if-a-gt-0-and-m-0-0-cos-mx-x-2-a-2-2-dx-

Question Number 201341 by mathlove last updated on 05/Dec/23

i f a > 0 a n d m ⩾ 0

\int_{0}^{\infty} \frac{c o s (m x)}{{(x^{2} + a^{2})}^{2}} d x = ?

Answered by Calculusboy last updated on 05/Dec/23

S o l u t i o n : s i n c e a > 0 a n d m \underline{>} 0 l e t a = 1 a n d m = 0

I = \int_{0}^{\infty} \frac{c o s (0 \times x)}{{(x^{2} + 1)}^{2}} d x \Leftrightarrow I = \int_{0}^{\infty} \frac{1}{{(x^{2} + 1)}^{2}} d x l e t x = t a n θ d x = {s e c}^{2} θ d θ

w h e n x = \infty θ = \frac{π}{2} a n d w h e n x = 0 θ = 0

I = \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} θ d θ}{{({t a n}^{2} θ + 1)}^{2}} \Leftrightarrow I = \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} θ d θ}{{({s e c}^{2} θ)}^{2}}

I = \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} θ}{{s e c}^{4} θ} d θ \Leftrightarrow I = \int_{0}^{\frac{π}{2}} \frac{1}{{s e c}^{2} θ} d θ

I = \int_{0}^{\frac{π}{2}} {c o s}^{2} θ d θ

b y u s i n g w a l l i s f o r m u l a (t h e p o w e r i s e v e n)

I_{n} = \frac{n - 1}{n} \times \frac{π}{2} (n = 2)

I = \frac{2 - 1}{2} \times \frac{π}{2}

I = \frac{π}{4}

Commented by mathlove last updated on 05/Dec/23

t h a n k s

Answered by Mathspace last updated on 05/Dec/23

f (a) = \int_{0}^{\infty} \frac{c o s (m x)}{x^{2} + a^{2}} d x

f^{^{'}} (a) = - 2 a \int_{0}^{\infty} \frac{c o s (m x)}{{(x^{2} + a^{2})}^{2}} d x \Rightarrow

\int_{0}^{\infty} \frac{c o s (m x)}{{(x^{2} + a^{2})}^{2}} d x = - \frac{1}{2 a} f^{^{'}} (a)

b u t f (a) = R e (\frac{1}{2} \int_{R} \frac{e^{i m x}}{x^{2} + a^{2}} d x)

f (z) = \frac{e^{i m z}}{z^{2} + a^{2}} = \frac{e^{i m z}}{(z - i a) (z + i a)}

t h e p o l e s o f f a r e i a a n d - i a

\int_{R} f (z) d z = 2 i π \sum_{a i \in p^{+}} {R e s a}_{i}

= 2 i π R e s (f, i a)

= 2 i π . \frac{e^{i m (i a)}}{2 i a} = \frac{π}{a} e^{- m a} \Rightarrow f (a) = \frac{π}{2 a} e^{- m a}

a n d f^{^{'}} (a) = - \frac{π}{2 a^{2}} e^{- m a} - \frac{m π}{a} e^{- m a}

f i n a l l y \int_{0}^{\infty} \frac{c o s (m x)}{(x^{2} + a^{2})} d x

= - \frac{1}{2 a} {- \frac{π}{a^{2}} - \frac{m π}{a}} e^{- m a}

= (\frac{π}{2 a^{3}} + \frac{m a}{2 a^{2}}) e^{- m a}

Commented by Mathspace last updated on 05/Dec/23

s o r r y \int_{0}^{\infty} \frac{c o s (m x)}{{(x^{2} + a^{2})}^{2}} d x

= - \frac{1}{2 a} {- \frac{π}{2 a^{2}} e^{- m a} - \frac{m π}{2 a}} e^{- m a}

= (\frac{π}{4 a^{3}} + \frac{m π}{4 a^{2}}) e^{- m a}

= \frac{π}{4 a^{3}} (1 + a m) e^{- m a}