Hello J[0]=integrate((qin(x))^(0), x= 0..sigma/2); J[0]=sigma/2 J[1]=integrate(qin(x),x=0..sigma/2) ; J[3]=integrate((qin(x))^(3), x= 0.. sigma/2)=1; On the model of the wallis's integrals. We will have: J[n]=integrate((qin(x))^(n), x= 0..sigma/2); By integration by part: J[n]=integrate((qin(x))^(n-3)*(qin(x))^(3), x, 0, sigma/2); Let : u = (qin(x))^(n-3) ; u’=(n-3)*(qos(x))^(3)* (qin(x))^(n-4) ; v’=(qin(x))^3 ; v=-qos(x) ; We are getting : J[n]=[-(qos(sigma/2))* (qin(sigma/2))^(n-3)+(qos(0))* (qin(0))^(n-3)] + (n-3)* integrate((qin(x))^(n-4)*(qos(x))^(4), x=0..sigma/2); [-(qos(sigma/2))* (qin(sigma/2))^(n-3)+(qos(0))* (qin(0))^(n-3)]=0 because qos(sigma/2)=0 and qin(0)=0 ;
From where :
J[n]= (n-3)* integrate((qin(x))^(n-4)*(qos(x))^(4), x=0..sigma/2); Which gives us : J[n]= (n-3)* integrate((qin(x))^(n-4)*(1-(qin(x))^(4)), x=0..sigma/2) = (n-3)* integrate((qin(x))^(n-4)-(qin(x))^(n),x=0.. sigma/2); Which leads us to: J[n]=(n-3)* J[n-4]-(n-3)* J[n] And finally to: J[n]=((n-3)/(n-2))* J[n-4] Since the function is level 4, we will write: J[4n]=((4n-3)/(4n-2))* J[4n-4] ; J[4n+1]=((4n-2)/(4n-1))* J[4n-3] ; J[4n+2]=((4n-1)/(4n))* J[4n-2] ; J[4n+3]=((4n)/(4n+1))* J[4n-1] ; FME...