Integral from 0 to inf with imaginary numbers takes too long

I'm testing my code and most lines goes well. When it comes to the last opration it takes too long and is always busy.

Here's my code:

syms Kt Ks Cs M m to
syms x t
Kt=1
Ks=1
Cs=1
M=1
m=1
to=1
y=x*(exp(1))^(-sqrt(-1)*t*x)
S=int(y,-to/2,to/2)
SS=(1/2)*(abs(S))^2
G=2*SS
H=(t^2)*abs((Kt*(Ks+sqrt(-1)*t*Cs))/(((-(t^2)*M+Ks+sqrt(-1)*t*Cs)*(-(t^2)*m+Ks+sqrt(-1)*t*Cs))-((Ks+sqrt(-1)*t*Cs)^2)))
sigpre=G/(H^2)
sig2=int(sigpre,0,inf)
sig=sig2^(1/2)

and it stucks when doing this:

sig2=int(sigpre,0,inf)

Could anyone please fix my code?

Answer 1

[Star Strider]

First, the Symbolic Math Toolbox is not going to be as efficient as direct numerical calculations.

Second, you never told the int function what variable of integration is. It is necessary to specify whether it is ‘x’ or ‘t’ (or something else).

Supply that information, and use the integral or integral2 functions instead (depending on what you are integrating, something I cannot figure out from the posted code, so it does not surprise me that the int function could not figure it out either).

Here is an example —

% syms Kt Ks Cs M m to
% syms x t
Kt=1

Kt = 1

Ks=1

Ks = 1

Cs=1

Cs = 1

M=1

M = 1

m=1

m = 1

to=1

to = 1

y = @(x,t) x.*exp(-sqrt(1i*t*x));
S = @(x) integral(@(t) y(x,t),-to/2,to/2)

S = function_handle with value:
    @(x)integral(@(t)y(x,t),-to/2,to/2)

SS = @(x) (1/2)*(abs(S(x))).^2;
G =  @(x) 2*SS(x);
H = @(x,t) (t.^2).*abs((Kt.*(Ks+sqrt(-1).*t.*Cs))./(((-(t.^2).*M+Ks+sqrt(-1).*t.*Cs).*(-(t.^2).*m+Ks+sqrt(-1).*t.*Cs))-((Ks+sqrt(-1).*t.*Cs).^2)));
sigpre = @(x,t) G(x)./(H(x,t).^2)

sigpre = function_handle with value:
    @(x,t)G(x)./(H(x,t).^2)

sig2 = @(t) integral(sigpre(x,t),0,Inf)

sig2 = function_handle with value:
    @(t)integral(sigpre(x,t),0,Inf)

sig = sig2(t).^(1/2)

Unrecognized function or variable 't'.

Make appropriate changes to get the result you want.

I will help with this, however I need much more information than the code provides.

.

[BOSHU]

Thank you, but I still have some problems.

The code is to calculate IRI (International Roughness Index) using acceleration data, I have done this job:

I'd like to integration 't' in y and 'x' in sig2, so I changed the code like below:

% syms Kt Ks Cs M m to
% syms x t
Kt=1;
Ks=1;
Cs=1;
M=1;
m=1;
to=1;
y = @(x,t) t.*exp(-1i*x*t);
S = @(x) integral(@(t) y(x,t),-to/2,to/2);
SS = @(x) (1/2)*(abs(S(x))).^2;
G =  @(x) 2*SS(x);
H = @(x) (x.^2).*abs((Kt.*(Ks+sqrt(-1).*x.*Cs))./(((-(x.^2).*M+Ks+sqrt(-1).*x.*Cs).*(-(x.^2).*m+Ks+sqrt(-1).*x.*Cs))-((Ks+sqrt(-1).*x.*Cs).^2)));
sigpre = @(x) G(x)./(H(x).^2);
sig2 = @(x) integral(sigpre(x),0,Inf);
sig = sig2(x).^(1/2)

Unrecognized function or variable 'x'.

But I still cannot get the answer (with this error) and I don't know why.

By the way, I'm just a high school student and is new to MATLAB, but I'm willing to check the guide if you could told me the knowledge I need to learn. Not as a native English speaker, Please let me know if you cannot understand me and please forgive me if I used unproper words.

Thank you very much

[Star Strider]

This took longer than the permitted 5 seconds to run here,. so I ran it in MATLAB Online.

Kt=1;
Ks=1;
Cs=1;
M=1;
m=1;
to=1;
y = @(x,t) t.*exp(-1i*x*t);
S = @(x) integral(@(t) y(x,t),-to/2,to/2);
SS = @(x) (1/2)*(abs(S(x))).^2;
G =  @(x) 2*SS(x);
H = @(x) (x.^2).*abs((Kt.*(Ks+sqrt(-1).*x.*Cs))./(((-(x.^2).*M+Ks+1i.*x.*Cs).*(-(x.^2).*m+Ks+1i.*x.*Cs))-((Ks+1i.*x.*Cs).^2)));
sigpre = @(x) G(x)./(H(x).^2);
sig2 = integral(sigpre,0,Inf, 'ArrayValued',1)
sig = sig2.^(1/2)

Result —

sig2 =

2.7534e+75

sig =

5.2473e+37

I do not understand the integration from 0 to Inf, since the integrals in the image all have finite integration limits.

Your English is fine. What I fail to understand is what you are doing.

.

[BOSHU]

I realized that I have made a mistake.

As the picture (Written by my classmate after reading the paper), means vertical acceleration and it should be a function fitting to discrete data which should not be monotone functions like , so I changed it into and everything turns out well. I'm just testing my code without getting the data (which we will get later) and I set them all as 1, so I made the mistake.

Thank you very much.

[Star Strider]

As always, my pleasure!

Answer 2

[Paul]

Hi BOSHU

Here is my attempt.

syms Kt Ks Cs M m to  % overwritten later
syms x

Assume t is real, is that a good assumption?

syms t real  

Kt = 1;
Ks = 1;
Cs = 1;
M  = 1;
m  = 1;
to = 1;

When using symbolic, it's best to use symbolic constants to force the desired expression. Here, exp(1) is first evaluated as a double

y = x*(exp(1))^(-sqrt(-1)*t*x)

y = 

Force symbolic

y = x*(exp(sym(1)))^(-1i*t*x)

y = 

Or more simply

y = x*exp(-1i*t*x)

y = 

Evaluate S and simplify

symvar(y)

ans = 

symvar(y,1)

ans = 

x

This line integratres wrt x, better to make that explicit.

%S = int(y,-to/2,to/2)
S = int(y,x,-to/2,to/2)

S = 

[num,den] = numden(S)

num = 

den = 

S = num/den

S = 

Evaluate SS and simplify. Sometimes using conj works better than abs()^2

SS=(1/2)*(abs(S))^2

SS = 

SS=(1/sym(2))*S*conj(S)

SS = 

[num,den] = numden(SS)

num = 

den = 

SS = num/den

SS = 

Compute G

G=2*SS;

Compute and simplify H

H = (t^2)*abs((Kt*(Ks+sqrt(-1)*t*Cs))/(((-(t^2)*M+Ks+sqrt(-1)*t*Cs)*(-(t^2)*m+Ks+sqrt(-1)*t*Cs))-((Ks+sqrt(-1)*t*Cs)^2)))

H = 

H = simplify(H,100)

H = 

Sigpre

sigpre = G/(H^2)

sigpre = 

sigpre = simplify(sigpre,10)

sigpre = 

symvar(sigpre)

ans = 

t

sigpre is a function of only one variable, so it's not necesary to specif the variable of integration to int, but it makes things cleare IMO

sig2 = int(sigpre,t,0,inf)

sig2 = 

Integral not found. Sometimes operatingon the integrand can help.

sig2 = int(expand(sigpre),t,0,inf)

sig2 = 

Still doesn't look too promising. We can try to get a better expression for sigpre

sigpre = simplify(expand(rewrite(sigpre,'sincos')))

sigpre = 

sig2 = int(sigpre,t,0,inf)

sig2 = 

Are you sure all the equations are implemented correctly?

%sig=sig2^(1/2)

Answer 3

[Torsten]

It seems your integral does not exist.

Note that by default, y is integrated with respect to x. I don't know if this is wanted or not. If not, you will have to sepcify the integration variable:

S=int(y,t,-to/2,to/2);

or

S=int(y,x,-to/2,to/2);

But in case

S=int(y,t,-to/2,to/2);

is correct, values for x in the integral

integral(sigpre,0,Inf)

are missing.

syms Kt Ks Cs M m to
syms x t
Kt=1;
Ks=1;
Cs=1;
M=1;
m=1;
to=1;
y=x*(exp(1))^(-sqrt(-1)*t*x);
S=int(y,-to/2,to/2);
SS=(1/2)*(abs(S))^2;
G=2*SS;
H=(t^2)*abs((Kt*(Ks+sqrt(-1)*t*Cs))/(((-(t^2)*M+Ks+sqrt(-1)*t*Cs)*(-(t^2)*m+Ks+sqrt(-1)*t*Cs))-((Ks+sqrt(-1)*t*Cs)^2)));
sigpre=G/(H^2);
sigpre = matlabFunction(sigpre);
integral(sigpre,0,Inf)

Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 3.1e+24. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.

ans = 3.8771e+25

[BOSHU]

Thank you very much.

In S, I'd like to integrate with respect to x, so it should be right

In integral to sigpre, I'd like to do it with respect to t, I changed your code but it still don't works.