Can anyone help me solve this Integration of three terms?

In summary, the conversation revolves around trying to solve an integration problem with various constants involved. The individual has tried multiple methods but has not been successful. Another person suggests simplifying the problem and introduces new constants to make the integral easier to solve. However, the integral still has multiple terms, and the suggested method is to solve it numerically or using infinite series. The conversation also discusses the importance of certain conditions for the integral to converge and avoid issues such as complex numbers and divergence.
  • #1
mahmud_dbm
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I have been trying to solve an integration that i have

upload_2015-3-3_0-54-32.png

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.
 
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  • #2
mahmud_dbm said:
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.)
 
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  • #3
Ray Vickson said:
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
upload_2015-3-3_1-45-55.png


then i don't know where to go.
 
  • #4
mahmud_dbm said:
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
[tex] f(t) = t^2 e^{-t^2/a} \frac{(A+mt)^{1/m}}{1+\alpha (A + mt)^{1/m}}, [/tex]
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
[tex] \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 [/tex]
If we further let ##1 + kv = y## (assuming ##k > 0##) we can get
[tex] \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 [/tex]
I doubt that there is a closed-form formula for this, but one can tackle it numerically if you are given the parameter values.
 
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  • #5
Ray Vickson said:
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
[tex] f(t) = t^2 e^{-t^2/a} \frac{(A+mt)^{1/m}}{1+\alpha (A + mt)^{1/m}}, [/tex]
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
[tex] \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 [/tex]
If we further let ##1 + kv = y## (assuming ##k > 0##) we can get
[tex] \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 [/tex]
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.
 
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  • #6
mahmud_dbm said:
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.
 
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  • #7
Ray Vickson said:
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.
 
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  • #8
mahmud_dbm said:
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?
 
  • #9
How can y be a function of t if the integral is over t?

Chet
 
  • #10
Does the answer have to be closed form, or can it be expressed as an infinite series?

Chet
 
  • #11
mahmud_dbm said:
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?
 
  • #12
Chestermiller said:
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.
 
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  • #13
Ray Vickson said:
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.
 
  • #14
mahmud_dbm said:
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"?
 
  • #15
Ray Vickson said:
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.
 
  • #16
mahmud_dbm said:
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.
 
  • #17
Ray Vickson said:
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.
 
  • #18
mahmud_dbm said:
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.
 
  • #19
Ray Vickson said:
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
upload_2015-3-3_23-31-27.png

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.
 
  • #20
Shouldn't those x's be t's?

Chet
 
  • #21
Chestermiller said:
Shouldn't those x's be t's?

Chet

Right Right, x's be t's, there is no x in the equation.
 
  • #22
mahmud_dbm said:
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.
 
  • #23
Ray Vickson said:
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.
 

1. How do I approach solving an integration of three terms?

When solving an integration of three terms, it is important to first identify the type of integration (e.g. trigonometric, logarithmic, etc.) and then use appropriate integration techniques such as u-substitution, integration by parts, or partial fractions.

2. What is the best way to simplify an integration of three terms?

Simplifying an integration of three terms can often be achieved by factoring out common terms or using algebraic manipulation to group terms with similar variables together.

3. What should I do if I get stuck on an integration of three terms?

If you get stuck on an integration of three terms, try breaking it down into smaller integrals or using online resources such as integration tables or calculus software to help guide you through the process.

4. Is there a specific order in which I should integrate the three terms?

Generally, there is no specific order in which you must integrate the three terms. However, it is important to pay attention to any special cases or restrictions and to always follow the rules of integration (e.g. the power rule, constant multiple rule, etc.).

5. How can I check if my solution to an integration of three terms is correct?

You can check your solution by taking the derivative of your final answer. If the derivative matches the original three terms, then your solution is correct. You can also use online integration calculators or ask a math tutor for assistance.

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