Multivariable Integration w.r.t to 1 variable: 2nd variable is a function of 1st

I need to find the integral of the function: , where g is an algebraic polynomial. I have been using the symbolic toolbox but the integral cant be evaluated and I obtain just the command as the output. The integration is w.r.t y, but there is a catch y itself is a function of x (my phyical model is such, but x and y represent two different quantitites). Moreover the limits of the integration w.r.t to y are functions of x. This is possible with the int function of sym toolbox but can't use as above mentioned.

I am trying using integral through a numerical approach which looks like this:

k = @(T) T;
w = @(k,T) log(1 -  exp(-k^2 + 1/T))
a = @(T) double(T);
b = @(T) double(2*T);
W = @(T) integral(@(k) w(k,T), a, b )

This time I get two errors during execution: i) a and b must be floating scalars

ii) W just returns the exact line of the command again viz: "integral(@(k) w(k,T), a, b )"

If any can suggest a way out, it will be really helpful.

Answer 1

[Torsten]

If y is a function of x, your integrand equals F(y(x~))*y'(x~) dx~. Thus your integral equals

integral_{x~=f_lower(x)}^{x~=f_upper(x)} F(y(x~))*y'(x~) dx~

and the result is a function of x.

So what is f_lower(x), f_upper(x) and y(x) in your case ?

[Aviral Srivastava]

The limits are . I have worked around a way in which I can use a specific numerical value of x in this integral. But for that I need the sample code I provided above to work. Could you please suggest a way around this.

[Torsten]

It's not possible that you choose the same variable as integration variable and as a variable that appears in the limits of your integral.

Further, k in your code above is a function handle. You can't use it at the same time as integration variable.

Could you write down in a mathematical notation the integral that you try to evaluate ?

[Aviral Srivastava]

Turns out in my model I can work around by rendering the limits as numerical values instead of them being functions os the variable. In that case the sample code becomes:

w = @(k,T) log(1 - exp(-k^2 + 1/T))*(at/k + bt*k)

a = @(T) 0;

b = @(T) 1;

W = @(T) integral(@(k) w(k,T), a, b )

I am not able to run this code without getting an error. The function w here represents the exact integral I want to carry out. If it can be done by int function it will greatly simplify the situation. If the above implementation works then also it is fine. The problem is I can't get head around either one.

[Torsten]

at = 1;
bt = 1;
w = @(k,T) log(1 -  exp(-k.^2 + 1/T)).*(at./k + bt*k);
a = @(T) eps;
b = @(T) 1;
W = @(T)integral(@(k)w(k,T),a(T),b(T))

W = function_handle with value:
    @(T)integral(@(k)w(k,T),a(T),b(T))

W(2)

ans = -1.7937e+01 + 1.1293e+02i

[Aviral Srivastava]

I think I forgot to mention at and bt are sym variables. In the end I have to solve for them after evaluating these integrals.

[Torsten]

syms at bt 
w1 = @(k,T) log(1 -  exp(-k.^2 + 1/T))./k ;
w2 = @(k,T) log(1 -  exp(-k.^2 + 1/T)).*k ;
a = @(T) eps;
b = @(T) 1;
W1 = @(T)integral(@(k)w1(k,T),a(T),b(T));
W2 = @(T)integral(@(k)w2(k,T),a(T),b(T));
W = @(T)at*W1(T)+bt*W2(T)

W = function_handle with value:
    @(T)at*W1(T)+bt*W2(T)

W(2)

ans = 

[Aviral Srivastava]

Thank you this helps a lot. Finally could you let me know if there is a way to approach this using the symbolic toolbox, as my existing framework is using that approach only.

[Dyuman Joshi]

The output expression looks a bit complicated -

syms at bt k T 
w1(k, T) = log(1 -  exp(-k^2 + 1/T))/k ;
w2(k,T) =  log(1 -  exp(-k^2 + 1/T))*k ;
a = sym('eps');
b = 1;
W1 = int(w1, k, a, b);
W2 = int(w2, k, a, b);
W(T) = at*W1+bt*W2(T);

out = W(2)

out = 

[Aviral Srivastava]

I must thank you for uptaking this effort, this is the exact problem that I am facing. Getting the functional form in the 2nd term but not sure about the 1st integral.

[Torsten]

Finally could you let me know if there is a way to approach this using the symbolic toolbox, as my existing framework is using that approach only.

W(2) is a symbolic expression - thus you are back in the realm of symbolic computations.

I don't think you can convert W = W(T) into a symbolic function since a numerical function (integral) is involved.

[Dyuman Joshi]

There's a good chance that the symbolic integral can not be computed.

If you want a numerical value, it will be better to go with numerical integration as Torsten has shown above.