I'm trying to integrate the following function from -inf to +inf with repsect to x.
fun = @(x) -3.*v.^2 -4.*v.*a./((x.^2+a.^2)*pi) - a.^2./((x.^2+a.^2)^2*pi^2)
fun =
function_handle with value:
@(x)-3.*v.^2-4.*v.*a./((x.^2+a.^2)*pi)-a.^2./((x.^2+a.^2)^2*pi^2)
>> integral(fun,-inf,inf)
Error using symengine
Not a square matrix.
Error in sym/privBinaryOp (line 1030)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in ^ (line 330)
B = privBinaryOp(A, p, 'symobj::mpower');
Error in @(x)-3.*v.^2-4.*v.*a./((x.^2+a.^2)*pi)-a.^2./((x.^2+a.^2)^2*pi^2)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 103)
[q,errbnd] = vadapt(@minusInfToInfInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
I don't know why i'm getting this error?
by the way v and a are both constants

 Akzeptierte Antwort

Ameer Hamza
Ameer Hamza am 4 Nov. 2020

0 Stimmen

You need use element-wise operator:
fun = @(x) -3.*v.^2 -4.*v.*a./((x.^2+a.^2)*pi) - a.^2./((x.^2+a.^2).^2*pi^2)
%^ put a dot here

4 Kommentare

Hi I just tried this, but I now have another error:
fun = @(x) -3.*v.^2 -4.*v.*a./((x.^2+a.^2)*pi) - a.^2./((x.^2+a.^2).^2*pi^2)
fun =
function_handle with value:
@(x)-3.*v.^2-4.*v.*a./((x.^2+a.^2)*pi)-a.^2./((x.^2+a.^2).^2*pi^2)
>> integral(fun,-inf,inf)
Error using integralCalc/finalInputChecks (line 522)
Input function must return 'double' or 'single' values. Found 'sym'.
Error in integralCalc/iterateScalarValued (line 315)
finalInputChecks(x,fx);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 103)
[q,errbnd] = vadapt(@minusInfToInfInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Randy Chen
Randy Chen am 4 Nov. 2020
i guess there's something wrong with my function in the first place?
The integral function is for numeric integration. If you have fixed values for a and v, define variables with those values before creating your fun function. If you want to perform the integration symbolically use the int function on a symbolic expression (not a function handle.)
syms a x
f = a*x^2;
int(f, x, 0, 5)
a = 3;
f = @(x) a*x.^2;
integral(f, 0, 5)
Randy Chen
Randy Chen am 4 Nov. 2020
thank you!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by