Question Number 66470 by mathmax by abdo last updated on 15/Aug/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{{n}} \:+\mathrm{8}\right)^{\mathrm{3}} }\:\:{withn}>\mathrm{1} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19
![let state ∀ a>0 f(a)=∫_0 ^∞ (dx/((x^n +a)^2 )) f(a)=∫_0 ^∞ (dx/(a^2 [((x/a^(1/n) ))^n +1]^2 )) when changing x=a^(1/n) u we get f(a)= a^((1/n) −2) ∫_0 ^∞ (du/((u^n +1)^2 )) now change v=(1/(u^n +1)) ⇒ u=((1/v) −1)^(1/n) ⇒ du=(1/n).(((−1)/v^2 ))((1/v) −1)^((1/n) −1) dv then f(a)=(1/n). a^((1/n)−2) ∫_0 ^1 v^2 .((1/v^2 ))((1/v) −1)^((1/n) −1) dv = (a^((1/n)−2) /n) ∫_0 ^1 v^(1−(1/n)) (1−v)^((1/n)−1) dv = (a^((1/n) −2) /n) B(2−(1/n) , (1/n)) = (a^((1/n) −2) /n) .((Γ(2−(1/n))Γ((1/n)))/(Γ(2))) as Γ(x+1)=xΓ(x) we have Γ(2−(1/n))=(1−(1/n))Γ(1−(1/n)) and Γ(2)=Γ(1)=1 So f(a)=(a^((1/n) −2) /n) .(1−(1/n))Γ(1−(1/n))Γ((1/n)) as Γ(1−z)Γ(z)=(π/(sin(πz))) we finally get f(a)= (1/n)(1−(1/n))a^((1/n) −2) .(π/(sin((π/n)))) Now we can deduce f(8) , f(3), f(n)](https://www.tinkutara.com/question/Q66488.png)
$${let}\:{state}\:\forall\:{a}>\mathrm{0}\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{{n}} +{a}\right)^{\mathrm{2}} }\: \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{a}^{\mathrm{2}} \left[\left(\frac{{x}}{{a}^{\frac{\mathrm{1}}{{n}}} }\right)^{{n}} +\mathrm{1}\right]^{\mathrm{2}} }\:\:\: \\ $$$${when}\:{changing}\:\:\:{x}={a}^{\frac{\mathrm{1}}{{n}}} {u}\:\:\:\:\:\:\:{we}\:{get} \\ $$$${f}\left({a}\right)=\:{a}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left({u}^{{n}} +\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$${now}\:{change}\:{v}=\frac{\mathrm{1}}{{u}^{{n}} \:+\mathrm{1}}\:\:\:\Rightarrow\:{u}=\left(\frac{\mathrm{1}}{{v}}\:−\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}} \Rightarrow\:{du}=\frac{\mathrm{1}}{{n}}.\left(\frac{−\mathrm{1}}{{v}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{{v}}\:−\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}\:−\mathrm{1}} {dv} \\ $$$${then} \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{{n}}.\:{a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{v}^{\mathrm{2}} .\left(\frac{\mathrm{1}}{{v}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{{v}}\:−\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}\:−\mathrm{1}} {dv} \\ $$$$\:\:\:\:=\:\frac{{a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{v}^{\mathrm{1}−\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}−{v}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dv} \\ $$$$\:\:\:\:\:\:=\:\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{2}} }{{n}}\:\:{B}\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\:,\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:=\:\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{2}} }{{n}}\:.\frac{\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right)}{\Gamma\left(\mathrm{2}\right)}\: \\ $$$${as}\:\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:\:{we}\:{have}\:\:\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)=\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\:\:\:{and}\:\Gamma\left(\mathrm{2}\right)=\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${So}\:\:{f}\left({a}\right)=\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{2}} }{{n}}\:.\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$${as}\:\:\Gamma\left(\mathrm{1}−{z}\right)\Gamma\left({z}\right)=\frac{\pi}{{sin}\left(\pi{z}\right)}\:\:{we}\:{finally}\:{get} \\ $$$$\:\: \\ $$$${f}\left({a}\right)=\:\frac{\mathrm{1}}{{n}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{2}} .\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}\: \\ $$$${Now}\:{we}\:{can}\:{deduce}\:\:{f}\left(\mathrm{8}\right)\:,\:{f}\left(\mathrm{3}\right),\:{f}\left({n}\right) \\ $$
Commented by ~ À ® @ 237 ~ last updated on 16/Aug/19
![if ∀ p>2 g_p (a)=∫_0 ^∞ (dx/((x^n +a)^p )) start as the previous and reach to g_p (a)=(a^((1/n) −p) /n) ∫_(0 ) ^1 v^p .((1/v^2 ))((1/v)−1)^((1/n)−1) dv = (a^((1/n) −p) /n) ∫_0 ^1 v^(p−2+1−(1/n)) (1−v)^((1/n)−1) dv =(a^((1/n)−p) /n) B(p−(1/n), (1/n)) =(a^((1/n)−p) /n) .((Γ(p−(1/n))Γ((1/n)))/(Γ(p))) as Γ(p+z)=z(z+1).......(z+p−1)Γ(z) we have Γ(p−(1/n))=Γ(p−1+(1−(1/n)))=Π_(k=0) ^(p−2) [(1−(1/n))+k] Γ(1−(1/n)) Now g_p (a)=(a^((1/n) −p) /(n.(p−1)!)) .(Π_(k=0) ^(p−2) [(1−(1/n))+k]) Γ(1−(1/n))Γ((1/n)) =(a^((1/n)−p) /(n.(p−1)!)) .(π/(sin((π/n)))) .Π_(k=0) ^(p−2) [(1−(1/n))+k] now we can deduce g_3 (8) ,∙.......](https://www.tinkutara.com/question/Q66490.png)
$$ \\ $$$$ \\ $$$$ \\ $$$${if}\:\:\forall\:{p}>\mathrm{2}\:\:\:\:{g}_{{p}} \left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{{n}} +{a}\right)^{{p}} }\: \\ $$$${start}\:{as}\:{the}\:{previous}\:{and}\:{reach}\:{to} \\ $$$${g}_{{p}} \left({a}\right)=\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−{p}} }{{n}}\:\:\int_{\mathrm{0}\:} ^{\mathrm{1}} \:\:{v}^{{p}} .\left(\frac{\mathrm{1}}{{v}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{{v}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dv} \\ $$$$\:\:\:\:\:\:\:=\:\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−{p}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{v}^{{p}−\mathrm{2}+\mathrm{1}−\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}−{v}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dv} \\ $$$$\:\:\:\:\:=\frac{{a}^{\frac{\mathrm{1}}{{n}}−{p}} }{{n}}\:{B}\left({p}−\frac{\mathrm{1}}{{n}},\:\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:\:\:=\frac{{a}^{\frac{\mathrm{1}}{{n}}−{p}} }{{n}}\:.\frac{\Gamma\left({p}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right)}{\Gamma\left({p}\right)}\:\: \\ $$$${as}\:\:\Gamma\left({p}+{z}\right)={z}\left({z}+\mathrm{1}\right)…….\left({z}+{p}−\mathrm{1}\right)\Gamma\left({z}\right)\:{we}\:{have}\:\Gamma\left({p}−\frac{\mathrm{1}}{{n}}\right)=\Gamma\left({p}−\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\right)=\underset{{k}=\mathrm{0}} {\overset{{p}−\mathrm{2}} {\prod}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)+{k}\right]\:\:\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right) \\ $$$${Now}\:\: \\ $$$${g}_{{p}} \left({a}\right)=\frac{{a}^{\frac{\mathrm{1}}{{n}}\:−{p}} }{{n}.\left({p}−\mathrm{1}\right)!}\:.\left(\underset{{k}=\mathrm{0}} {\overset{{p}−\mathrm{2}} {\prod}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)+{k}\right]\right)\:\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$\:\:=\frac{{a}^{\frac{\mathrm{1}}{{n}}−{p}} }{{n}.\left({p}−\mathrm{1}\right)!}\:.\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}\:.\underset{{k}=\mathrm{0}} {\overset{{p}−\mathrm{2}} {\prod}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)+{k}\right] \\ $$$${now}\:{we}\:{can}\:{deduce}\:\:{g}_{\mathrm{3}} \left(\mathrm{8}\right)\:,\centerdot……. \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
$${thank}\:{you}\:{sir}. \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
$${let}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:\:{with}\:{a}>\mathrm{0}\:\:{we}\:{have}\:{proved}\:{that} \\ $$$${g}\left({a}\right)\:=\frac{\pi\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} }{{nsin}\left(\frac{\pi}{{n}}\right)}\:\:\:{also}\:{we}\:{have}\: \\ $$$${g}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}\left({a}+{x}^{{n}} \right)}{\left({a}+{x}^{{n}} \right)^{\mathrm{4}} }{dx}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}{g}^{'} \left({a}\right)\:\:{we}\:{have}\: \\ $$$${g}^{'} \left({a}\right)\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\frac{\mathrm{1}}{{n}}−\mathrm{2}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{3}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{2}\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{3}} \\ $$$${a}=\mathrm{8}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{8}+{x}^{{n}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{2}\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)\mathrm{8}^{\frac{\mathrm{1}}{{n}}−\mathrm{3}} \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
$${error}\:{at}\:{final}\:{line}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{3}} }\:=\frac{\pi}{\mathrm{2}{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{3}} \\ $$$${a}=\mathrm{8}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{8}+{x}^{{n}} \right)^{\mathrm{3}} }\:=\frac{\pi}{\mathrm{2}{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)\mathrm{8}^{\frac{\mathrm{1}}{{n}}−\mathrm{3}} \\ $$