Question-213138

Question Number 213138 by Spillover last updated on 31/Oct/24

Commented by Frix last updated on 31/Oct/24

We must first solve \int \sin t^{3} d t which might

be possible using

\sin t^{3} = \frac{e^{i t^{3}} - e^{- i t^{3}}}{2 i} = \frac{1 - e^{2 t^{3}}}{{2 e}^{t^{3}}} i

But how ?

Then \int \sin t^{2} d t (Fresnel - Integral)

I {don}^{'} t think many here among us can solve

this .

Approximating I get

\underset{0}{\int^{1}} \underset{0}{\int^{\frac{1}{2}}} \underset{0}{\int^{\frac{1}{3}}} \sin (x + y^{2} + z^{3}) d z d y d x \approx . 088997

Answered by MrGaster last updated on 31/Oct/24

= \int_{0}^{1} \int_{0}^{\frac{1}{2}} {[- \frac{\cos (x + y^{2} + z^{3})}{3 z^{2}}]}_{0}^{\frac{1}{3}} d y d x

= \int_{0}^{1} \int_{0}^{\frac{1}{2}} (- \frac{\cos (x + y^{2} + \frac{1}{27}}{3 \cdot \frac{1}{9}} + \frac{\cos (x + y^{2})}{3 \cdot 0^{2}}) d y d x

= \int_{0}^{1} \int_{0}^{\frac{1}{2}} (- 3 \cos (x + y^{2} + \frac{1}{27}) + \cos 2 (x + y^{2})) d y d x

= \int_{0}^{1} {[\sin (x + y^{2}) - 3 \sin (x + y^{2} + \frac{1}{27})]}_{0}^{\frac{1}{2}} d x

= \int_{0}^{1} (\sin (x + \frac{1}{4}) - 3 \sin (x + \frac{1}{4} + \frac{1}{27}) - \sin (x) + 3 \sin (x + \frac{1}{27})) d x

= {[- \cos (x + \frac{1}{4}) - 3 \sin (x + \frac{1}{4} + \frac{1}{27}) + \cos (x) - 3 \cos (x + \frac{1}{27})]}_{0}^{1}

= - \cos (1 + \frac{1}{4}) + 3 \cos (1 + \frac{1}{4} + \frac{1}{27}) + \cos (1) - 3 \cos (1 + \frac{1}{27}) + \cos (\frac{1}{4}) - 3 \cos (\frac{1}{4} + \frac{1}{27}) - \cos (0) + \cos (\frac{1}{27})

Commented by Frix last updated on 31/Oct/24

No .

Answered by Berbere last updated on 31/Oct/24

I m \int_{0}^{1} \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{3}} e^{i (x + y^{2} + z^{3})} d x d y d z

= I m \int_{0}^{1} e^{i x} \int_{0}^{\frac{1}{2}} e^{{i y}^{2}} d y \int_{0}^{\frac{1}{3}} e^{{i z}^{3}} d z

= I m [- {i e}^{i} + i] . A . B

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

\int e^{x^{n}} d x = \int e^{- y} . \frac{1}{n} e^{i \frac{π}{n}} y^{\frac{1}{n} - 1} d y = - \frac{1}{n} e^{i \frac{π}{n}} Γ (\frac{1}{n}, y)

Γ (a, z) = \int_{z}^{\infty} t^{a - 1} e^{- t} d t [i n c o m p l e t e G a m m a f u n c t i o n]

- \frac{1}{n} e^{i \frac{π}{n}} Γ (\frac{1}{n}, - x^{n}) + c

\Rightarrow A = \int_{0}^{\frac{1}{2}} e^{{i y}^{2}} d y = \frac{1}{2} \int_{0} e^{- z} i^{\frac{1}{2}} z^{\frac{1}{2} - 1} d z

= - \frac{1}{2} i^{\frac{1}{2}} Γ (\frac{1}{2}, z)

{A = - \frac{1}{2} i^{\frac{1}{2}} Γ (\frac{1}{2}, - {i y}^{2})]}_{0}^{\frac{1}{2}} = - \frac{1}{2} e^{\frac{i π}{4}} Γ (\frac{1}{2}, - \frac{i}{4}] + \frac{1}{2} e^{i \frac{π}{4}} Γ (\frac{1}{2}, 0]

B = - \frac{1}{3} i^{\frac{1}{3}} Γ (\frac{1}{3}, - \frac{i}{27}) + \frac{i^{\frac{1}{3}}}{3} Γ (\frac{1}{3}, 0)

\int_{0}^{1} \int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{3}} s i n (x + y^{2} + z^{3}) d x d y d z

= TM (i (1 - e^{i}) (- \frac{1}{2} e^{i \frac{π}{4}} (Γ (\frac{1}{2}, - \frac{i}{4}) + Γ (\frac{1}{2}, 0)) . (- \frac{1}{3} e^{i \frac{π}{6}} Γ (\frac{1}{3}, - \frac{i}{27},) + \frac{e^{i \frac{π}{6}}}{3} Γ (\frac{1}{3}, 0))

IM i m a g i n a r y P a r t

Commented by Frix last updated on 31/Oct/24

I was hoping you show us the solution,

thank you!

Commented by Berbere last updated on 01/Nov/24

w i t h e p l e a s u r G o d b l e s s Y o u

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