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Question Number 81427 by abdomathmax last updated on 13/Feb/20

1) c a l c u l a t e \int_{0}^{\infty} c o s (x^{3}) d x

2) f i n d t h e v a l u e o f \int_{0}^{\infty} c o s (x^{n}) d x w i t h n ⩾ 2

Commented by mind is power last updated on 13/Feb/20

2) x^{n} = y \Rightarrow d x = \frac{d y}{n} y^{\frac{1}{n} - 1}

\Rightarrow \frac{1}{n} \int_{0}^{+ \infty} c o s (y) y^{\frac{1}{n} - 1} d y

= \frac{1}{n} R e {\int_{0}^{+ \infty} e^{i y} y^{\frac{1}{n} - 1} d y

= \frac{1}{n} R e {\int_{0}^{i \infty} e^{- z} {(i z)}^{\frac{1}{n} - 1} . i d z}

= \frac{1}{n} R e {i^{\frac{1}{n}} \int_{0}^{i \infty} e^{- z} z^{\frac{1}{n} - 1}}

= \frac{1}{n} . c o s (\frac{π}{2 n}) . Γ (\frac{1}{n})

Commented by abdomathmax last updated on 13/Feb/20

t h a n k y o u s i r .

Commented by abdomathmax last updated on 13/Feb/20

1) \int_{0}^{\infty} c o s (x^{3}) d x = R e (\int_{0}^{\infty} e^{{i x}^{3}} d x)

c h a n g e m r n t {i x}^{3} = - t t g i v e x^{3} = - \frac{t}{i} \Rightarrow x = {(i t)}^{\frac{1}{3}}

= e^{\frac{i π}{6}} t^{\frac{1}{3}} \Rightarrow

\int_{0}^{\infty} e^{{i x}^{3}} d x = \int_{0}^{\infty} e^{- t} e^{\frac{i π}{6}} \times \frac{1}{3} t^{\frac{1}{3} - 1} d x

= \frac{1}{3} e^{\frac{i π}{6}} \int_{0}^{\infty} t^{\frac{1}{3} - 1} e^{- t} d t = \frac{1}{3} e^{\frac{i π}{6}} Γ (\frac{1}{3})

\Rightarrow \int_{0}^{\infty} c o s (x^{3}) d x = \frac{1}{3} c o s (\frac{π}{6}) Γ (\frac{1}{3})

= \frac{1}{3} \frac{\sqrt{3}}{2} Γ (\frac{1}{3}) = \frac{1}{2 \sqrt{3}} \times Γ (\frac{1}{3})

Commented by abdomathmax last updated on 13/Feb/20

2) \int_{0}^{\infty} c o s (x^{n}) d x = R e (\int_{0}^{\infty} e^{- {i x}^{n}} d x) c h a n g e m e n t

{i x}^{n} = t \Rightarrow x^{n} = \frac{t}{i} \Rightarrow x = {(\frac{t}{i})}^{\frac{1}{n}} = \frac{t^{\frac{1}{n}}}{e^{\frac{i π}{2 n}}} = e^{- \frac{i π}{2 n}} t^{\frac{1}{n}} \Rightarrow

\int_{0}^{\infty} e^{- {i x}^{n}} d x = \int_{0}^{\infty} e^{- t} e^{- \frac{i π}{2 n}} \times \frac{1}{n} t^{\frac{1}{n} - 1} d t

= \frac{1}{n} e^{- \frac{i π}{2 n}} \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{- t} d t

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

Answered by mind is power last updated on 13/Feb/20

1)

u = x^{3} \Rightarrow \int_{0}^{+ \infty} \frac{1}{3} \frac{c o s (u)}{u^{\frac{2}{3}}} d u

= \int_{0}^{+ \infty} \frac{c o s (u)}{u^{\frac{2}{3}}} d u

= R e \int_{0}^{+ \infty} e^{i u} u^{- \frac{2}{3}} d u

- y = i u \Rightarrow

d u = i d y

R e {i^{\frac{1}{3}} \int_{0}^{i \infty} e^{- y} y^{- \frac{2}{3}} d y}

R e {e^{\frac{i π}{6}} \int_{0}^{i \infty} e^{- y} y^{- \frac{2}{3}} d y}

\int_{r}^{R} e^{- y} y^{- \frac{2}{3}} d y + \int_{{R e}^{i θ}} e^{- y} y^{- \frac{2}{3}} d y + \int_{i R}^{i r} e^{- y} y^{- \frac{2}{3}} d y + \int_{C_{r}} e^{- y} y^{- \frac{2}{3}} d y

C_{r} = {r e}^{i t}, t \in [\frac{π}{2}, 0] r \to 0

\int_{C_{R}} e^{- y} y^{- \frac{2}{3}} d y = 0 R \to \infty

\int c_{r} e^{- y} y^{- \frac{2 .}{3}} d y \to 0

\Rightarrow \int_{i R}^{i r} e^{- y} y^{- \frac{2}{3}} d y = - \int_{r}^{R} e^{- y} y^{- \frac{2}{3}} d y

\Leftrightarrow \int_{0}^{i \infty} e^{- y} y^{- \frac{2}{3}} d y = \int_{0}^{+ \infty} e^{- y} y^{- \frac{2}{3}} d y = Γ (\frac{1}{3})

\Rightarrow \int_{0}^{+ \infty} c o s (x^{3}) d x = \frac{1}{3} Γ (\frac{1}{3}) c o s (\frac{π}{6}) = \frac{\sqrt{3}}{6} . Γ (\frac{1}{3})