TO SOLVE AN INTEGRAL WHERE A FUNCTION CONTAINS A COMPLEX SUM

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RB 2017-5-14

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https://ww2.mathworks.cn/matlabcentral/answers/340189-to-solve-an-integral-where-a-function-contains-a-complex-sum

评论： RB 2017-5-16

Hello! I am stuck up with a problem where my function contains a sum of certain quantities indeed matrices and scalars that are computed in a loop. The matrices are 2 by 2.

delta=0.75;
for i=2:n
           APHNEW=expm([(i-1).*H,(i-1).*(2*(Q'*Q));(i-1).*(R+M),(i-1).*(-(H'))]);
           APHNEW11=APHNEW(1:2,1:2); %That is good
           APHNEW21=APHNEW(3:4,1:2);
           APHNEW12=APHNEW(1:2,3:4);
           APHNEW22=APHNEW(3:4,3:4);
           BPHNEW22=inv(APHNEW22);
           PSIi=BPHNEW22*APHNEW21;
           SPSI1=SPSI1+PSIi;
           PHIi=beta*(log(det(APHNEW22))+(i-1)*trace(H'))/2;          
           S0i(i)=exp(-((rbar+mubar)*(i-1)+PHIi));
           S0i2(i)=(-((rbar+mubar)*(i-1)+PHIi));
           Yn0=Yn0+S0i2(i);
           fun0=@(Gamma) exp(trace(1i*(THETA1*inv(eye(2)-2i*SIGMA*Gamma)*SIGMA*Gamma)))/((det(eye(2)-2i*SIGMA*Gamma))^(beta/2));
           %In the next line Gamma1 is a scalar and is indeed the parameter
           %of the Fourier Transform
           %the next two lines are separate functions
           fun1 = @(Gamma1) exp((1i*Gamma1+delta)*Yn0)*S0i(i)*fun0(1i*PSIi-(Gamma1-1i*delta)*SPSI); 
           fun3 = @(Gamma1) exp((1i*Gamma1+delta)*Yn0)*(S0i(i)*fun0(1i*PSIi-(Gamma1-1i*delta)*SPSI)-K*fun0(-(Gamma1-1i*delta)*SPSI)); 
  end

I need to compute fun1 on sum of

S0i(i)*fun0(1i*PSIi-(Gamma1-1i*delta)*SPSI);

and then use this in fun3 which I am writing without sum.

I then use it to define a new function which I have to integrate using quadgk between 0 to infinity. However I am not being able to sum the arguments since Gamma1 is unknown and used in quadgk.

Any help would be great.

Thanks

Warm regards,

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RB 2017-5-14

Actually I am still stuck up the fact that the function involves a sum of arguments is creating a problem. I canot run a loop to sum the arguments of the function.

P.S. The other 'i' appearing in exponent and in functions is iota.

Any help would be appreciated.

Thanks.

RB 2017-5-16