This integral (Fresnel Sine Integral) does not have a closed form
F r e s n e l i S i n e i I n t e g r a l ( S ( z ) ) = ∫ 0 z s i n ( π x 2 2 ) d x Fresnel \phantom{i}Sine\phantom{i} Integral(S(z))=\int_{0}^{z}sin(\frac{\pi x^{2}}{2})dx F res n e l i S in e i I n t e g r a l ( S ( z )) = ∫ 0 z s in ( 2 π x 2 ) d x
hence:
∫ s i n ( x 2 ) d x = [ 2 π 2 S ( 2 x π ) + c \int sin(x^2)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}x}{\sqrt{\pi}})+c ∫ s in ( x 2 ) d x = [ 2 2 π S ( π 2 x ) + c
therefore:
∫ 1 2 ( s i n ( x 2 ) + x y ) d x = [ 2 π 2 S ( 2 x π ) + x 2 y 2 ] 1 2 \int^{2}_{1}(sin(x^2)+xy)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}x}{\sqrt{\pi}})+\frac{x^{2}y}{2}]^{2}_{1}\\ ∫ 1 2 ( s in ( x 2 ) + x y ) d x = [ 2 2 π S ( π 2 x ) + 2 x 2 y ] 1 2
∫ 1 2 ( s i n ( x 2 ) + x y ) d x = [ 2 π 2 S ( 2 ( 2 ) π ) + ( 2 ) 2 y 2 ] − [ 2 π 2 S ( 2 ( 1 ) π ) + ( 1 ) 2 y 2 ] \int^{2}_{1}(sin(x^2)+xy)dx=[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(2)}{\sqrt{\pi}})+\frac{(2)^{2}y}{2}]-[\frac{\sqrt{2}\sqrt{\pi}}{2}S(\frac{\sqrt{2}(1)}{\sqrt{\pi}})+\frac{(1)^{2}y}{2}]\\ ∫ 1 2 ( s in ( x 2 ) + x y ) d x = [ 2 2 π S ( π 2 ( 2 ) ) + 2 ( 2 ) 2 y ] − [ 2 2 π S ( π 2 ( 1 ) ) + 2 ( 1 ) 2 y ]
∫ 1 2 ( s i n ( x 2 ) + x y ) d x = 0.4945081876 + 1.5 y \int^{2}_{1}(sin(x^2)+xy)dx=0.4945081876+1.5y ∫ 1 2 ( s in ( x 2 ) + x y ) d x = 0.4945081876 + 1.5 y
#339914 on Dec 2023