Question-65729 – Tinku Tara

Question Number 65729 by aliesam last updated on 02/Aug/19

Commented by mathmax by abdo last updated on 03/Aug/19

a) l e t f (α) = \int_{0}^{\infty} e^{i x} x^{- α} d x \Rightarrow f (α) =_{i x = t} \int_{0}^{\infty} e^{t} {(\frac{t}{i})}^{- α} \frac{d t}{i}

= \frac{1}{i^{1 - α}} \int_{0}^{\infty} e^{t} t^{1 - α - 1} d t = \frac{1}{i^{1 - α}} Γ (1 - α) t h e c o n v e r g e n c e o f f (α)

i s a s s u r e d

b) \int_{0}^{\infty} c o s x x^{- α} d x = R e (\int_{0}^{\infty} e^{i x} x^{- α} d x) = R e (\frac{1}{i^{α - 1}} Γ (1 - α)) b u t

\frac{1}{i^{α - 1}} Γ (1 - α) = \frac{1}{{(e^{i \frac{π}{2}})}^{α - 1}} Γ (1 - α) = e^{- \frac{i π}{2} (α - 1)} Γ (1 - α)

= e^{\frac{i π}{2}} e^{- \frac{i π}{2} α} Γ (1 - α) = i (c o s (\frac{π α}{2}) - i s i n (\frac{π α}{2})) Γ (1 - α)

= s i n (\frac{π α}{2}) Γ (1 - α) + i c o s (\frac{π α}{2}) Γ (1 - α) \Rightarrow

\int_{0}^{\infty} c o s x x^{- α} d x = s i n (\frac{π α}{2}) Γ (1 - α)

Commented by aliesam last updated on 03/Aug/19

t h a n k y o u s i r . b u t t h e f i r s t q u e s t i o n

i n e e d t o p r o v e t h e i n t e g r a l i s i n e e d

c o n v e r g e n t b e f o r e c o m p u t i n g i t s v a l u e

c o n c e r n i n g . t h e s e c o n d q u e s t i o n, i m e a n t b y c o m p l e x i n t e g r a t i o n

i s b y u s i n g t h e r e s i d u e t h e o r e m

Commented by aliesam last updated on 04/Aug/19

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