0-sinx-x-dx-

Question Number 142338 by rs4089 last updated on 30/May/21

\int_{0}^{\infty} \frac{s i n x}{x^{μ}} d x = ?

Answered by Dwaipayan Shikari last updated on 30/May/21

\frac{1}{x^{μ}} = \frac{1}{Γ (μ)} \int_{0}^{\infty} e^{- x t} t^{μ - 1} d t

\int_{0}^{\infty} \frac{s i n (x)}{x^{μ}} d x = \frac{1}{Γ (μ)} \int_{0}^{\infty} \int_{0}^{\infty} e^{- x t} t^{μ - 1} s i n (x) d x d t

= \frac{1}{2 i Γ (μ)} \int_{0}^{\infty} \frac{t^{μ - 1}}{t - i} - \frac{t^{μ - 1}}{t + i} d t

= \frac{1}{2 Γ (μ)} \int_{0}^{\infty} \frac{g^{\frac{μ}{2} - 1}}{1 + g} d g = \frac{1}{2 Γ (μ)} B (\frac{μ}{2}, 1 - \frac{μ}{2})

= \frac{π}{2 Γ (μ) s i n (\frac{π μ}{2})}

Commented by rs4089 last updated on 30/May/21

t h a n k s a l o t s i r

Answered by mathmax by abdo last updated on 30/May/21

f (a) = \int_{0}^{\infty} \frac{sinx}{x^{a}} dx \Rightarrow f (a) = - Im (\int_{0}^{\infty} x^{- a} e^{- ix} dx) we have

\int_{0}^{\infty} x^{- a} . e^{- ix} dx =_{ix = t} \int_{0}^{\infty} {(- it)}^{- a} e^{- t} \frac{dt}{i}

= {(- i)}^{- a + 1} \int_{0}^{\infty} t^{- a} e^{- t} dt = {(e^{- \frac{i π}{2}})}^{1 - a} . Γ (1 - a)

= Γ (1 - a) . e^{(a - 1) \frac{π}{2}} = Γ (1 - a) {\cos (a - 1) \frac{π}{2} + isin (a - 1) \frac{π}{2}}

= Γ (1 - a) {\sin (\frac{π a}{2}) - icos (\frac{π a}{2})} \Rightarrow

f (a) = \cos (\frac{π a}{2}) . Γ (1 - a) we know Γ (a) . Γ (1 - a) = \frac{π}{\sin (π a)} \Rightarrow

Γ (1 - a) = \frac{π}{\sin (π a) . Γ (a)} \Rightarrow f (a) = \frac{π \cos (\frac{π a}{2})}{\sin (π a) . Γ (a)}

= \frac{π}{Γ (a)} . \frac{\cos (\frac{π a}{2})}{\sin (π a)} with 0 < a < 1

Commented by mathmax by abdo last updated on 30/May/21

but \sin (π a) = 2 \sin (\frac{π a}{2}) \cos (\frac{π a}{2}) \Rightarrow

\int_{0}^{\infty} \frac{sinx}{x^{a}} dx = \frac{π}{2 Γ (a) \sin (\frac{π a}{2})}