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how to do definite integration when the integrand has a constant term

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Akriti Raj 2021-7-22

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评论： Akriti Raj 2021-7-23

How to find the value of this integral in matlab ? q should remain as it is in the final answer as I is a function of q

sigma= 1/(1.5e-12)

L=6.4747e3

k=2*pi*1.8331/(390e-12)

D=2.3167e-4

omega sholud me from 379.62 to 389.62

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Walter Roberson 2021-7-22

编辑：Walter Roberson 2021-7-22

在 MATLAB Online 中打开

format long g

Q = @(v) sym(v); %convert to rational

Pi = Q(pi);

sigma = 1/(Q(1.5)*10^-12)

sigma =

L = Q(64747)/10

L =

k = 2 * Pi* Q(18331)/10^4/(Q(390)*10^-12)

k =

D = Q(23167)*10^-8

D =

omega_min = Q(37962)/100

omega_min =

omega_max = Q(38962)/100

omega_max =

syms q

syms omega

inner = exp((omega/sigma)^2/2) * sinc(L*q^2/k + D*omega*L/2).^2

inner =

I = 1/(sigma * sqrt(2*Pi)) * int(inner, omega, omega_min, omega_max)

I =

simplify(I)

ans =

inner_approx = taylor(inner, omega, (omega_min+omega_max)/2, 'order', 4)

inner_approx =

I_approx = 1/(sigma * sqrt(2*Pi)) * int(inner_approx, omega, omega_min, omega_max)

I_approx =

simplify(I_approx)

ans =

fplot(I, [-5 5]); title('exact')

double(subs(I,q,[-5 -1 0 1 5]))

ans = 1×5

3.49318454188821e-18 3.49318582811878e-18 3.4931858817153e-18 3.49318582811878e-18 3.49318454188821e-18

vpa(subs(I,q,[-5 -1 0 1 5]*10^9))

ans =

fplot(I_approx, [-5 5]); title('approximated')

double(subs(I_approx,q,[-5 -1 0 1 5]))

ans = 1×5

-3.2139136293569e-16 -3.21388912778066e-16 -3.21388810680452e-16 -3.21388912778066e-16 -3.2139136293569e-16

vpa(subs(I_approx,q,[-5 -1 0 1 5]*10^9))

ans =

David Goodmanson 2021-7-22

thanks Walter I missed that

Akriti Raj 2021-7-23

@walter q should be between -1 to 1

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