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Question Number 63664 by mathmax by abdo last updated on 07/Jul/19

let f(x)=∫_0 ^∞ (t^(a−1) /(x+t^n )) dt with 0<a<1 and x>0 and n≥2 1) determine a explicit form of f(x) 2) calculate g(x) =∫_0 ^∞ (t^(a−1) /((x+t^n )^2 )) dt 3) find f^((k)) (x) at form of integrals 4) calculate ∫_0 ^∞ (t^(a−1) /(9+t^2 )) dt and ∫_0 ^∞ (t^(a−1) /((9+t^2 )^2 )) 5) calculate U_n =∫_0 ^∞ (t^((1/n)−1) /(2^n +t^n )) dt and study the convergence of Σ U_n

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\:{dt}\:\:\:{with}\:\mathrm{0}<{a}<\mathrm{1}\:\:{and}\:\:{x}>\mathrm{0}\:{and}\:{n}\geqslant\mathrm{2} \\ $$ $$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }\:{dt} \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{f}^{\left({k}\right)} \left({x}\right)\:{at}\:{form}\:{of}\:{integrals} \\ $$ $$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{9}+{t}^{\mathrm{2}} }\:{dt}\:\:\:\:{and}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{9}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$\left.\mathrm{5}\right)\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{2}^{{n}} \:+{t}^{{n}} }\:{dt}\:\:{and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

1)we have f(x)=∫_0 ^∞ (t^(a−1) /(x +t^n ))dt changement t =^n (√x)u give f(x) =∫_0 ^∞ (((^n (√x)u)^(a−1) )/(x +xu^n ))^n (√x)du =(x^(((a−1)/n)+(1/n)) /x) ∫_0 ^∞ (u^(a−1) /(1+u^n )) du =x^((a/n)−1) ∫_0 ^∞ (u^(a−1) /(1+u^n )) =_(u=α^(1/n) ) x^((a/n)−1) ∫_0 ^∞ (α^((a−1)/n) /(1+α)) (1/n) α^((1/n)−1) dα = x^((a/n)−1) ∫_0 ^∞ (α^(((a−1)/n)+(1/n)−1) /(1+α)) dα = x^((a/n)−1) ∫_0 ^∞ (α^((a/n)−1) /(1+α)) dα ⇒ f(x)=x^((a/n)−1) (π/(sin(((πa)/n))))

$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}\:+{t}^{{n}} }{dt}\:\:{changement}\:{t}\:=^{{n}} \sqrt{{x}}{u}\:{give} \\ $$ $${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(^{{n}} \sqrt{{x}}{u}\right)^{{a}−\mathrm{1}} }{{x}\:+{xu}^{{n}} }\:^{{n}} \sqrt{{x}}{du} \\ $$ $$=\frac{{x}^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}} }{{x}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{{a}−\mathrm{1}} }{\mathrm{1}+{u}^{{n}} }\:{du}\:={x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{{a}−\mathrm{1}} }{\mathrm{1}+{u}^{{n}} } \\ $$ $$=_{{u}=\alpha^{\frac{\mathrm{1}}{{n}}} } \:\:\:\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{{a}−\mathrm{1}}{{n}}} }{\mathrm{1}+\alpha}\:\frac{\mathrm{1}}{{n}}\:\alpha^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{d}\alpha \\ $$ $$=\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\alpha^{\frac{{a}−\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha\:=\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{{a}}{{n}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha\:\Rightarrow \\ $$ $${f}\left({x}\right)={x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\:\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)} \\ $$ $$ \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

2) we have by derivation f^′ (x) =∫_0 ^∞ (∂/∂x)((t^(a−1) /(x+t^n )))dt =−∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =−g(x) ⇒g(x)=−f^′ (x) f(x)=(π/(sin(((πa)/n)))) x^((a/n)−1) ⇒f^′ (x) =((π((a/n)−1))/(sin(((πa)/n)))) x^((a/n)−2) ⇒ g(x)=∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =((π(1−(a/n)))/(sin(((πa)/n)))) x^((a/n)−2)

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{by}\:{derivation}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{x}}\left(\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\right){dt} \\ $$ $$=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=−{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)=−{f}^{'} \left({x}\right) \\ $$ $${f}\left({x}\right)=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\pi\left(\frac{{a}}{{n}}−\mathrm{1}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \:\Rightarrow \\ $$ $${g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{{n}}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

3) we have f(x)=∫_0 ^∞ (t^(a−1) /(x +t^n )) dt ⇒f^((k)) (x) =∫_0 ^∞ (∂^k /∂x^k )((t^(a−1) /(x+t^n )))dt =∫_0 ^∞ (((−1)^k k!)/((x+t^n )^(k+1) )) t^(a−1) dt ⇒f^((k)) (x) =(−1)^k k! ∫_0 ^∞ (t^(a−1) /((x+t^n )^(k+1) ))dt

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}\:+{t}^{{n}} }\:{dt}\:\Rightarrow{f}^{\left({k}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial^{{k}} }{\partial{x}^{{k}} }\left(\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\right){dt} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}+{t}^{{n}} \right)^{{k}+\mathrm{1}} }\:{t}^{{a}−\mathrm{1}} \:{dt}\:\:\:\Rightarrow{f}^{\left({k}\right)} \left({x}\right)\:=\left(−\mathrm{1}\right)^{{k}} {k}!\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{{k}+\mathrm{1}} }{dt} \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

4) we have proved that ∫_0 ^∞ (t^(a−1) /(x+t^n )) dt =(π/(sin(((πa)/n)))) x^((a/n)−1) and ∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =((π(1−(a/n)))/(sin(((πa)/n)))) x^((a/n)−2) x=9 and n=2 ⇒∫_0 ^∞ (t^(a−1) /(9+t^2 ))dt =(π/(sin(((πa)/2))))×9^((a/2)−1) ∫_0 ^∞ (t^(a−1) /((9+t^2 )^2 ))dt =((π(1−(a/2)))/(sin(((πa)/2))))×9^((a/2)−2)

$$\left.\mathrm{4}\right)\:{we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }\:{dt}\:\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \:{and} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}^{{n}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{{n}}\right)}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{2}} \\ $$ $${x}=\mathrm{9}\:{and}\:{n}=\mathrm{2}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{9}+{t}^{\mathrm{2}} }{dt}\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}×\mathrm{9}^{\frac{{a}}{\mathrm{2}}−\mathrm{1}} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{9}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\frac{\pi\left(\mathrm{1}−\frac{{a}}{\mathrm{2}}\right)}{{sin}\left(\frac{\pi{a}}{\mathrm{2}}\right)}×\mathrm{9}^{\frac{{a}}{\mathrm{2}}−\mathrm{2}} \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

5) we have ∫_0 ^∞ (t^(a−1) /(x+t^n ))dt =(π/(sin(((πa)/n)))) x^((a/n)−1) a=(1/n) and x =2^n ⇒∫_0 ^∞ (t^((1/n)−1) /(2^n +t^n ))dt = (π/(sin((π/n^2 ))))(2^n )^((1/n^2 )−1) =(π/(sin((π/n^2 ))))×2^((1/n)−n) =(π/(2^(n−(1/n)) sin((π/n^2 )))) .

$$\left.\mathrm{5}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}^{{n}} }{dt}\:=\frac{\pi}{{sin}\left(\frac{\pi{a}}{{n}}\right)}\:{x}^{\frac{{a}}{{n}}−\mathrm{1}} \\ $$ $${a}=\frac{\mathrm{1}}{{n}}\:\:{and}\:{x}\:=\mathrm{2}^{{n}} \:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{2}^{{n}} \:+{t}^{{n}} }{dt}\:=\:\frac{\pi}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\left(\mathrm{2}^{{n}} \right)^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}} \\ $$ $$=\frac{\pi}{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}×\mathrm{2}^{\frac{\mathrm{1}}{{n}}−{n}} \:=\frac{\pi}{\mathrm{2}^{{n}−\frac{\mathrm{1}}{{n}}} \:\:\:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)}\:. \\ $$

Commented bymathmax by abdo last updated on 09/Jul/19

⇒ U_n =(π/(2^(n−(1/n)) sin((π/n^2 ))))

$$\Rightarrow\:{U}_{{n}} =\frac{\pi}{\mathrm{2}^{{n}−\frac{\mathrm{1}}{{n}}} \:{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} \\ $$

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