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Question Number 29003 by abdo imad last updated on 03/Feb/18
find ∫_0 ^∞ (dx/(1+x^3 )) .
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:. \\ $$
Commented by abdo imad last updated on 03/Feb/18
let put I= ∫_0 ^∞ (dx/(1+x^3 )) let put x^3 =t ⇔x=t^(1/3) and I= ∫_0 ^∞ (1/(1+t)) (1/3)t^((1/3)−1) dt=(1/3) ∫_0 ^∞ (t^((1/3)−1) /(1+t))dt but we know that ∫_0 ^∞ (t^(a−1) /(1+t))dt= (π/(sin(πa))) with0<a<1 due to that ∫_0 ^∞ (t^((1/3)−1) /(1+t))dt= (π/(sin((π/3))))= (π/((√3)/2))=((2π)/( (√3))) ⇒ I= ((2π)/(3(√3))) .
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\:{let}\:{put}\:{x}^{\mathrm{3}} ={t}\:\Leftrightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:{and} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} \:{dt}=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:{but}\:{we}\:{know}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\mathrm{0}<{a}<\mathrm{1}\:{due}\:{to}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}=\:\frac{\pi}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{I}=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\:. \\ $$
Answered by sma3l2996 last updated on 03/Feb/18
we have (1/(1+x^3 ))=(1/((x+1)(x^2 −x+1)))=(a/(1+x))+((bx+c)/(x^2 −x+1)) a=(1/3) , c=(2/3) , b=−(1/3) (1/(1+x^3 ))=(1/3)((1/(1+x))−((x−2)/(x^2 −x+1))) so ∫_0 ^∞ (dx/(1+x^3 ))=(1/3)∫_0 ^∞ ((1/(x+1))−((2x−4)/(2(x^2 −x+1))))dx =(1/3)(lim_(x→∞) ln(x+1)−(1/2)∫_0 ^∞ ((2x−1−3)/(x^2 −x+1))dx) =(1/3)lim_(x→∞) (ln(x+1)−(1/2)ln(x^2 −x+1))−(1/2)∫_0 ^∞ (dx/(x^2 −x+1)) ∫_0 ^∞ (dx/(x^2 −x+1))=∫_0 ^∞ (dx/((x−(1/2))^2 +(3/4)))=(4/3)∫_0 ^∞ (dx/((((2x−1)/( (√3))))^2 +1)) t=((2x−1)/( (√3)))⇒dx=((√3)/2)dt ∫_0 ^∞ (dx/(x^2 −x+1))=((2(√3))/3)∫_0 ^∞ (dt/(t^2 +1))=((2(√3))/3)(lim_(x→∞) arctant−0) =(((√3)π)/3) A=lim_(x→∞) (ln(x+1)−(1/2)ln(x^2 −x+1)) =lim_(x→∞) (ln(x(1+(1/x)))−(1/2)ln(x^2 (1−(1/x)+(1/x^2 ))) =lim_(x→∞) (lnx+ln(1+1/x)−ln(x)−(1/2)ln(1−1/x+1/x^2 )) =lim_(x→∞) (ln(1+(1/x))−(1/2)ln(1−(1/x)+(1/x^2 )))=0 so ∫_0 ^∞ (dx/(1+x^3 ))=(1/3)×(0)−(1/2)∫_0 ^∞ (dx/(x^2 −x+1))=−(1/2)×(((√3)π)/( (√3))) =−((2(√3)π)/3)
$${we}\:{have}\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}=\frac{{a}}{\mathrm{1}+{x}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}}\:\:,\:\:{c}=\frac{\mathrm{2}}{\mathrm{3}}\:,\:{b}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\right) \\ $$$${so}\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\underset{{x}\rightarrow\infty} {{lim}ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\underset{{x}\rightarrow\infty} {{lim}}\left({ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${t}=\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\underset{{x}\rightarrow\infty} {{lim}arctant}−\mathrm{0}\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{3}} \\ $$$${A}=\underset{{x}\rightarrow\infty} {{lim}}\left({ln}\left({x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\right) \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left({ln}\left({x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)\right. \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left({lnx}+{ln}\left(\mathrm{1}+\mathrm{1}/{x}\right)−{ln}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\mathrm{1}/{x}+\mathrm{1}/{x}^{\mathrm{2}} \right)\right) \\ $$$$=\underset{{x}\rightarrow\infty} {{lim}}\left({ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\right)=\mathrm{0} \\ $$$${so}\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}×\left(\mathrm{0}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$=−\frac{\mathrm{2}\sqrt{\mathrm{3}}\pi}{\mathrm{3}} \\ $$
Commented by abdo imad last updated on 03/Feb/18
we have (1/(1+x^3 ))>0 ∀ x≥0 so ∫_0 ^∞ (dx/(1+x^3 ))>0 so how do you find a ngative value sir ?...
$${we}\:{have}\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }>\mathrm{0}\:\:\forall\:{x}\geqslant\mathrm{0}\:\:\:{so}\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }>\mathrm{0}\:{so}\:{how}\:{do}\:{you} \\ $$$${find}\:{a}\:{ngative}\:{value}\:{sir}\:?… \\ $$

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