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Question Number 40126 by maxmathsup by imad last updated on 16/Jul/18
calculate ∫_0 ^1 ((tdt)/(1+t^4 ))
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$
Commented by maxmathsup by imad last updated on 21/Jul/18
we have ∫_0 ^∞ (t/(1+t^4 ))dt = ∫_0 ^1 (t/(1+t^4 ))dt + ∫_1 ^(+∞) (t/(1+t^4 ))dt but ∫_1 ^(+∞) (t/(1+t^4 ))dt =_(t=(1/x)) ∫_0 ^1 (1/(x(1+(1/x^4 )))) (dx/x^2 ) = ∫_0 ^1 (dx/(x^3 (1+(1/x^4 )))) = ∫_0 ^1 (dx/(x^3 +(1/x))) dx =∫_0 ^1 ((xdx)/(x^4 +1)) ⇒ ∫_0 ^∞ (t/(1+t^4 ))dt =2 ∫_0 ^1 ((tdt)/(1+t^4 )) ⇒ ∫_0 ^1 ((tdt)/(1+t^4 )) =(1/2) ∫_0 ^∞ (t/(1+t^4 ))dt changement t=u^(1/4) give ∫_0 ^∞ (t/(1+t^4 ))dt =∫_0 ^∞ ((u^(1/4) )/(1+u)) (1/4) u^((1/4)−1) du =(1/4) ∫_0 ^∞ (u^((1/2)−1) /(1+u))du =(1/4) (π/(sin((π/2)))) =(π/4) ⇒ ∫_0 ^1 ((tdt)/(1+t^4 )) = (π/8) .
$${we}\:{have}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:+\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=_{{t}=\frac{\mathrm{1}}{{x}}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)}\:\frac{{dx}}{{x}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{3}} \:+\frac{\mathrm{1}}{{x}}}\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xdx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:{changement} \\ $$$${t}={u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:\:{give}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:}{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{4}}\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{u}}{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\:\frac{\pi}{\mathrm{8}}\:. \\ $$$$ \\ $$

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