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Question Number 62220 by maxmathsup by imad last updated on 17/Jun/19

let f(x) =∫_0 ^∞ (t^2 /(x^6 +t^6 )) dt with x>0 1) calculate f(x) 2) calculate g(x) =∫_0 ^∞ (t^2 /((x^6 +t^6 )^2 ))dt 3) find values of integrals ∫_0 ^∞ (t^2 /(t^6 +8))dt and ∫_0 ^∞ (t^2 /((t^6 +8)^2 ))dt .

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{t}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:\:+{t}^{\mathrm{6}} }\:{dt}\:\:\:\:\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\left({x}^{\mathrm{6}} \:+{t}^{\mathrm{6}} \right)^{\mathrm{2}} }{dt} \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} \:+\mathrm{8}}{dt}\:\:\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{6}} +\mathrm{8}\right)^{\mathrm{2}} }{dt}\:. \\ $$

Commented bymaxmathsup by imad last updated on 19/Jun/19

1) f(x)=∫_0 ^∞ (t^2 /(x^6 +t^6 ))dt changement t =xu give f(x) =∫_0 ^∞ ((x^2 u^2 )/(x^6 +x^6 u^6 )) xdu =(1/x^3 ) ∫_0 ^∞ (u^2 /(1+u^6 ))du changement u^6 =α give u=α^(1/6) x^3 f(x) = ∫_0 ^∞ (((α^(1/6) )^2 )/(1+α)) (1/6) α^((1/6)−1) du = ∫_0 ^∞ (α^((1/3)+(1/6)−1) /(1+α)) du =(1/6)∫_0 ^∞ (α^((1/2)−1) /(1+α)) dα =(1/6) (π/(sin((π/2)))) =(π/6) ⇒f(x) =(π/(6x^3 )) (x>0) 2) we have f^′ (x) = −∫_0 ^∞ ((6x^5 t^2 )/((x^6 +t^6 )^2 )) dt =−6x^5 ∫_0 ^∞ (t^2 /((x^6 +t^6 )^2 )) dt =−6x^5 g(x) ⇒ g(x) =−((f^′ (x))/(6x^5 )) but f(x) =(π/6) x^(−3) ⇒f^′ (x) =(π/6) (−3)x^(−4) =−(π/(2x^4 )) ⇒ g(x) =−(1/(6x^5 ))(−(π/(2x^4 ))) =(π/(12x^9 )) 3)∫_0 ^∞ (t^2 /(t^6 +8)) dt = ∫_0 ^∞ (t^2 /(((√2))^6 +t^6 )) f((√2)) =(π/(6((√2))^3 )) =(π/(12(√2))) =((π(√2))/(24)) ∫_0 ^∞ (t^2 /((t^6 +8)^2 )) dt =g((√2)) = (π/(12((√2))^9 )) =(π/(12.(√2)(2^4 ))) =(π/(12×16×(√2))) =...

$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:+{t}^{\mathrm{6}} }{dt}\:\:{changement}\:{t}\:={xu}\:{give}\:\: \\ $$ $${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}} {u}^{\mathrm{2}} }{{x}^{\mathrm{6}} \:+{x}^{\mathrm{6}} {u}^{\mathrm{6}} }\:{xdu}\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{6}} }{du}\:\:\:{changement}\:\:{u}^{\mathrm{6}} \:=\alpha\:{give}\:{u}=\alpha^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$ $${x}^{\mathrm{3}} {f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\alpha^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\mathrm{2}} }{\mathrm{1}+\alpha}\:\frac{\mathrm{1}}{\mathrm{6}}\:\alpha^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:{du}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{du}\:=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha \\ $$ $$=\frac{\mathrm{1}}{\mathrm{6}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{6}}\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{6}{x}^{\mathrm{3}} }\:\:\:\:\:\left({x}>\mathrm{0}\right) \\ $$ $$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{6}{x}^{\mathrm{5}} {t}^{\mathrm{2}} }{\left({x}^{\mathrm{6}} \:+{t}^{\mathrm{6}} \right)^{\mathrm{2}} }\:{dt}\:=−\mathrm{6}{x}^{\mathrm{5}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left({x}^{\mathrm{6}} \:+{t}^{\mathrm{6}} \right)^{\mathrm{2}} }\:{dt}\:=−\mathrm{6}{x}^{\mathrm{5}} \:{g}\left({x}\right)\:\Rightarrow \\ $$ $${g}\left({x}\right)\:=−\frac{{f}^{'} \left({x}\right)}{\mathrm{6}{x}^{\mathrm{5}} }\:\:\:\:\:\:\:\:\:{but}\:{f}\left({x}\right)\:=\frac{\pi}{\mathrm{6}}\:{x}^{−\mathrm{3}} \:\Rightarrow{f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{6}}\:\left(−\mathrm{3}\right){x}^{−\mathrm{4}} \:\:=−\frac{\pi}{\mathrm{2}{x}^{\mathrm{4}} }\:\Rightarrow \\ $$ $${g}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}{x}^{\mathrm{5}} }\left(−\frac{\pi}{\mathrm{2}{x}^{\mathrm{4}} }\right)\:=\frac{\pi}{\mathrm{12}{x}^{\mathrm{9}} } \\ $$ $$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{6}} \:+\mathrm{8}}\:{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}} \:+{t}^{\mathrm{6}} }\:{f}\left(\sqrt{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{6}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{3}} }\:=\frac{\pi}{\mathrm{12}\sqrt{\mathrm{2}}}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{24}} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{6}} \:+\mathrm{8}\right)^{\mathrm{2}} }\:{dt}\:={g}\left(\sqrt{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{12}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{9}} }\:=\frac{\pi}{\mathrm{12}.\sqrt{\mathrm{2}}\left(\mathrm{2}^{\mathrm{4}} \right)}\:=\frac{\pi}{\mathrm{12}×\mathrm{16}×\sqrt{\mathrm{2}}}\:=... \\ $$

Commented bymaxmathsup by imad last updated on 19/Jun/19

i have used the result ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) if 0<a<1

$${i}\:{have}\:{used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)}\:{if}\:\mathrm{0}<{a}<\mathrm{1} \\ $$

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