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 44587 by maxmathsup by imad last updated on 01/Oct/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^{2018}}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

Commented by prof Abdo imad last updated on 02/Oct/18

$$changementt=x^{\frac{1}{2018}}give$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^{2018}}={\int }_0^{\mathrm{\infty }}\frac{1}{2018}\frac{x^{\frac{1}{2018}-1}}{1+x}dx$$$$=\frac{1}{2018}{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{2018}-1}}{1+x}dx=\frac{1}{2018}\frac{\pi }{sin\left(\frac{\pi }{2018}\right)}$$$$=\frac{\pi }{2018sin\left(\frac{\pi }{2018}\right)}$$$$ihaveusedtheresult{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}$$$$with0<a<1.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

$$=areaunderthecurve=2\times 1=2$$$$soansis\frac{1}{2}\times 2=1becauselimitofintregalfrom$$$$0to\mathrm{\infty }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com