Evaluate-0-sinx-x-a-dx-

Question Number 134474 by rs4089 last updated on 04/Mar/21

E v a l u a t e \int_{0}^{\infty} \frac{s i n x}{x^{a}} d x

Answered by Dwaipayan Shikari last updated on 04/Mar/21

\int_{0}^{\infty} \frac{s i n x}{x^{a}} d x

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

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

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

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

= \frac{1}{2 Γ (a)} . \frac{π}{s i n (\frac{π}{2} a)}

Answered by mnjuly1970 last updated on 04/Mar/21

ϕ = - i m \int_{0}^{\infty} \frac{e^{- i x}}{x^{a}} d x = - i m \int_{0}^{\infty} e^{- i x} x^{- a} d x

= - i m (L [x^{- a}]) ∣_{s = i}

= - i m {\frac{Γ (1 - a)}{s^{1 - a}} ∣_{s = i}}

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

= - i m {\frac{π}{Γ (a) s i n (π a) (s i n (\frac{π}{2} a) + i c o s (\frac{π}{2} a))}

= - i m {\frac{π (s i n (\frac{π}{2} a) - i c o s (\frac{π}{2} a))}{Γ (a) s i n (π a)}}

= \frac{π c o s (\frac{π}{2} a)}{2 Γ (a) s i n (\frac{π}{2} a) c o s (\frac{π}{2} a)} = \frac{π}{2 Γ (a) s i n (\frac{π}{2} a)} \dots

Answered by mathmax by abdo last updated on 27/Mar/21

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

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

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

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

Φ = - Γ (1 - a) \sin (\frac{π (a - 1)}{2}) = \sin (\frac{π}{2} (1 - a)) \times Γ (1 - a)