Menu Close

let-f-a-0-dx-x-n-a-n-with-n-integr-2-and-a-gt-0-1-calculate-f-a-intems-of-a-2-let-g-a-0-dx-x-n-a-n-2-calculate-g-a-interms-of-a-3-find-the-values-of-i

Tinku Tara 0 Comments
Pin




Question Number 56345 by maxmathsup by imad last updated on 14/Mar/19
let f(a) =∫_0 ^∞ (dx/(x^n +a^n )) with n integr ≥2 and a>0 1) calculate f(a) intems of a 2) let g(a) =∫_0 ^∞ (dx/((x^n +a^n )^2 )) calculate g(a) interms of a 3) find the values of integrals ∫_0 ^∞ (dx/(x^8 +16)) and ∫_0 ^∞ (dx/((x^8 +16)^2 ))
$${let}\:\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{x}^{{n}} \:+{a}^{{n}} }\:\:\:{with}\:{n}\:{integr}\:\geqslant\mathrm{2}\:\:{and}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:{intems}\:{of}\:{a} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{{n}} \:+{a}^{{n}} \right)^{\mathrm{2}} }\:\:{calculate}\:{g}\left({a}\right)\:{interms}\:{of}\:{a} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{8}} +\mathrm{16}}\:\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{8}} \:+\mathrm{16}\right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 17/Mar/19
1) changement x=at give f(a) =∫_0 ^∞ ((adt)/(a^n t^n +a^n )) =(1/a^(n−1) ) ∫_0 ^∞ (dt/(t^n +1)) ∫_0 ^∞ (dt/(t^n +1)) =_(t=u^(1/n) ) ∫_0 ^∞ (1/(n(1+u))) u^((1/n)−1) du =(1/n) ∫_0 ^∞ (u^((1/n)−1) /(1+u)) du =(1/n) (π/(sin((π/n)))) =(π/(nsin((π/n)))) (we have proved that ∫_0 ^∞ (x^(a−1) /(1+x)) dx =(π/(sin(πa))) with 0<a<1) ⇒ ★f(a) =(π/(n a^(n−1) sin((π/n))))★ with n≥2 2)we have f^′ (a) =∫_0 ^∞ (∂/∂a)((1/(x^n +a^n )))dx =−∫_0 ^∞ ((na^(n−1) )/((x^n +a^n )^2 ))dx =−na^(n−1) g(a) ⇒g(a) =−((f(a))/(na^(n−1) )) =−(1/(na^(n−1) )) (π/(n a^(n−1) sin((π/n)))) ⇒ ★g(a)=−(π/(n^2 a^(2n−2) sin((π/n)))) ★ 3) we have ∫_0 ^∞ (dx/(x^8 +16)) =∫_0 ^∞ (dx/(x^8 +((√2))^8 )) ⇒n=8 and a=(√2) ⇒ ∫_0 ^∞ (dx/(x^8 +16)) = (π/(8((√2))^7 sin((π/8)))) also we have ∫_0 ^∞ (dx/((x^8 +16)^2 )) =∫_0 ^∞ (dx/((x^2 +((√2))^8 )^2 )) ⇒n=8 and a=(√2) at g(a) ⇒ ∫_0 ^∞ (dx/((x^8 +16)^2 )) =(π/(64 ((√2))^(14) sin((π/8)))) =(π/(64 .2^7 sin((π/8)))) with sin((π/8)) =((√(2−(√2)))/2) .
$$\left.\mathrm{1}\right)\:{changement}\:{x}={at}\:{give}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{adt}}{{a}^{{n}} {t}^{{n}} \:+{a}^{{n}} }\:=\frac{\mathrm{1}}{{a}^{{n}−\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{{n}} \:+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{{n}} \:+\mathrm{1}}\:=_{{t}={u}^{\frac{\mathrm{1}}{{n}}} } \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{n}\left(\mathrm{1}+{u}\right)}\:{u}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{du}\:=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\:\:\:\:\left({we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}}\:{dx}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\:\mathrm{0}<{a}<\mathrm{1}\right)\:\Rightarrow \\ $$$$\bigstar{f}\left({a}\right)\:=\frac{\pi}{{n}\:{a}^{{n}−\mathrm{1}} {sin}\left(\frac{\pi}{{n}}\right)}\bigstar\:\:{with}\:{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\partial}{\partial{a}}\left(\frac{\mathrm{1}}{{x}^{{n}} \:+{a}^{{n}} }\right){dx}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{na}^{{n}−\mathrm{1}} }{\left({x}^{{n}} \:+{a}^{{n}} \right)^{\mathrm{2}} }{dx} \\ $$$$=−{na}^{{n}−\mathrm{1}} \:{g}\left({a}\right)\:\Rightarrow{g}\left({a}\right)\:=−\frac{{f}\left({a}\right)}{{na}^{{n}−\mathrm{1}} }\:=−\frac{\mathrm{1}}{{na}^{{n}−\mathrm{1}} }\:\frac{\pi}{{n}\:{a}^{{n}−\mathrm{1}} {sin}\left(\frac{\pi}{{n}}\right)}\:\Rightarrow \\ $$$$\bigstar{g}\left({a}\right)=−\frac{\pi}{{n}^{\mathrm{2}} \:{a}^{\mathrm{2}{n}−\mathrm{2}} \:{sin}\left(\frac{\pi}{{n}}\right)}\:\bigstar \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{8}} \:+\mathrm{16}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{8}} \:+\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }\:\:\Rightarrow{n}=\mathrm{8}\:{and}\:{a}=\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{8}} \:+\mathrm{16}}\:=\:\frac{\pi}{\mathrm{8}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} \:{sin}\left(\frac{\pi}{\mathrm{8}}\right)}\:\:{also}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{8}} \:+\mathrm{16}\right)^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\left(\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \right)^{\mathrm{2}} }\:\Rightarrow{n}=\mathrm{8}\:{and}\:{a}=\sqrt{\mathrm{2}}\:\:{at}\:{g}\left({a}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{8}} \:+\mathrm{16}\right)^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{64}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{14}} \:{sin}\left(\frac{\pi}{\mathrm{8}}\right)}\:=\frac{\pi}{\mathrm{64}\:.\mathrm{2}^{\mathrm{7}} \:{sin}\left(\frac{\pi}{\mathrm{8}}\right)}\:\:\:{with} \\ $$$${sin}\left(\frac{\pi}{\mathrm{8}}\right)\:=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:. \\ $$

Pin

Leave a Reply

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