solve-the-integral-sinx-x-1-3-dx-

Question Number 110175 by mathdave last updated on 27/Aug/20

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

\int_{- \infty}^{+ \infty} \frac{\sin x}{\sqrt[3]{x}} d x

Answered by mathmax by abdo last updated on 28/Aug/20

I = \int_{- \infty}^{+ \infty} \frac{sinx}{x^{\frac{1}{3}}} dx = 2 \int_{0}^{\infty} \frac{sinx}{x^{\frac{1}{3}}} dx = - 2 Im (\int_{0}^{\infty} x^{- \frac{1}{3}} e^{- ix} dx)

changement ix = t give - x = it \Rightarrow x = - it \Rightarrow

\int_{0}^{\infty} x^{- \frac{1}{3}} e^{- ix} dx = \int_{0}^{\infty} {(- it)}^{- \frac{1}{3}} e^{- t} (- i) dt

= {(- i)}^{- \frac{1}{3} + 1} \int_{0}^{\infty} t^{- \frac{1}{3}} e^{- t} dt = {(e^{- \frac{i π}{2}})}^{\frac{2}{3}} \int_{0}^{\infty} t^{\frac{2}{3} - 1} e^{- t} dt

= e^{- \frac{i π}{3}} Γ (\frac{2}{3}) = Γ (\frac{2}{3}) . (\cos (\frac{π}{3}) - isin (\frac{π}{3})) \Rightarrow

I = 2 (Γ (\frac{2}{3}) \sin (\frac{π}{3}) = \sqrt{3} \times Γ (\frac{2}{3})

Answered by mathdave last updated on 28/Aug/20

s o l u t i o n

l e t I = \int_{- \infty}^{+ \infty} \frac{\sin x}{\sqrt[3]{x}} d x = 2 \int_{0}^{\infty} \frac{\sin x}{x^{\frac{1}{3}}} d x

f r o m m a z i d e n t i t y i n t e g r a l

\int_{0}^{\infty} \frac{\sin (a x)}{x^{n}} d x = \frac{π a^{n - 1}}{2 Γ (n) \sin (\frac{n π}{2})}

t a k i n g a = 1 a n d n = \frac{1}{3}

∵ I = 2 \int_{0}^{\infty} \frac{\sin}{x^{\frac{1}{3}}} d x = 2 ∙ \frac{π \times 1^{\frac{1}{3} - 1}}{2 Γ (\frac{1}{3}) \sin (\frac{π}{6})} = \frac{π}{\frac{1}{2} Γ (\frac{1}{3})}

∵ \int_{- \infty}^{+ \infty} \frac{\sin x}{\sqrt[3]{x}} d x = \frac{2 π}{Γ (\frac{1}{3})} = 2.3454

b y m a t h d a v e (28 / 08 / 2020)

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