hi-every-one-is-it-right-if-we-use-tylor-in-this-integration-and-if-there-were-another-way-that-will-be-very-cool-sin-x-4-dx-

Question Number 91615 by M±th+et+s last updated on 01/May/20

h i e v e r y o n e i s i t r i g h t i f w e u s e t y l o r

i n t h i s i n t e g r a t i o n a n d i f t h e r e w e r e

a n o t h e r w a y t h a t w i l l b e v e r y c o o l

\int s i n (x^{4}) d x

Commented by MJS last updated on 02/May/20

you can use \sin θ = \frac{e^{i θ} - e^{- i θ}}{2 i} and then the

incomplete Γ - function

Commented by M±th+et+s last updated on 02/May/20

t h a n k y o u s i r m j s

l e t m e t r y

I = \int s i n (x^{4}) d x = \int \frac{e^{{i x}^{4}} - e^{- {i x}^{4}}}{2 i} d x

= \frac{1}{2 i} \int e^{{i x}^{4}} d x - \frac{1}{2 i} \int e^{- {i x}^{4}} d x

{i x}^{4} = - k {i x}^{4} = p

x = i^{\frac{1}{4}} k^{\frac{1}{4}} x = (- i^{\frac{1}{4}}) p^{\frac{1}{4}}

d x = \frac{i^{\frac{1}{4}}}{4} k^{\frac{- 3}{4}} d k d x = \frac{(- i^{\frac{1}{4}})}{4} p^{- \frac{3}{4}} d p

I = \frac{1}{2 i} \int e^{- k} \frac{i^{\frac{1}{4}}}{4} k^{- \frac{3}{4}} d k - \frac{1}{2 i} \int e^{- p} \frac{(- i^{\frac{1}{4}})}{4} p^{\frac{- 3}{4}} d p

I = \frac{i^{- \frac{3}{4}}}{8} \int e^{- k} k^{(- \frac{3}{4} + 1) - 1} d k - \frac{{(- 1)}^{\frac{1}{4}} {(i)}^{\frac{- 3}{4}}}{8} \int e^{- p} p^{\frac{1}{4} - 1} d p

I = \frac{{(i)}^{\frac{- 3}{4}}}{8} [- Γ (\frac{1}{4}, k)] - \frac{{(- 1)}^{\frac{1}{4}} {(i)}^{- \frac{3}{4}}}{8} [- Γ (\frac{1}{4}, p)] + c

I = \frac{- {(i)}^{\frac{- 3}{4}}}{8} Γ (\frac{1}{4}, - {i x}^{4}) + \frac{{(- 1)}^{\frac{1}{4}} {(i)}^{\frac{- 3}{4}}}{8} Γ (\frac{1}{4}, {i x}^{4}) + c

Γ (a, x) i s i n c o m p l e t e g a m m a f u n c t i o n