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Question Number 35983 by abdo mathsup 649 cc last updated on 26/May/18
calculate ∫_0 ^∞ ((2(√t) +1)/(t^5 +3))dt .
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}\sqrt{{t}}\:+\mathrm{1}}{{t}^{\mathrm{5}} \:\:\:+\mathrm{3}}{dt}\:\:. \\ $$
Commented by prof Abdo imad last updated on 27/May/18
let put I = ∫_0 ^∞ ((2(√t) +1)/(t^5 +3))dt changement (√t) =x give I = ∫_0 ^∞ ((2x +1)/(x^(10) +3))2xdx = 4 ∫_0 ^∞ (x^2 /(x^(10) +3))dx +2∫_0 ^∞ (x/(x^(10) +3))dx but changement x=3^(1/(10)) u give ∫_0 ^∞ (x^2 /(x^(10) +3))dx = ∫_0 ^∞ ((3^(1/5) u^2 )/(3u^(10) +3)) 3^(1/(10)) du = (3^(3/(10)) /3) ∫_0 ^∞ (u^2 /(1+u^(10) ))du then we use the chang. u^(10) =t ⇒u=t^(1/(10)) ⇒ ∫_0 ^∞ (u^2 /(1+u^(10) ))du =∫_0 ^∞ (t^(1/5) /(1+t)) (1/(10))t^((1/(10))−1) dt =(1/(10))∫_0 ^∞ (t^((3/(10))−1) /(1+t))dt =(1/(10)) (π/(sin(((3π)/(10))))) by the same chang. we get ∫_0 ^∞ (x/(x^(10) +3))dx = ∫_0 ^∞ ((3^(1/(10)) u)/(3u^(10) +3)) 3^(1/(10)) du =(3^(1/5) /3) ∫_0 ^∞ ((u du)/(1+u^(10) )) =3^((−4)/5) ∫_0 ^∞ (t^(1/(10)) /(1+t)) (1/(10)) t^((1/(10))−1) dt = (3^(−(4/5)) /(10)) ∫_0 ^∞ (t^((2/5)−1) /(1+t))dt = (3^(−(4/5)) /(10)) (π/(sin(((2π)/5)))) ⇒ I =(4/(10)) (π/(sin(((3π)/(10))))).(3^(3/(10)) /3) + 2 (3^(−(4/5)) /(10)) (π/(sin(((2π)/5)))) I = ((2π)/5) 3^(−(7/(10))) (1/(sin(((3π)/(10))))) +(π/5) 3^(−(4/5)) .(1/(sin(((2π)/5)))) I have used the formula ∫_0 ^∞ (t^(a−1) /(1+t))dt = (π/(sin(πa))) with 0<a<1 .
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}\sqrt{{t}}\:+\mathrm{1}}{{t}^{\mathrm{5}} \:+\mathrm{3}}{dt}\:{changement}\:\sqrt{{t}}\:={x} \\ $$$${give}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{x}\:+\mathrm{1}}{{x}^{\mathrm{10}} \:+\mathrm{3}}\mathrm{2}{xdx}\:=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{10}} \:+\mathrm{3}}{dx} \\ $$$$\:+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}}{{x}^{\mathrm{10}} \:+\mathrm{3}}{dx}\:\:\:{but}\:{changement}\:{x}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{10}}} \:{u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{10}} \:+\mathrm{3}}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{5}}} \:{u}^{\mathrm{2}} }{\mathrm{3}{u}^{\mathrm{10}} \:+\mathrm{3}}\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{10}}} \:{du} \\ $$$$=\:\frac{\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{10}}} }{\mathrm{3}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{10}} }{du}\:\:{then}\:{we}\:{use}\:{the}\:{chang}.\: \\ $$$${u}^{\mathrm{10}} \:={t}\:\Rightarrow{u}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} \:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{10}} }{du}\: \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{5}}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{10}}{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{3}}{\mathrm{10}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)}\:\:\:{by}\:{the}\:{same}\:{chang}.\:{we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}}{{x}^{\mathrm{10}} \:+\mathrm{3}}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{10}}} \:{u}}{\mathrm{3}{u}^{\mathrm{10}} \:+\mathrm{3}}\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{10}}} \:{du} \\ $$$$=\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{5}}} }{\mathrm{3}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{u}\:{du}}{\mathrm{1}+{u}^{\mathrm{10}} }\:\:=\mathrm{3}^{\frac{−\mathrm{4}}{\mathrm{5}}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{10}}} }{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{10}}\:{t}^{\frac{\mathrm{1}}{\mathrm{10}}−\mathrm{1}} \:{dt} \\ $$$$=\:\frac{\mathrm{3}^{−\frac{\mathrm{4}}{\mathrm{5}}} }{\mathrm{10}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\:\frac{\mathrm{3}^{−\frac{\mathrm{4}}{\mathrm{5}}} }{\mathrm{10}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{4}}{\mathrm{10}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)}.\frac{\mathrm{3}^{\frac{\mathrm{3}}{\mathrm{10}}} }{\mathrm{3}}\:\:+\:\mathrm{2}\:\frac{\mathrm{3}^{−\frac{\mathrm{4}}{\mathrm{5}}} }{\mathrm{10}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)} \\ $$$${I}\:=\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{3}^{−\frac{\mathrm{7}}{\mathrm{10}}} \:\:\:\:\frac{\mathrm{1}}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)}\:\:+\frac{\pi}{\mathrm{5}}\:\mathrm{3}^{−\frac{\mathrm{4}}{\mathrm{5}}} \:.\frac{\mathrm{1}}{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)} \\ $$$${I}\:{have}\:{used}\:{the}\:{formula}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:=\:\frac{\pi}{{sin}\left(\pi{a}\right)}\:{with}\:\mathrm{0}<{a}<\mathrm{1}\:. \\ $$

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