Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 63510 by turbo msup by abdo last updated on 05/Jul/19

let f(x)=∫_0 ^∞ (t^(a−1) /(x+t)) dt with x>0 and 0<a<1 1)calculate f(x) 2)calculate g(x)=∫_0 ^∞ (t^(a−1) /((x+t)^2 ))dt 3)find the value of∫_0 ^∞ (t^(a−1) /((1+t)^2 ))dt

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}}\:{dt}\:{with}\:{x}>\mathrm{0} \\ $$ $${and}\:\mathrm{0}<{a}<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right){calculate}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right){calculate}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}\right)^{\mathrm{2}} }{dt} \\ $$ $$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$

Commented bymathmax by abdo last updated on 06/Jul/19

1) changement t=xu give f(x)=∫_0 ^∞ (((xu)^(a−1) )/(x+xu)) xdu =x^(a−1) ∫_0 ^∞ (u^(a−1) /(1+u))du =x^(a−1) (π/(sin(πa))) ⇒f(x)=((π x^(a−1) )/(sin(πa))) 2)we have f^′ (x) =∫_0 ^∞ (∂/∂x)((t^(a−1) /(x+t)))dt =−∫_0 ^∞ (t^(a−1) /((x+t)^2 )) dt =−g(x) ⇒ g(x)=−f^′ (x) but f(x) =(π/(sin(πa))) e^((a−1)ln(x)) ⇒ f^′ (x) =((π(a−1))/(xsin(πa))) x^(a−1) =((π(a−1) x^(a−2) )/(sin(πa))) ⇒g(x)=((π(1−a))/(sin(πa))) x^(a−2) 3) ∫_0 ^∞ (t^(a−1) /((1+t)^2 )) dt =g(1) =((π(1−a))/(sin(πa))) (0<a<1) . remark we have for all integr n f^((n)) (x) =∫_0 ^∞ (((−1)^n n! t^(a−1) )/((x+t)^(n+1) )) dt ⇒ ∫_0 ^∞ (t^(a−1) /((x+t)^(n+1) ))dt =(((−1)^n )/(n!)) f^((n)) (x) we have f(x) =(π/(sin(πa))) e^((a−1)ln(x)) let find (e^(λln(x)) )^((n)) (e^(λlnx) )^((1)) = (λ/x) e^(λln(x)) ⇒(e^(λlnx) )^((2)) =(λ^2 /x^2 ) e^(λln(x)) ⇒(e^(λln(x)) )^((n)) =(λ^n /x^n ) e^(λlnx) ⇒ f^((n)) (x) =(π/(sin(πa))) (((a−1)^n )/x^n ) x^(a−1) =((π (a−1)^n )/(sin(πa))) x^(a−n−1) ⇒ ∫_0 ^∞ (t^(a−1) /((x+t)^(n+1) ))dt =((π(1−a)^n )/(n! sin(πa))) x^(a−n−1) special case x=1 ⇒∫_0 ^∞ (t^(a−1) /((1+t)^(n+1) ))dt =((π(1−a)^n )/(n!sin(πa))) .

$$\left.\mathrm{1}\right)\:{changement}\:{t}={xu}\:{give}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left({xu}\right)^{{a}−\mathrm{1}} }{{x}+{xu}}\:{xdu}\:={x}^{{a}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\frac{{u}^{{a}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$ $$={x}^{{a}−\mathrm{1}} \:\frac{\pi}{{sin}\left(\pi{a}\right)}\:\Rightarrow{f}\left({x}\right)=\frac{\pi\:{x}^{{a}−\mathrm{1}} }{{sin}\left(\pi{a}\right)} \\ $$ $$\left.\mathrm{2}\right){we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial}{\partial{x}}\left(\frac{{t}^{{a}−\mathrm{1}} }{{x}+{t}}\right){dt}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}\right)^{\mathrm{2}} }\:{dt}\:=−{g}\left({x}\right)\:\Rightarrow \\ $$ $${g}\left({x}\right)=−{f}^{'} \left({x}\right)\:\:\:{but}\:\:{f}\left({x}\right)\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{e}^{\left({a}−\mathrm{1}\right){ln}\left({x}\right)} \:\Rightarrow \\ $$ $${f}^{'} \left({x}\right)\:=\frac{\pi\left({a}−\mathrm{1}\right)}{{xsin}\left(\pi{a}\right)}\:{x}^{{a}−\mathrm{1}} \:\:=\frac{\pi\left({a}−\mathrm{1}\right)\:{x}^{{a}−\mathrm{2}} }{{sin}\left(\pi{a}\right)}\:\Rightarrow{g}\left({x}\right)=\frac{\pi\left(\mathrm{1}−{a}\right)}{{sin}\left(\pi{a}\right)}\:{x}^{{a}−\mathrm{2}} \\ $$ $$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:{dt}\:={g}\left(\mathrm{1}\right)\:=\frac{\pi\left(\mathrm{1}−{a}\right)}{{sin}\left(\pi{a}\right)}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0}<{a}<\mathrm{1}\right)\:. \\ $$ $${remark}\:\:\:{we}\:{have}\:{for}\:{all}\:{integr}\:{n} \\ $$ $${f}^{\left({n}\right)} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!\:{t}^{{a}−\mathrm{1}} }{\left({x}+{t}\right)^{{n}+\mathrm{1}} }\:{dt}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}\right)^{{n}+\mathrm{1}} }{dt}\:=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{f}^{\left({n}\right)} \left({x}\right)\: \\ $$ $${we}\:{have}\:{f}\left({x}\right)\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{e}^{\left({a}−\mathrm{1}\right){ln}\left({x}\right)} \:\:{let}\:{find}\:\:\left({e}^{\lambda{ln}\left({x}\right)} \right)^{\left({n}\right)} \\ $$ $$\left({e}^{\lambda{lnx}} \right)^{\left(\mathrm{1}\right)} \:=\:\frac{\lambda}{{x}}\:{e}^{\lambda{ln}\left({x}\right)} \:\Rightarrow\left({e}^{\lambda{lnx}} \right)^{\left(\mathrm{2}\right)} \:=\frac{\lambda^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:{e}^{\lambda{ln}\left({x}\right)} \:\Rightarrow\left({e}^{\lambda{ln}\left({x}\right)} \right)^{\left({n}\right)} \:=\frac{\lambda^{{n}} }{{x}^{{n}} }\:{e}^{\lambda{lnx}} \:\Rightarrow \\ $$ $${f}^{\left({n}\right)} \left({x}\right)\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:\frac{\left({a}−\mathrm{1}\right)^{{n}} }{{x}^{{n}} }\:{x}^{{a}−\mathrm{1}} \:=\frac{\pi\:\left({a}−\mathrm{1}\right)^{{n}} }{{sin}\left(\pi{a}\right)}\:{x}^{{a}−{n}−\mathrm{1}} \:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left({x}+{t}\right)^{{n}+\mathrm{1}} }{dt}\:=\frac{\pi\left(\mathrm{1}−{a}\right)^{{n}} }{{n}!\:{sin}\left(\pi{a}\right)}\:{x}^{{a}−{n}−\mathrm{1}} \\ $$ $${special}\:{case}\:\:{x}=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{n}+\mathrm{1}} }{dt}\:=\frac{\pi\left(\mathrm{1}−{a}\right)^{{n}} }{{n}!{sin}\left(\pi{a}\right)}\:. \\ $$ $$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com