a−nser to question 25765...we put I=∫_0 ^∞ (cos(x^(2n) )(1+x^2 )^(−1) dx and J=∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx...we have 2(I+iJ)=∫_R e^(ix^(2n) ) (1+x^2 )^(−1) dx...let f(z)=e^(ix^(2n) ) (1+x^2 )^(−1) by residus therem ∫_R f(z)dz = 2iπRes(f.i) but Res(f.i)= lim_(z−i) (z−i)f(z)=e^(i(i)^(2n) ) = e^(i(−1)_ ^n ) (2i)^(−1) then ∫_R f(z)dz = πe^((−1)^n ) = π( cos(−1)^n +isin(−1)^n ) then ∫_0 ^∞ cos(x^(2n) )(1+x^2 )^(−1) dx =π2^(−1) cos(−1)^n and ∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx=π.2^(−1) sin(−1)^n


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Question Number 25769 by abdo imad last updated on 14/Dec/17

$a - n s e r t o q u e s t i o n 25765 . . . w e p u t I = \int_{0}^{\infty} (c o s (x^{2 n}) {(1 + x^{2})}^{- 1} d x a n d J = \int_{0}^{\infty} s i n (x^{2 n}) {(1 + x^{2})}^{- 1} d x . . . w e h a v e 2 (I + i J) = \int_{R} e^{{i x}^{2 n}} {(1 + x^{2})}^{- 1} d x . . . l e t f (z) = e^{{i x}^{2 n}} {(1 + x^{2})}^{- 1} b y r e s i d u s t h e r e m \int_{R} f (z) d z = 2 i π R e s (f . i) b u t R e s (f . i) = {l i m}_{z - i} (z - i) f (z) = e^{i {(i)}^{2 n}} = e^{i {(- 1)}_{}^{n}} {(2 i)}^{- 1} t h e n \int_{R} f (z) d z = π e^{{(- 1)}^{n}} = π (c o s {(- 1)}^{n} + i s i n {(- 1)}^{n}) t h e n \int_{0}^{\infty} c o s (x^{2 n}) {(1 + x^{2})}^{- 1} d x = π 2^{- 1} c o s {(- 1)}^{n} a n d \int_{0}^{\infty} s i n (x^{2 n}) {(1 + x^{2})}^{- 1} d x = π . 2^{- 1} s i n {(- 1)}^{n_{}} <> . . . .$
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