calculate-0-dt-1-t-2018-

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

c a l c u l a t e \int_{0}^{\infty} \frac{d t}{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

c h a n g e m e n t t = x^{\frac{1}{2018}} g i v e

\int_{0}^{\infty} \frac{d t}{1 + t^{2018}} = \int_{0}^{\infty} \frac{1}{2018} \frac{x^{\frac{1}{2018} - 1}}{1 + x} d x

= \frac{1}{2018} \int_{0}^{\infty} \frac{x^{\frac{1}{2018} - 1}}{1 + x} d x = \frac{1}{2018} \frac{π}{s i n (\frac{π}{2018})}

= \frac{π}{2018 s i n (\frac{π}{2018})}

i h a v e u s e d t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)}

w i t h 0 < a < 1 .

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

= a r e a u n d e r t h e c u r v e = 2 \times 1 = 2

s o a n s i s \frac{1}{2} \times 2 = 1 b e c a u s e l i m i t o f i n t r e g a l f r o m

0 t o \infty