Menu Close

Question-203385

Tinku Tara 0 Comments
Pin




Question Number 203385 by patrice last updated on 18/Jan/24
Answered by Mathspace last updated on 18/Jan/24
I=∫_0 ^1 (dx/(1+x^3 )) ⇒I=∫_0 ^1 Σ_(n=0) ^∞ (−1)^n x^(3n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(3n) dx =Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =Σ_(n=0) ^∞ (1/(6n+1))−Σ_(n=0) ^∞ (1/(3(2n+1)+1)) =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+4))) =Σ_(n=0) ^∞ (3/((6n+1)(6n+4))) =(1/(12))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(4/6)))) =(1/(12)){((Ψ((4/6))−Ψ((1/6)))/((4/6)−(1/6)))} =(1/(12))×2(Ψ((4/6))−Ψ((1/6))) =(1/6)(Ψ((2/3))−Ψ((1/6))) now use Ψ(s)=−γ+∫_0 ^1 ((1−x^(s−1) )/(1−x))dx to calculate Ψ((2/3))and Ψ((1/6))
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{3}{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{3}{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{1}}−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{4}}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{3}}{\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\mathrm{6}{n}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{6}}\right)\left({n}+\frac{\mathrm{4}}{\mathrm{6}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left\{\frac{\Psi\left(\frac{\mathrm{4}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\frac{\mathrm{4}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}×\mathrm{2}\left(\Psi\left(\frac{\mathrm{4}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right) \\ $$$${now}\:{use}\:\Psi\left({s}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}−\mathrm{1}} }{\mathrm{1}−{x}}{dx} \\ $$$${to}\:{calculate}\:\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right){and}\:\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$
Commented by MathematicalUser2357 last updated on 04/Mar/24
(1/6)[(−γ+∫_0 ^1 ((1−(1/( (x)^(1/3) )))/(1−x))dx)−{−γ+∫_0 ^1 ((1−(1/( ((x)^(1/6) )^5 )))/(1−x))dx}]
$$\frac{\mathrm{1}}{\mathrm{6}}\left[\left(−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{x}}}}{\mathrm{1}−{x}}{dx}\right)−\left\{−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{\:\left(\sqrt[{\mathrm{6}}]{{x}}\right)^{\mathrm{5}} }}{\mathrm{1}−{x}}{dx}\right\}\right] \\ $$
Answered by Mathspace last updated on 18/Jan/24
classic method f(x)=(1/(x^3 +1))=(1/((x+1)(x^2 −x+1))) =(a/(x+1))+((bx+c)/(x^2 −x+1)) a=(1/3) lim_(x→+∞) xf(x)=0=a+b ⇒ b=−(1/3) f(0)=1=a+c ⇒c=1−a =1−(1/3)=(2/3) ⇒ f(x)=(1/(3(x+1)))+((−(1/3)x+(2/3))/(x^2 −x+1)) ⇒∫f(x)dx=∫(dx/(3(x+1))) −(1/3)∫((x−2)/(x^2 −x+1))dx =(1/3)ln∣x+1∣−(1/6)∫((2x−1−3)/(x^2 −x+1))dx =(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1) +(1/2)∫(dx/(x^2 −x+1)) and ∫(dx/(x^2 −x+1))=∫(dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) ((√3)/2)∫(dt/((3/4)(1+t^2 ))) =(4/3).((√3)/2)∫(dt/(1+t^2 ))=(2/( (√3)))arctan(((2x−1)/( (√3)))) ⇒∫f(x)dx=(1/3)ln∣x+1∣ −(1/6)ln(x^2 −x+1)+(1/( (√3)))arctan(((2x−1)/( (√3)))) I=∫_0 ^1 f(x)dx =[(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1) +(1/( (√3)))arctan(((2x−1)/( (√3))))]_0 ^1 =(1/3)ln(2)+(1/( (√3)))arctan((1/( (√3)))) +(1/( (√3)))arctan((1/( (√3)))) =((ln2)/3)+(2/( (√3)))×(π/3)=((ln2)/2)+((2π)/(3(√3)))
$${classic}\:{method} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{{x}+\mathrm{1}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xf}\left({x}\right)=\mathrm{0}={a}+{b}\:\Rightarrow \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}={a}+{c}\:\Rightarrow{c}=\mathrm{1}−{a} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{2}}{\mathrm{3}}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\Rightarrow\int{f}\left({x}\right){dx}=\int\frac{{dx}}{\mathrm{3}\left({x}+\mathrm{1}\right)} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:{and} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{dt}}{\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow\int{f}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\right. \\ $$$$\left.+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\frac{{ln}\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\pi}{\mathrm{3}}=\frac{{ln}\mathrm{2}}{\mathrm{2}}+\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Pin

Leave a Reply

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