How to Integrate Lognormal PDF multiplied by a function.

Hi,

I am trying to ∫ D(V)*F(V)dV eq1

min=_(0.8e-26) and max=(1.35e-26)

D(t,V)=2*Ps*(1-exp(-(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300))))) and F(V)= Lognormal PDF=(1./(V.*sigma.*sqrt(2.*pi)).*exp(-log(V)-mu).^2/(2.*sigma^2)).*V

Find below, the code I wrote to compute the integral of eq1:

Dp=@(V)(1./(V.*sigma.*sqrt(2.*pi)).*exp(-log(V)-mu).^2/(2.*sigma^2)).*V.*2*Ps*(1-exp(-(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300)))))

Dp =
  function_handle with value:
 @(V)(1./(V.*sigma.*sqrt(2.*pi)).*exp(-log(V)-mu).^2/(2.*sigma^2)).*V.*2*Ps*(1-exp(-(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300)))))
  
>> integral(Dp,0.8e-26,1.35e-26)

Error message I got were:

Matrix dimensions must agree.

Error in
@(V)(1./(V.*sigma.*sqrt(2.*pi)).*exp(-log(V)-mu).^2/(2.*sigma^2)).*V.*2*Ps*(1-exp(-(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300)))))

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 75)
        [q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);

Please can anyone kindly assist me or advice me?

Thank you

[Tommy]

Would you be able to provide all of your code, or at least the values you are using for all variables (sigma, mu, Ps, t, Vo, Wb, Kb)?

[David Hill]

Yes, and place you code in a proper code block.

[Mudaga Andrew Nomuoja]

Thanks Tommy, here we go:

Kb=1.38e-23;Ps=0.74;Wb=1.44e-19;Vo=1.00e-13:V=1.00e-26: sigma=0.05:Mu=0;

[Tommy]

Thanks - how about t?

Answer 1

[Tommy]

The error (Matrix dimensions must agree) occurs because your t is an array of length 3503185 and the V passed into Dp() by integral() is an array of length 150. Therefore, the following portion of Dp() throws the error:

>> -(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300)))
Matrix dimensions must agree.

You can find the integral of Dp() over your range of V by specifying a scalar value for t. If you want to see how the integral varies over a range of log(t), then you can evaluate the integral at each value of t independently and combine the results:

Kb=1.38e-23;Ps=0.74;Wb=1.44e-19;Vo=1.00e-13;V=1.00e-26;sigma=0.05;mu=0;
t=logspace(log10(0.00000226),log10(55),5000);
Dp=@(V,t)(1./(V.*sigma.*sqrt(2.*pi)).*exp(-log(V)-mu).^2/(2.*sigma^2)).*V.*2*Ps.*(1-exp(-(t)./(Vo*exp((Ps*10000000-Wb)*V/(Kb*300)))));
y = arrayfun(@(t) integral(@(V) Dp(V, t), 0.8e-26, 1.35e-26), t);
plot(log(t), y);

I decreased the length of t considerably because it would take a very long time to evaluate the integral 3503185 different times, and I changed t from a linearly-spaced vector to one spaced logarithmically.

[Mudaga Andrew Nomuoja]

Thanks Tommy, I will run the code and let you the outcome. I am immensely grateful for your feedback!!!

[Mudaga Andrew Nomuoja]

The code worked fine and generated a graph for further analysis. I most grateful.

Answer 2

[Mudaga Andrew Nomuoja]

Hi Tommy,

t=0.00000226:0.0000157:55;x=log(t);

I was hoping to plot the integral Dp as a function of log(t). Hence I used the above parameter.