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Question Number 117437 by frc2crc last updated on 11/Oct/20

∫_0 ^∞ (dx/(a^3 +x^3 )) generaly ∫_0 ^∞ (dx/(p+x^n ))

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{a}^{\mathrm{3}} +{x}^{\mathrm{3}} } \\ $$$$ \\ $$$${generaly} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{p}+{x}^{{n}} } \\ $$

Answered by AbduraufKodiriy last updated on 11/Oct/20

Commented by AbduraufKodiriy last updated on 11/Oct/20

Sorry for my mistakes

$$\boldsymbol{{Sorry}}\:\boldsymbol{{for}}\:\boldsymbol{{my}}\:\boldsymbol{{mistakes}} \\ $$

Answered by Bird last updated on 12/Oct/20

A_p =∫_0 ^∞ (dx/(p+x^n )) if p>0 we hsve A_p =∫_0 ^∞ (dx/((p^(1/n) )^n +x^n )) =(1/p)∫_0 ^∞ (dx/(1+((x/p^(1/n) ))^n )) we do the changement (x/p^(1/n) ) =t ⇒A_p =(1/p)∫_0 ^∞ (1/(1+t^n ))p^(1/(n )) dt =p^((1/n)−1) ∫_0 ^∞ (dt/(1+t^n )) =_(t =z^(1/n) ) =p^((1/n)−1) ∫_0 ^∞ (1/(1+z))(1/n)z^((1/n)−1) dz =(1/n)p^((1/n)−1) ∫_0 ^∞ (z^((1/n)−1) /(1+z))dz A_p =(1/n)p^((1/n)−1) ×(π/(sin((π/n)))) ⇒ A_p =((π p^((1/n)−1) )/(nsin((π/n))))

$${A}_{{p}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{p}+{x}^{{n}} }\:\:{if}\:{p}>\mathrm{0}\:{we}\:{hsve} \\ $$$${A}_{{p}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({p}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}} \:+{x}^{{n}} } \\ $$$$=\frac{\mathrm{1}}{{p}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+\left(\frac{{x}}{{p}^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} }\:{we}\:{do}\:{the}\:{changement} \\ $$$$\frac{{x}}{{p}^{\frac{\mathrm{1}}{{n}}} }\:={t}\:\Rightarrow{A}_{{p}} =\frac{\mathrm{1}}{{p}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{{n}} }{p}^{\frac{\mathrm{1}}{{n}\:}} \:\:{dt} \\ $$$$={p}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{{n}} }\:=_{{t}\:={z}^{\frac{\mathrm{1}}{{n}}} } \\ $$$$={p}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}+{z}}\frac{\mathrm{1}}{{n}}{z}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dz} \\ $$$$=\frac{\mathrm{1}}{{n}}{p}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\frac{{z}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{z}}{dz} \\ $$$${A}_{{p}} =\frac{\mathrm{1}}{{n}}{p}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} ×\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}\:\Rightarrow \\ $$$${A}_{{p}} =\frac{\pi\:{p}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{{nsin}\left(\frac{\pi}{{n}}\right)} \\ $$

Commented by Bird last updated on 12/Oct/20

∫_0 ^∞ (dx/(a^3 +x^3 )) we take a^3 =p and n=3 ⇒∫_0 ^∞ (dx/(a^3 +x^3 )) =((π (a)^(3((1/3)−1)) )/(3sin((π/3)))) =((π a^(−2) )/(3.((√3)/2))) =((2πa^(−2) )/(3(√3))) =((2π)/(3a^2 (√3)))

$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{a}^{\mathrm{3}} +{x}^{\mathrm{3}} }\:\:{we}\:{take}\:{a}^{\mathrm{3}} ={p}\:{and}\:{n}=\mathrm{3} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{a}^{\mathrm{3}} \:+{x}^{\mathrm{3}} }\:=\frac{\pi\:\left({a}\right)^{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)} }{\mathrm{3}{sin}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$$=\frac{\pi\:{a}^{−\mathrm{2}} }{\mathrm{3}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\frac{\mathrm{2}\pi{a}^{−\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}^{\mathrm{2}} \sqrt{\mathrm{3}}} \\ $$

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