fn = c1/((lamda​^5)*((exp(​c2/(lamda*​T))) - 1)); I want to integrate this equation but getting Explicit Integral warning. here C1,C2,T are all unknown, integration has to be done in terms of lamda under the limits 0 to infinity. can anyone help me

fn = c1/((lamda^5)*((exp(c2/(lamda*T))) - 1)); I want to integrate this equation but getting Explicit Integral warning. Here C1,C2,T are all known, integration has to be done in terms of lamda under the limits 0 to infinity. can anyone help me

[Star Strider]

It will help if you post (or attach) your code, including the values of your constants. (If you choose to attach it, use the ‘paperclip’ button, and complete both the ‘Choose file’ and ‘Attach file’ steps as necessary.)

We have no idea what the problem is with your code if we don’t have it to work with.

[Dhruv Sangal]

 syms lamda;
 T = 300;
 sigma = 5.67*10^(-8);

 c1 = 3.742 * 10^8;
 c2 = 1.439 * 10^4;
 eml = 0.9;

 wave = xlsread('emivswave.xlsx',1);

 wavelength = wave(:,1);
 emissivity = wave(:,2);

 a(:,1) = 1:0.1:104;

 w = interp1(wavelength,a);
 e = interp1(emissivity,a);

 fn = c1/((lamda^5)*((exp(c2/(lamda*T))) - 1));
 fn1 = fn*eml;
 em = int(fn1,lamda,0,Inf);

 Emst= em/(sigma*T^4);

[Dhruv Sangal]

            374200000 
  ------------------------------ * eml
       5 /    /   1439   \     \ 
  lamda  | exp| -------- | - 1 | 

         \    \ 30 lamda /     /
Integration of this

[Star Strider]

Could you describe what you are doing and the result you expect? A PDF reference would be appropriate to help describe it. The reason is that we don‘t know if you have encoded your equations correctly. The best we can otherwise do with code is to proofread it for obvious errors.

[Dhruv Sangal]

Its Planck's distribution. Attached screenshot has the equation. Thanks

[Dhruv Sangal]

Complete equation

[Star Strider]

Not complete. We’re still missing ε(λ,T).

[Walter Roberson]

In Maple, this integral has some weird behavior if you leave T as symbolic. The behavior right at lamda = 0 is not clear: the limit from the left of the expression is infinity and the limit from the right is 0.

[Roger Stafford]

Since lambda = 0 is the lower limit of integration it doesn't matter what the limit from the left is. Only the limit from the right matters. Also, by my reasoning below, the integral is precisely proportional to the fourth power of T, and there is no need to go the symbolic route to obtain a general expression for the result.

[Walter Roberson]

The value at 0 matters, and the value at 0 has two divisions by 0, so the limit of the expression matters at 0 matters. But the limit is not defined because there is a non-removable discontinuity. Every non-empty compact region around lamda = 0 contains both an infinity and a small value, so there is no limit at 0.

[Roger Stafford]

I have to disagree with you, Walter. For the purposes of this integration as defined by Dhruv, the only thing that matters at λ = 0 is the limit from the right. This integral is in fact convergent at this lower integration limit because there is a finite limit to the integral as its lower limit of integration approaches zero from the right. That there is a zero-divided-by-zero situation right at the point where λ is equal to zero, or that there is a discontinuity there, do not matter for this integral to be considered convergent, only its behavior as the lower limit of integration approaches zero from the right. Yes, it is an “improper” integral, but nevertheless it is convergent. I refer you to the Wikipedia article on the subject:

   https://en.wikipedia.org/wiki/Improper_integral

Answer 1

[Star Strider]

With the constants and the additional information, this runs:

C1 = 3.742E+8;
C2 = 1.439E+4;
E = @(L,T) C1./(L.^5 .* exp(C2./(L.*T)) - 1);
ET = @(T) integral(@(L) E(L,T), 0, Inf);

T = linspace(0, 310, 50);

for k1 = 1:length(T)
    ETT(k1) = ET(T(k1));
end

figure(1)
plot(T, ETT)
grid
xlabel('T (°K)')
ylabel('Spectral Blackbody Emmisive Power')

The loop is unavoidable.

I will let you determine if this does what you want.

[Roger Stafford]

@Star: According to my reasoning, the requested integral is exactly proportional to the fourth power of T, and given that fact, it is unnecessary to do numerical integration more than once. Do you agree with my reasoning on this?

[Star Strider]

I certainly won’t argue with you! You’re a much more capable mathematician than I am.

I am just posting code to do the integration, if that is what ‘Dhruv Sangal’ wants demonstrated.

[Dhruv Sangal]

Thanks a lot for the help everyone. The code works fine now though I am still little bit confused with the reasoning. But for now its fine. I will discuss with my professor on reasoning part. Thanks a lot

[Star Strider]

Our collective pleasure.

Answer 2

[Roger Stafford]

Your integral can be transformed into one that doesn’t have any parameters as follows. Change your variable of integration from lambda to x in accordance with this equation:

   x = C2/(T*lambda)

The integral then is transformed to this:

   C1*T^4/C2^4 * integral of x^3/(exp(x)-1) from x = 0 to x = inf.

As you see, this integral doesn’t depend on any parameters. Once it is solved, you can evaluate your original integral by multiplying it by C1*T^4/C2^4.

[Roger Stafford]

Note: I have assumed here that C2/T is positive. If it is negative, the range of integration for x is from x = 0 to x = -inf, which would be divergent.