How to symbolically integrate f(T) = exp (E/(8.314*T) through integration by parts or with trapezoidal method?

I have a function in T which must be integrated trapezoidally. Eα is a constant. Can someone suggest how to perform such an integration in matlab?

Since this is an exponential inteagral with no unique solution, another option is to integrate by parts upto n times, ignore the nth integral and assume the terms upto n-1 as the solution. can this be done in matlab?

Thanks in advance :)

[J. Alex Lee]

is this a homework or exercise for implementing trapezoidal integration, or do you need to evaluate the integral for some purpose?

i think your nomenclature needs to be clarified...trapezoidal integration is like a pedagogical tool to introduce to numerical integration techniques and probably isn't so useful in the real world except as a really fast tool for approximate work.

i think typically "integration by parts" is reserved for a strategy unrelated to numerical integration...

edit: trapezoidal isn't so useful when you have a well-defined kernel function like you have here...if you only had data and no underlying function for the kernel, it may be one of few choices you have to approximate an "integral"

[Rahul Ramesh]

Hi

This is part of my research work. I am trying to minimize a temperature integral function. The minimum of this function in a temperatureb(T) intervel gives the value of E (activation energy). This integral function has multiple integral terms as the ones mentioned in this question. Only temperature values are known. E is the unknown. 8.314 is the universal gas constant.

Yes. You are correct. The only way is to approximate this integral. Hence, I am looking into trapezoidal integration. Or integration by parts to expand this integral into atleast 10 terms and then approximate.

The complete mathematical problem is as below:

Thanks @J. Alex Lee

[Star Strider]

I doubt that the integrateByParts function is appropriate here, however the series, taylor, or pade functions could work.

.

[John D'Errico]

I just remembered that I recall seeing you ask the same question before. PLEASE STOP ASKING THE SAME QUESTION MULTIPLE TIMES.

[Rahul Ramesh]

Hi John. Sorry. The last question I thought I deleted it before asking this one. I have deleted it now. Thanks for pointing it out.

Answer 1

[John D'Errico]

It is simply not true that there is no UNIQUE solution. There is no EXACT, analytical solution that can be written in a simple algebrac form using only addition, multiplication, etc. That is a very different thing.

It is also not true that it MUST be integrated using trapezoidal methods. For example, it is trivial to accomplish much more accurately using a tool such as integral. However, Ea MUST be given a value. Just saying that Ea is a constant is not sufficient. Any tool such as trapezoidal rule here is also meaningless if Ea is not a known constant. You cannot possibly know how to approximate the solution if you do not know the value of that constant. For example, if Ea is a relatively large number versus a very small number, then to have any degree of accuracy in the result, you will need to use very different approximations.

Doing some sort of integration by parts is silly, when you will decide to drop some terms arbitrarily. I'm sorry, but it is. I think that simply misunderstands tools like integration by parts.

But to do the computation is trivial, as long as Ea can be specified.

format long g
K = @(T,Ea) exp(Ea./(8.314*T));
intval = integral(@(T) K(T,3),270,290)

intval = 
          20.0258016842926

So fairly accurately, when Ea == 3, the integral is as shown. There is nothing non-unique about that solution. And there was no need to use trapezoidal rule. I could have done so. But the result would be of far lower accuracy unless we used a large number of points in the trapezoidal rule approoximation. Is the above correct? How close is it?

syms T
Ea = sym(3);
vpaintegral(exp(Ea./(sym('8.314')*T)),270,290,'reltol',1e-20)

ans = 

20.025801684292594502

So the result provided by integral seems accurate to as close as double precision arithmetic can report. That more than 5 significant digits is required is silly, since 8.314 is surely not known as EXACTLY that value.

What can you do if Ea is an unknown constant? The solution uses a function called ei, the exponential integral. It is a well defined function, much as is sin(x) or cos(x). Perhaps you know about sin and cos, but not about ei. That does not mean it does not exist.

syms Ea T
vpa(int(exp(Ea./(sym('8.314')*T)),270,290),5)

ans = 

intEa = matlabFunction(int(exp(Ea./(sym('8.314')*T)),270,290));

And now we can evaluate that function for any value of Ea.

intEa([1 3 10 100])

ans = 1×4

20.0085968640238 20.0258016842926 20.0861351906965 20.8782458224677

Thre is no need at all to use anything fancy. Just use MATLAB.

[J. Alex Lee]

and even without symbolic toolbox, you can still get some mileage out of base matlab as i start to show in a separate answer

[Rahul Ramesh]

@John D'Errico

Thanks. It worked. However, when integrating with lower limit as 0, say 0 to 293, it returns an expression with limits in it. Guess, the limit should be removed before minimizing for Ea?

Answer 2

[J. Alex Lee]

The application still is unclear to me...

If you just want numerical integration, look at the function "integral", as implemented below.

It may be worth noting that the "exact" indefinite integral can be expressed

where Ei is the exponential integral, which seems to have a related in-built numerical algorithm in matlab: expint.

If you have crazy values of Ea, the tops and bottoms in your double sum expression may get crazy too, and then you'll have to deal with finite precision problems in the ratio. If that expression is already the result of some kind of discretization of soemthing else, may be worth going back to that and looking at the problem mathematically in more detail.

Ta = 270;
Tb = 290;

R = 8.314;
Ea = 1e+3;

num_soln = integral(@(T)intKrnl(T,R,Ea),Ta,Tb,"AbsTol",1e-16,"RelTol",1e-16);
san_soln = indefInt(Tb,R,Ea)-indefInt(Ta,R,Ea);

fplot(@(T)intKrnl(T,R,Ea),[Ta,Tb])

function out = indefInt(T,R,Ea)
Ei = -(expint(-Ea/(R*T))+1i*pi);
out = T*exp(Ea/(R*T)) - Ea/R*Ei;
end

function out = intKrnl(T,R,Ea)
out = exp(Ea./(R*T));
end