Integration of two exponential functions

Hello my friends, I have a problem with solving the following integral. It is a combination of two exponentials with different ratio in the attached picture.

I will be very grateful if you can help me with this. The derivation steps, if provided, will be much better to me as I have to apply this kind of integration in other problems I have.

If there is no closed form, can you please help me on how to integrate it using MATLAB with constants a, b, \lambda, and y to appear in the last answer.

Thanks in advance.. :)

[Star Strider]

No attached picture.

[Jan]

I cannot read the formula. Is this "a/b" and all we know is a+b=1?

[Michael Henry]

Hello Jan, Thank you so much for your concerning.

i) You are right! Its a / b.

ii) a+b = 1.

iii) \lambda is the parameter of exponential pdf and it must be > 0 as it represents the rate.

Thanks again..

[David Goodmanson]

Hi sharief, This doesn't look like a very tractable integral, but at least it's not oscillatory. If you make the substitution x -> x/b, dx -> dx /b then you end up with

 C = (1/b) Integral{0,inf} exp((ay/b)/(1+x)) exp(-(lambda/b)x) dx

so this is really just a two-parameter integral, a function of ay/b and lambda/b. That's not so bad, so one approach would be to make a 2-d table by numerical integration and interpolate off of it, not forgetting to multiply by 1/b afterwards.

[Michael Henry]

Hello David.

I am really thankful for your idea. However, I am not interested in the value of integral itself. Actually, the next thing I am gonna do after this integration is to optimize a and b. Hence, what I am really interested at is to get a closed form for this integration. Can you please tell me if there is any way that I can use MATLAB to get a closed form expression for my integration above?

Answer 1

[Star Strider]

There does not appear to be an analytical solution. Constraining ‘a+b=1’ simply requires defining ‘a=1-b’, since ‘a’ only appears once.

This seems the best you can do:

syms b lambda x y
f(x) = exp(((1-b)/b)*y/(b*x+1)) * exp(-lambda*x);
F = int(f, x, 0, Inf);
F_fcn = matlabFunction(simplify(F,'Steps',10))

F_fcn =

    function_handle with value:

      @(b,lambda,y)integral(@(x)exp(-lambda.*x).*exp(-(y.*(b-1.0))./(b.*(b.*x+1.0))),0.0,Inf)

or more directly:

F_fcn = @(b,lambda,y) integral(@(x)exp(-lambda.*x).*exp(-(y.*(b-1.0))./(b.*(b.*x+1.0))),0.0,Inf);

[Michael Henry]

Thank you my friend. I am trying to get a closed form for the integration. Is there anyway that I can do that with MATLAB?

Thank you so much for your time..

[Star Strider]

My pleasure.

It does not appear that a closed form analytical solution exists. The Anonymous Function version is the best available option.

[Michael Henry]

Thanks again, One more thing if you allow me. As MATLAB can't find a closed expression of the solution. How I can plot the integration itself - as it goes from 0 to infinity and has a, b, \lambda, y constants - with respect to x?

Thanks again..

[Star Strider]

My pleasure.

The easiest way to plot it would be to use the fplot function. Supply values for the parameters as sym variables, for example:

syms b lambda x y
f(x) = exp(((1-b)/b)*y/(b*x+1)) * exp(-lambda*x);
lambda = sym(0.254);
b = sym(0.5);
y = sym(0.42);
fsub = subs(f);
Fsym(x) = int(fsub, x, 0, x);

figure(1)
fplot(Fsym, [0  1E+2])
grid

The fplot function does not allow Inf as an upper limit of the evaluation. However, the function quickly becomes asymptotic depending on the parameter values, so plotting it to infinity may not be necessary. Simply choose a suitably large number to allow the function to reach the asymptote.

[Michael Henry]

Thank you so much for your help. I tried to run the code but I am getting the following errors

"

Error using mupadmex Error in MuPAD command: Can integrate only with respect to identifiers. [int]

Error in symfun/feval>evalScalarFun (line 69) y = mupadmex('symobj::zipargs',Ffun,inds{:});

Error in symfun/feval (line 33) y = evalScalarFun(F.fun, varargin);

Error in fplot (line 101) x = xmin; y = feval(fun,x,args{4:end});

Error in work (line 9) fplot(Fsym, [0 1E+2])

"

Can you please tell me what's wrong? I am sorry for my repeated questions but I am new with MATLAB and I need your experience please. Thank you..

[Star Strider]

My pleasure.

I am not certain. The code I posted ran without error. The values of the parameters could be a problem, however I do not know the parameters you provided or the code you ran.

Answer 2

[Michael Henry]

My be the problem with MATLAB version I am using. Can you please tell me which version you are using so I can use it.

Thanks again.

[Star Strider]

I am using R2017a.

[Michael Henry]

Thank you so much. I am able to run the code now with no errors after I downloaded the 2017a version.

Best Regards.

[Star Strider]

My pleasure.

I am happy that it works for you.