find-the-value-of-0-cos-x-x-2-1-x-2-2-x-2-3-dx-2-calculate-0-dx-x-2-1-x-2-2-x-2-3-

Question Number 34021 by prof Abdo imad last updated on 29/Apr/18

f i n d t h e v a l u e o f \int_{0}^{+ \infty} \frac{c o s (α x)}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} d x

2) c a l c u l a t e \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)}

Commented by prof Abdo imad last updated on 30/Apr/18

l e t p u t I = \int_{0}^{\infty} \frac{c o s (α x)}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3 {} d x

2 I = \int_{- \infty}^{+ \infty} \frac{c o s (α x)}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} d x

= R e (\int_{- \infty}^{+ \infty} \frac{e^{i α x}}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} d x) l e t

c o n s i d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{e^{i α z}}{(z^{2} + 1) (z^{2} + 2) (z^{2} + 3)} t h e p o l e s o f φ a r e

i, - i, i \sqrt[]{2}, - i \sqrt{2}, i \sqrt{3}, - i \sqrt{3} a n d R e s i d u s t h e o r e m

g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, i \sqrt{2}) + R e s (φ, i \sqrt{3})

φ (z) = \frac{e^{i α z}}{(z - i) (z + i) (z - i \sqrt{2}) (z + i \sqrt{2}) (z - i \sqrt{3}) (z + i \sqrt{3})}

R e s (φ, i) = \frac{e^{- α}}{2 i (- 1 + 2) (- 1 + 3)} = \frac{e^{- α}}{4 i}

R e s (φ, i \sqrt{2}) = \frac{e^{- α \sqrt{2}}}{(2 i \sqrt{2}) (- 2 + 1) (- 2 + 3)}

= - \frac{e^{- α \sqrt{2}}}{2 i \sqrt{2}}

R e s (φ, i \sqrt{3}) = \frac{e^{- α \sqrt{3}}}{2 i \sqrt{3} (- 3 + 1) (- 3 + 2)}

= \frac{e^{- α \sqrt{3}}}{4 i \sqrt{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{e^{- α}}{4 i} - \frac{e^{- α \sqrt{2}}}{2 i \sqrt{2}} + \frac{e^{- α \sqrt{3}}}{4 i \sqrt{3}})

= \frac{π}{2} e^{- α} - \frac{π}{\sqrt{2}} e^{- α \sqrt{2}} + \frac{π}{2 \sqrt{3}} e^{- α \sqrt{3}} \Rightarrow

I = \frac{π}{4} e^{- α} - \frac{π}{2 \sqrt{2}} e^{- α \sqrt{2}} + \frac{π}{4 \sqrt{3}} e^{- α \sqrt{3}} .

Commented by abdo imad last updated on 30/Apr/18

2) l e t t a k e α = 0 \Rightarrow

\int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} = \frac{π}{4} - \frac{π}{2 \sqrt{2}} + \frac{π}{4 \sqrt{3}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

\underset{0}{\int^{\infty}} \frac{1}{(x^{2} + 3)} . \frac{(x^{2} + 2) - (x^{2} + 1)}{(x^{2} + 2) (x^{2} + 1)} d x

\underset{0}{\int^{\infty}} \frac{1}{(x^{2} + 3)} . \frac{1}{(x^{2} + 1)} d x - \underset{0}{\int^{\infty}} \frac{1}{(x^{2} + 3)} . \frac{1}{(x^{2} + 2)} d x

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

1 / 2 \underset{0}{\int^{\infty}} {\frac{1}{(x^{2} + 1)} - \frac{1}{(x^{2} + 3)}} d x - \underset{0}{\int^{\infty}} (\frac{1}{x^{2} + 2} - \frac{1}{x^{2} + 3}) ._{} d x

f i r s t i n t r e g l 1 / 2 {{t a n}^{- 1} x - \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{x}{\sqrt{3}})}

s e c o n d 1 / \sqrt{2} {t a n}^{- 1} (x / \sqrt{2)} - 1 / \sqrt{3} {t a n}^{- 1} (x / \sqrt{3})

a d d t h e m \dots