Can anyone help me solve this Integration of three terms?

[mahmud_dbm]

I have been trying to solve an integration that i have

I am not even sure if it's possible. Here, A, m, alpha, a these are constants. I have tried few methods, but couldn't find any way out. I would appreciate any help.

 

[Ray Vickson]

I have been trying to solve an integration that i have

View attachment 79831
I am not even sure if it's possible. Here, A, m, alpha, a these are constants. I have tried few methods, but couldn't find any way out. I would appreciate any help.

Please show us first some of the methods you HAVE tried, whether or not they failed. (Those are PF requirements.)

 

[mahmud_dbm]

Please show us first some of the methods you HAVE tried, whether or not they failed. (Those are PF requirements.)

Thank you so much for replying.
Like is said, i tried with multiple integration method but individual terms were also complicated, like

then i don't know where to go.

 

[Ray Vickson]

Thank you so much for replying.
Like is said, i tried with multiple integration method but individual terms were also complicated, likeView attachment 79834

then i don't know where to go.

If I were doing it I would simplify it first, by re-defining some of the variables and introducing some new constants. If we start with
f(t) = t^2 e^{-t^2/a} \frac{(A+mt)^{1/m}}{1+\alpha (A + mt)^{1/m}},
and assume that $a > 0$, we can let $t/\sqrt{a} = v$, $C = a^{3/2} A^{1/m}$, $k = m \sqrt{a}/A$, $b = \alpha A^{1/m}$ and, finally, $p = 1/m$. We then have the cleaner-looking result that
\int_0^{\infty} f(t) \, dt = C \int_0^{\infty} v^2 e^{-v^2} \frac{(1+k v)^p}{1 + b(1+k v)^p} \, dv
If we further let $1 + kv = y$ (assuming $k > 0$) we can get
\text{integral} = \frac{C e^{1/k}}{k^3} \int_1^{\infty} \frac{(y-1)^2 y^p}{1+b y^p} e^{-y/k} \, dy
I doubt that there is a closed-form formula for this, but one can tackle it numerically if you are given the parameter values.

 

[mahmud_dbm]

If I were doing it I would simplify it first, by re-defining some of the variables and introducing some new constants. If we start with
f(t) = t^2 e^{-t^2/a} \frac{(A+mt)^{1/m}}{1+\alpha (A + mt)^{1/m}},
and assume that $a > 0$, we can let $t/\sqrt{a} = v$, $C = a^{3/2} A^{1/m}$, $k = m \sqrt{a}/A$, $b = \alpha A^{1/m}$ and, finally, $p = 1/m$. We then have the cleaner-looking result that
\int_0^{\infty} f(t) \, dt = C \int_0^{\infty} v^2 e^{-v} \frac{(1+k v)^p}{1 + b(1+k v)^p} \, dv
If we further let $1 + kv = y$ (assuming $k > 0$) we can get
\text{integral} = \frac{C e^{1/k}}{k^3} \int_1^{\infty} \frac{(y-1)^2 y^p}{1+b y^p} e^{-y/k} \, dy
I doubt that there is a closed-form formula for this, but one can tackle it numerically if you are given the parameter values.

Dear Ray

Thank you so much.

I already see some lights.

Now as you said, if the parameter values were given, then it could be solved. Now, you were right in assuming a>0, because a = 0.1, m = 1, 2, 3, 4, 5 upto 20. So, yes k>0. But I want to understand how did you come up with all the right assumptions?? By the way, so now i have the integration, how do we do the integration ? Still i have multiple terms in the integration.

 

[Ray Vickson]

Dear Ray

Thank you so much.

I already see some lights.

Now as you said, if the parameter values were given, then it could be solved. Now, you were right in assuming a>0, because a = 0.1, m = 1, 2, 3, 4, 5 upto 20. So, yes k>0. But I want to understand how did you come up with all the right assumptions?? By the way, so now i have the integration, how do we do the integration ? Still i have multiple terms in the integration.

Looking at the original integral, we need a > 0 in order to have a convergent integral. Why is that? Well, the integrand is not defined at all if a = 0, and the integral diverges exponentially if a < 0. Next, we need A > 0 in order that (A + mt) be ≥ 0 for all t > 0. This is needed because if (A + mt) < 0 in a range of t-values, then (A + mt)^(1/m) will involve roots of negative numbers, so will involve complex numbers, etc. Finally, we need α ≥ 0 in order to not have the denominator go through 0 at t increases up from 0. If the denominator did go through 0, we would need to worry about the possible divergence of the integral, and whether to interpret as a principle-value integral, or whatever. All these issues are avoided if we just assume all your original parameters a, A,m,α are > 0.

As to how to integrate: NUMERICALLY! Or, perhaps you can expand things in convergent infinite series and then integrate term-by-term.

 

[mahmud_dbm]

Looking at the original integral, we need a > 0 in order to have a convergent integral. Why is that? Well, the integrand is not defined at all if a = 0, and the integral diverges exponentially if a < 0. Next, we need A > 0 in order that (A + mt) be ≥ 0 for all t > 0. This is needed because if (A + mt) < 0 in a range of t-values, then (A + mt)^(1/m) will involve roots of negative numbers, so will involve complex numbers, etc. Finally, we need α ≥ 0 in order to not have the denominator go through 0 at t increases up from 0. If the denominator did go through 0, we would need to worry about the possible divergence of the integral, and whether to interpret as a principle-value integral, or whatever. All these issues are avoided if we just assume all your original parameters a, A,m,α are > 0.

As to how to integrate: NUMERICALLY! Or, perhaps you can expand things in convergent infinite series and then integrate term-by-term.

Dear Sir

Thank you so much for you reply and thorough analysis.

Sir, in the 4th line, should it be exp(-v^2) ?
And sir, i tried with MATLAB, it's showing values. I don't know how they got the value.

and what did you mean by "expand things in convergent infinite series", how do i expand that term, especially that y^p/(1+b*y^p) format?

Thanks in advance.

 

[mahmud_dbm]

Dear Sir

Thank you so much for you reply and thorough analysis.

Sir, in the 4th line, should it be exp(-v^2) ?
And sir, i tried with MATLAB, it's showing values. I don't know how they got the value.

and what did you mean by "expand things in convergent infinite series", how do i expand that term, especially that y^p/(1+b*y^p) format?

Thanks in advance.

Ray

I mean, as i said earlier m>1, so that means p is fractional. How do i expand that fractional power thing?

 

[Chestermiller]

How can y be a function of t if the integral is over t?

Chet

 

[Chestermiller]

Does the answer have to be closed form, or can it be expressed as an infinite series?

Chet

 

[Ray Vickson]

Dear Sir

Thank you so much for you reply and thorough analysis.

Sir, in the 4th line, should it be exp(-v^2) ?
And sir, i tried with MATLAB, it's showing values. I don't know how they got the value.

and what did you mean by "expand things in convergent infinite series", how do i expand that term, especially that y^p/(1+b*y^p) format?

Thanks in advance.

Yes, the 4th line should have exp(-v^2), not exp(-v). I have edited out this typo.

Exactly what input did you use with MATLAB, and what type of answer did MATLAB deliver?

 

[mahmud_dbm]

Does the answer have to be closed form, or can it be expressed as an infinite series?

Chet

Dear Chet

The answer should be just a value, for the given parameters, so it should be in closed form, but now i doubt if it's possible. So, finite or infinite solution, please do enlighten me.

 

[mahmud_dbm]

Yes, the 4th line should have exp(-v^2), not exp(-v). I have edited out this typo.

Exactly what input did you use with MATLAB, and what type of answer did MATLAB deliver?

Ray

I simply wrote the function, integrated using built-in integration function.

bt=0.02;
s=1;
g=3;
a=0.1;
k=g*s/bt;
p=1/g;
b=a*bt^p;
C=2*s*bt^p;

fun=@(y) (C/k^3)*((y-1).^2.*exp(-(y-1)./k).^2.*y.^p)./(1+b.*y.^p);
m=integral(fun,0,Inf);

the m is now 0.6839.

 

[Ray Vickson]

Ray

I simply wrote the function, integrated using built-in integration function.

bt=0.02;
s=1;
g=3;
a=0.1;
k=g*s/bt;
p=1/g;
b=a*bt^p;
C=2*s*bt^p;

fun=@(y) (C/k^3)*((y-1).^2.*exp(-(y-1)./k).^2.*y.^p)./(1+b.*y.^p);
m=integral(fun,0,Inf);

the m is now 0.6839.

So, did MATLAB give you a numerical answer, or did it give you a "formula"?

 

[mahmud_dbm]

So, did MATLAB give you a numerical answer, or did it give you a "formula"?

MATLAB gave me a numerical answer i need to know the how and the formula.

 

[Ray Vickson]

MATLAB gave me a numerical answer i need to know the how and the formula.

I suspect that there is NO possible formula. I did say---more than once---that I guessed you would need to do numerical integration when looking at a numerical example (that is, with numerical values of the inputs). MATLAB probably tried some "analytical" methods, found it could not be done, and then applied one of its built-in numerical integration routines.

 

[mahmud_dbm]

I suspect that there is NO possible formula. I did say---more than once---that I guessed you would need to do numerical integration when looking at a numerical example (that is, with numerical values of the inputs). MATLAB probably tried some "analytical" methods, found it could not be done, and then applied one of its built-in numerical integration routines.

Ray

Surely your reply helped me a lot, now i understand the problem at least. Thank you so much for your time.
Using MATLAB, i would probably get the numerical value but i won't be able to explain how i got that.
That's why i needed the explanation. I will share if i can find anything.

Thank you all.

 

[Ray Vickson]

Ray

I simply wrote the function, integrated using built-in integration function.

bt=0.02;
s=1;
g=3;
a=0.1;
k=g*s/bt;
p=1/g;
b=a*bt^p;
C=2*s*bt^p;

fun=@(y) (C/k^3)*((y-1).^2.*exp(-(y-1)./k).^2.*y.^p)./(1+b.*y.^p);
m=integral(fun,0,Inf);

the m is now 0.6839.

Please give the original parameters in your first post. Here you have bt, s and g, none of which were present in the original post. Just tell us the values of m, a, A, and α, as they appeared in post #1. I want to check your numerical value, but to do that I need to know exactly what you computed.

Anyway, it looks like you have entered (in part) exp(-(y-1)/k)^2 , which is = exp(-2(y-1)/k), instead of the correct form exp(-[ (y-1)/k ls]^2) = exp(-(y-1)^2/k^2), which is very different.

 

[mahmud_dbm]

Please give the original parameters in your first post. Here you have bt, s and g, none of which were present in the original post. Just tell us the values of m, a, A, and α, as they appeared in post #1. I want to check your numerical value, but to do that I need to know exactly what you computed.

Anyway, it looks like you have entered (in part) exp(-(y-1)/k)^2 , which is = exp(-2(y-1)/k), instead of the correct form exp(-[ (y-1)/k ls]^2) = exp(-(y-1)^2/k^2), which is very different.

Dear Ray

Another constant can be multiplied sometimes which is (2/a^4), this is optional.

There is a little change from the 1st post, it's a^2 instead of a. It's a constant, so doesn't make any difference.
Other values are like alpha = 0.1, a=1, m=3 (usually >1), A = 0.02.

You are right about the exponential term, my mistake.

 

[Chestermiller]

Shouldn't those x's be t's?

Chet

 

[mahmud_dbm]

Shouldn't those x's be t's?

Chet

Right Right, x's be t's, there is no x in the equation.

 

[Ray Vickson]

Dear RayView attachment 79893
Another constant can be multiplied sometimes which is (2/a^4), this is optional.

There is a little change from the 1st post, it's a^2 instead of a. It's a constant, so doesn't make any difference.
Other values are like alpha = 0.1, a=1, m=3 (usually >1), A = 0.02.

You are right about the exponential term, my mistake.

You say "...so it doesn't make any difference". It DOES make a difference when you are telling other people what you did; they need to know exactly what your parameters are. Of course, when a = 1 it really does not make a difference, since a^2 = a in that case; but when a ≠ 1 this no longer holds. In general, you should not change the notation in your posts without telling anyone.

 

[mahmud_dbm]

You say "...so it doesn't make any difference". It DOES make a difference when you are telling other people what you did; they need to know exactly what your parameters are. Of course, when a = 1 it really does not make a difference, since a^2 = a in that case; but when a ≠ 1 this no longer holds. In general, you should not change the notation in your posts without telling anyone.

Okay so, now you can omit that optional constant part, which i haven't included, and it's a^2, not a, it was a mistake in the first post. But, like i said, a=1, so it wouldn't make any difference, in my case.