{\alpha}\:=\:\boldsymbol{\mathrm{cos}\beta}\:\:\:\Rightarrow\:\:\boldsymbol{{d}\alpha}\:\:=\:−\boldsymbol{\mathrm{sin}}\:\boldsymbol{\beta}\:\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{{I}}\:\:\:=\:\:\:−\mathrm{2}\int\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−\boldsymbol{\mathrm{cos}}\:\boldsymbol{\beta}}{\mathrm{1}+\boldsymbol{\mathrm{cos}}\:\boldsymbol{\beta}}}\:\right)\boldsymbol{{d}\beta} \\ $$$$\boldsymbol{{I}}\:\:=\:\:−\mathrm{2}\int\frac{\boldsymbol{\beta}}{\mathrm{2}}\:\boldsymbol{{d}\beta} \\ $$$$\:\:\boldsymbol{{I}}\:\:\:=\:\:\:−\frac{\boldsymbol{\beta}^{\mathrm{2}} }{\mathrm{2}}\:+\boldsymbol{{c}} \\ $$$$\boldsymbol{{I}}\:=\:\:−\frac{\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \boldsymbol{\alpha}\right)^{\mathrm{2}} \:}{\mathrm{2}}\:+\:\boldsymbol{{c}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{\alpha}\:=\:\mathrm{0}\:\:\boldsymbol{{then}}\:\boldsymbol{{I}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{c}}\:\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\:\:\boldsymbol{{I}}\:\:=\:\:\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}}\:−\:\frac{\left(\boldsymbol{\mathrm{cos}}^{−\mathrm{1}} \boldsymbol{\alpha}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\boldsymbol{{now}}\:\boldsymbol{{put}}\:\boldsymbol{\alpha}\:=\:\mathrm{1} \\ $$$$\boldsymbol{{I}}\:\:=\:\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{8}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *