Menu Close

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
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
$${hi}\:{every}\:{one}\:{is}\:{it}\:{right}\:{if}\:{we}\:{use}\:{tylor} \\ $$$${in}\:{this}\:{integration}\:{and}\:{if}\:{there}\:{were} \\ $$$${another}\:{way}\:{that}\:{will}\:{be}\:{very}\:{cool} \\ $$$$\int{sin}\left({x}^{\mathrm{4}} \right){dx}\: \\ $$$$ \\ $$$$ \\ $$
Commented by MJS last updated on 02/May/20
you can use sin θ =((e^(iθ) −e^(−iθ) )/(2i)) and then the  incomplete Γ−function
$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{sin}\:\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} −\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2i}}\:\mathrm{and}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{incomplete}\:\Gamma−\mathrm{function} \\ $$
Commented by  M±th+et+s last updated on 02/May/20
thank you sir mjs   let me try    I=∫sin(x^4 )dx=∫((e^(ix^4 ) −e^(−ix^4 ) )/(2i))dx  =(1/(2i))∫e^(ix^4 ) dx−(1/(2i))∫e^(−ix^4 ) dx            ix^4 =−k         ix^4 =p            x=i^(1/4) k^(1/4)        x=(−i^(1/4) )p^(1/4)        dx=(i^(1/4) /4)k^((−3)/4) dk   dx=(((−i^(1/4) ))/4)p^(−(3/4)) dp         I=(1/(2i))∫e^(−k ) (i^(1/4) /4)k^(−(3/4)) dk −(1/(2i))∫e^(−p) (((−i^(1/4) ))/4)p^((−3)/4) dp  I=(i^(−(3/4)) /8)∫e^(−k)  k^((−(3/4)+1)−1) dk −(((−1)^(1/4) (i)^((−3)/4) )/8)∫e^(−p)  p^((1/4)−1) dp  I=(((i)^((−3)/4) )/8)[−Γ((1/4),k)]−(((−1)^(1/4) (i)^(−(3/4)) )/8)[−Γ((1/4),p)]+c  I=((−(i)^((−3)/4) )/8)Γ((1/4),−ix^4 )+(((−1)^(1/4) (i)^((−3)/4) )/8)Γ((1/4),ix^4 )+c    Γ(a,x) is incomplete gamma function
$${thank}\:{you}\:{sir}\:{mjs}\: \\ $$$${let}\:{me}\:{try} \\ $$$$ \\ $$$${I}=\int{sin}\left({x}^{\mathrm{4}} \right){dx}=\int\frac{{e}^{{ix}^{\mathrm{4}} } −{e}^{−{ix}^{\mathrm{4}} } }{\mathrm{2}{i}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int{e}^{{ix}^{\mathrm{4}} } {dx}−\frac{\mathrm{1}}{\mathrm{2}{i}}\int{e}^{−{ix}^{\mathrm{4}} } {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:{ix}^{\mathrm{4}} =−{k}\:\:\:\:\:\:\:\:\:{ix}^{\mathrm{4}} ={p} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}={i}^{\frac{\mathrm{1}}{\mathrm{4}}} {k}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:\:\:\:\:\:{x}=\left(−{i}^{\frac{\mathrm{1}}{\mathrm{4}}} \right){p}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\:\:\:\:\:{dx}=\frac{{i}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{4}}{k}^{\frac{−\mathrm{3}}{\mathrm{4}}} {dk}\:\:\:{dx}=\frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}{\mathrm{4}}{p}^{−\frac{\mathrm{3}}{\mathrm{4}}} {dp} \\ $$$$\:\:\:\:\: \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}{i}}\int{e}^{−{k}\:} \frac{{i}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{4}}{k}^{−\frac{\mathrm{3}}{\mathrm{4}}} {dk}\:−\frac{\mathrm{1}}{\mathrm{2}{i}}\int{e}^{−{p}} \frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}{\mathrm{4}}{p}^{\frac{−\mathrm{3}}{\mathrm{4}}} {dp} \\ $$$${I}=\frac{{i}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\int{e}^{−{k}} \:{k}^{\left(−\frac{\mathrm{3}}{\mathrm{4}}+\mathrm{1}\right)−\mathrm{1}} {dk}\:−\frac{\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \left({i}\right)^{\frac{−\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\int{e}^{−{p}} \:{p}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dp} \\ $$$${I}=\frac{\left({i}\right)^{\frac{−\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\left[−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}},{k}\right)\right]−\frac{\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \left({i}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\left[−\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}},{p}\right)\right]+{c} \\ $$$${I}=\frac{−\left({i}\right)^{\frac{−\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}},−{ix}^{\mathrm{4}} \right)+\frac{\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \left({i}\right)^{\frac{−\mathrm{3}}{\mathrm{4}}} }{\mathrm{8}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}},{ix}^{\mathrm{4}} \right)+{c} \\ $$$$ \\ $$$$\Gamma\left({a},{x}\right)\:{is}\:{incomplete}\:{gamma}\:{function} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *