Can I get the integral formula using Matlab?

I have the following integration:

and m and n are integers.

Can I get the formula for this integration using Matlab?

Here is my attempt:

clear all
syms x m n 
f = (1/pi)*sin(n*x/2)*(pi-x)^2*sin(m*x/2);
Fint = int(f,x,[0 2*pi]);

I got this long result which is ok for me in case it is correct:

(5734161139222659*(8*m^3*cos(pi*m)*sin(pi*n) - 8*n^3*cos(pi*n)*sin(pi*m) - 8*pi*m*n^3 + 8*pi*m^3*n + 4*m^4*pi*sin(pi*m)*sin(pi*n) - 4*n^4*pi*sin(pi*m)*sin(pi*n) + 24*m*n^2*cos(pi*m)*sin(pi*n) - 24*m^2*n*cos(pi*n)*sin(pi*m) - m^5*pi^2*cos(pi*m)*sin(pi*n) + n^5*pi^2*cos(pi*n)*sin(pi*m) - m*n^4*pi^2*cos(pi*m)*sin(pi*n) + m^4*n*pi^2*cos(pi*n)*sin(pi*m) - 2*m^2*n^3*pi^2*cos(pi*n)*sin(pi*m) + 2*m^3*n^2*pi^2*cos(pi*m)*sin(pi*n) - 8*m*n^3*pi*cos(pi*m)*cos(pi*n) + 8*m^3*n*pi*cos(pi*m)*cos(pi*n)))/(9007199254740992*(m^2 - n^2)^3)

when I assume then the result is undefied..

I appreciate any help..

[Stephen23]

A much more accurate approach is to use SYM(PI):

syms x m n 
f = (1/sym(pi))*sin(n*x/2)*(sym(pi)-x)^2*sin(m*x/2);
Fint = int(f,x,[0,2*pi])

Fint = 

"when I assume then the result is undefied.."

Because the integral is undefined for those values:

fsurf(Fint)

fsurf(Fint,[0.5,1.5])

subs(Fint,{m,n},{1,1})

Error using sym/subs (line 156)
Division by zero.

[Paul]

Hi Stephen,

In addition to using sym(pi),

syms x m n 
f = (1/sym(pi))*sin(n*x/2)*(sym(pi)-x)^2*sin(m*x/2);
Fint = int(f,x,[0,2*sym(pi)])

Fint = 

One "feature" of int is that it (almost) never recognizes special cases of the parameters of the integrand. For the particular case at hand, if we set m = n = 1 first, then the integral follows

int(subs(f,[m,n],[1,1]),x,[0,2*sym(pi)])

ans = 

This result can also be obtained by taking the limit of Fint (though, TBH I'm not sure if this sequential limit approach is strictly correct)

simplify(limit(limit(Fint,n,1),m,1))

ans = 

If m and n are both integers (not stated by the OP, but might be the case based on the notation), then Fint simplifies further

assumeAlso([m,n],'integer');
simplify(Fint,100)

ans = 

Answer 1

[David Goodmanson]

Hello Haya,

The integral is very much defined for m=n. In that case you are integrating the function

(1/pi)*sin(n*x/2)^2*(x-pi)^2                 

which is well behaved and has a calculable integral. The syms expression contains a denominator

1/(m^2-n^2)^3

which has a factor of

1/(n-m)^3

and is a cubic zero when m-->n. When m--> n the numerator also must contain a factor of (n-m)^3 so that the entire expression will be a (yet to be determined) 0/0 form. To get the result you don't have to actually suss out the (n-m)^3 factor in the numerator. It's enough to know that it exists. Taking the m=n expression at the top above,

clear all
syms n x
assume(n,{'positive','integer'})
pi = sym('pi')
f = (1/pi)*sin(n*x/2)^2*(pi-x)^2;
Fint = int(f,x,[0 2*pi])

you obtain

Fint = 
pi^2/3 - (n*(pi + pi*cos(2*n*pi)) - sin(2*n*pi) + (n^2*pi^2*sin(2*n*pi))/2)/(n^3*pi)

and after setting

sin(2*n*pi) = 0    cos(2*n*pi) = 1  

(there must be a way to get syms to do this but I don't know what it is) this reduces to

Fint = pi^2/3 - 2/n^2

which doesn't work for n = 0, but that case can be ignored since the integrand is 0.

[Paul]

Hi David,

clear all
syms n x
assume(n,{'positive','integer'})

Here, Pi1 is not the sympolic constant pi. Its just a symbolic variable that renders with the pi symbol.

Pi1 = sym('pi')

Pi1 = 

π

which is why, e.g., sin(Pi1) is not known.

sin(Pi1)

ans = 

f = (1/Pi1)*sin(n*x/2)^2*(Pi1-x)^2;
Fint = int(f,x,[0 2*Pi1])

Fint = 

Consequently, Fint doesn't simplify

simplify(Fint)

ans = 

Instead, define the symbolic constant this way

Pi2 = sym(pi)

Pi2 = 

π

Which is not the same as Pi1

Pi2 - Pi1

ans = 

sin(Pi2)

ans = 

0

Now we get the same-looking integral

f = (1/Pi2)*sin(n*x/2)^2*(Pi2-x)^2;
Fint = int(f,x,[0 2*Pi2])

Fint = 

but it simplifies as expected.

simplify(Fint)

ans = 

IIRC, sym('pi') used to return the symbolic constant pi, but that behavior changed a few releases ago.

[David Goodmanson]

Hi Paul, thanks for the explanation

Answer 2

[Walter Roberson]

syms x m n 
f = (1/sym(pi))*sin(n*x/2)*(pi-x)^2*sin(m*x/2);
Fint = int(f,x,[0 2*pi]);
limit(Fint, m, n)

ans = 

subs(ans, n, 1)

ans = 

simplify(ans)

ans = 

●

[David Goodmanson]

Hi Walter, thanks for the clarification