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Question Number 44587 by maxmathsup by imad last updated on 01/Oct/18
calculate ∫_0 ^∞ (dt/(1+t^(2018) ))
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2018}} } \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
Commented by prof Abdo imad last updated on 02/Oct/18
changement t =x^(1/(2018)) give ∫_0 ^∞ (dt/(1+t^(2018) )) =∫_0 ^∞ (1/(2018)) (x^((1/(2018))−1) /(1+x))dx =(1/(2018)) ∫_0 ^∞ (x^((1/(2018))−1) /(1+x))dx =(1/(2018)) (π/(sin((π/(2018))))) =(π/(2018sin((π/(2018))))) i have used the result ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) with0<a<1 .
$${changement}\:{t}\:={x}^{\frac{\mathrm{1}}{\mathrm{2018}}} \:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2018}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2018}}\:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2018}}−\mathrm{1}} }{\mathrm{1}+{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2018}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{2018}}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2018}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2018}}\right)} \\ $$$$=\frac{\pi}{\mathrm{2018}{sin}\left(\frac{\pi}{\mathrm{2018}}\right)} \\ $$$${i}\:{have}\:{used}\:{the}\:{result}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$$${with}\mathrm{0}<{a}<\mathrm{1}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
=area under the curve=2×1=2 so ans is (1/2)×2=1 because limit of intregal from 0 to ∞
$$={area}\:{under}\:{the}\:{curve}=\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$${so}\:{ans}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}=\mathrm{1}\:{because}\:{limit}\:{of}\:{intregal}\:{from} \\ $$$$\mathrm{0}\:{to}\:\infty \\ $$

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