Closed-form solution for an integral

In summary: A_k = A^k, B_k = A^{k-1}B##So for the original integral we have## \int_0^z \frac{1}{(1+z-x)(1+x)^K} dx = \int_0^z \frac{1}{(1+z-x)(1+x)^{K-1}}dx + \frac{A^{K-1}B}{K-1} \int_0^z \frac{1}{1+x}dx##The first term is just the integral for ##K-1##. The second term can be computed by hand## \int_0^z \frac{1}{1
  • #1
EngWiPy
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Hello all,

Is there a closed form solution for the following integral

[tex]\int_0^z\frac{1}{1+z-x}\frac{1}{(1+x)^K}\,dx[/tex]

for a positive integer ##K\geq 1##, and ##z\geq 0##? I searched the table of integrals, but couldn't find something similar.

Thanks in advance for any hint
 
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  • #3
Thanks. I tried to compute the definite integral from 0 to 1, and from 0 to 0.5, but gave me "Standard computation time exceeded ...". I think this is because it is an online version of Mathematica with limited resources. I guess I can use the indefinite integral to find the definite one, but this is for an academic paper, and I need to reference the solution. I don't think saying that Mathematica gave this solution is acceptable, isn't it?
 
  • #4
  1. EngWiPy said:
    I don't think saying that Mathematica gave this solution is acceptable, isn't it?
    No but you should be able to prove that the integral is correct by differentiating it and matching it to what you started with. You will probably need to use some facts about Hypergeometric numbers to complete that, but I imagine the ones you need will be available in the above-linked wiki article on those numbers. There are formulas for differentiation of the function wrt its fourth argument, which will be needed, here.
 
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  • #5
andrewkirk said:
  1. No but you should be able to prove that the integral is correct by differentiating it and matching it to what you started with. You will probably need to use some facts about Hypergeometric numbers to complete that, but I imagine the ones you need will be available in the above-linked wiki article on those numbers. There are formulas for differentiation of the function wrt its fourth argument, which will be needed, here.

I am sorry, but this means that I somehow found the solution, and then proved it is the correct solution backwardly. Reviewers will ask how I obtained the solution in the first place. I think it will be hard to convince them without a forward solution.

I was thinking that the above integral can be written as

[tex]
A\int_0^z(1+z-x)^{-1}\,dx +\sum_{k=1}^KB_k\int_0^z(1+x)^{-k}\,dx
[/tex]

using partial fraction decomposition. The coefficients can be found by solving ##K+1## linear equations, and I think the resulted integrals have solutions.
 
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  • #6
EngWiPy said:
Reviewers will ask how I obtained the solution in the first place.
What journal are you writing for? I can't think of any journal that sets that standard. What reviewers are concerned about are that mathematical results are proved to be correct, not how you came up with the result. Differentiation and matching constitutes a proof that the integral is correct, and that is all that any reviewer I have ever known would be concerned with.
 
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  • #7
I have reviewed papers in my field, and if encountered with a similar situation, I probably would question the method, because it is not systematic and doesn't provide guidance on how to solve the integral. It seems like guessing, then testing the guess, which, although correct, isn't systematic.
 
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  • #8
EngWiPy said:
I have reviewed papers in my field, and if encountered with a similar situation, I probably would question the method, because it is not systematic and doesn't provide guidance on how to solve the integral.

That criticism is understandable if your field investigates "How to solve integrals". However, papers on many scientific topics simply state solutions to integrations because they are a side issue. They often leave verification of the integrations to the referees and readers.
 
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  • #9
Stephen Tashi said:
That criticism is understandable if your field investigates "How to solve integrals". However, papers on many scientific topics simply state solutions to integrations because they are a side issue. They often leave verification of the integrations to the referees and readers.

I don't know if this is different across fields, but in my field I haven't seen a paper that doesn't reference a solution to an integral, if it isn't derived in the paper itself, and I have learned through the years to reference any equation that I am not deriving. Usually researchers reference the Table of Integrals if the solution is there, which is a reliable reference to reviewers. You don't have to show how to reach to the solutions, but you need to cite a reliable reference.
 
  • #10
EngWiPy said:
I don't know if this is different across fields, but in my field I haven't seen a paper that doesn't reference a solution to an integral, if it isn't derived in the paper itself, and I have learned through the years to reference any equation that I am not deriving. Usually researchers reference the Table of Integrals if the solution is there, which is a reliable reference to reviewers. You don't have to show how to reach to the solutions, but you need to cite a reliable reference.
It's very common to just say that a result can be verified. Integrals, solutions to differential equations, etc. are full of solutions that are just known to be true due to a "lucky" coincidence. All you are responsible for is due diligence in showing that your statements are true and verifyable.
 
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  • #11
Interesting. I was talking the other day with a researcher with a math background, and he says the same thing (he uses Mathematica to find solution to integrations, without even verifying). It seems that this is a common and acceptable practice in mathematics and maybe in physics as well.
 
  • #12
EngWiPy said:
[tex]
A\int_0^z(1+z-x)^{-1}\,dx +\sum_{k=1}^KB_k\int_0^z(1+x)^{-k}\,dx
[/tex]
using partial fraction decomposition. The coefficients can be found by solving ##K+1##" linear equations,

Maybe we can build up the solutions to the equations rather than starting from scratch for each value of ##k##

Begin with ##k = 1## by finding ##A,B## that give the expansion
##f(x) = \frac{1}{(1 + z - x)(1+x)} = \frac{A}{1+z-x} + \frac{B}{1+x}##

##A = B = \frac{1}{z+2}##

Then
## \frac{1}{(1 + z - x)(1+x)^2}= f(x)\frac{1}{(1+x)} = \frac{A}{(1+z-x)(1+x)} + \frac{B}{(1+x)(1+x)} ##
## \ = A ( \frac {1}{(1+z-x)(1+x)}) + \frac{B}{(1+x)^2} ##
## \ = A (\frac{A}{(1+z-x)} + \frac{B}{(1+x)} ) + \frac{B}{(1+x)^2}##
##\ = A^2 \frac{1}{1+z-x} + AB\frac{1}{1+x} + B \frac{1}{(1+x)^2}##

By a similar process
##\frac{1}{(1+z-x)(1+x)^3} = A^3 \frac{1}{1+z-x} + A^2B \frac{1}{1+x} + AB\frac{1}{(1+x)^2} + B\frac{1}{(1+x)^3}##
 
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  • #13
Stephen Tashi said:
Maybe we can build up the solutions to the equations rather than starting from scratch for each value of ##k##

Begin with ##k = 1## by finding ##A,B## that give the expansion
##f(x) = \frac{1}{(1 + z - x)(1+x)} = \frac{A}{1+z-x} + \frac{B}{1+x}##

##A = B = \frac{1}{z+2}##

Then
## \frac{1}{(1 + z - x)(1+x)^2}= f(x)\frac{1}{(1+x)} = \frac{A}{(1+z-x)(1+x)} + \frac{B}{(1+x)(1+x)} ##
## \ = A ( \frac {1}{(1+z-x)(1+x)}) + \frac{B}{(1+x)^2} ##
## \ = A (\frac{A}{(1+z-x)} + \frac{B}{(1+x)} ) + \frac{B}{(1+x)^2}##
##\ = A^2 \frac{1}{1+z-x} + AB\frac{1}{1+x} + B \frac{1}{(1+x)^2}##

By a similar process
##\frac{1}{(1+z-x)(1+x)^3} = A^3 \frac{1}{1+z-x} + A^2B \frac{1}{1+x} + AB\frac{1}{(1+x)^2} + B\frac{1}{(1+x)^3}##

Very elegant. In general we can write

[tex]\frac{1}{(1+z-x)(1+x)^k} = \frac{A^k}{(1+z-x)}+B\sum_{m=1}^k\frac{A^{k-m}}{(1+x)^m}[/tex]

All we have to calculate in this case is just ##A## and ##B##. Thanks
 

Related to Closed-form solution for an integral

1. What is a closed-form solution for an integral?

A closed-form solution for an integral is a mathematical expression that gives the exact value of an integral without the need for numerical approximation. It is a solution that can be written in a finite number of standard mathematical operations, such as addition, subtraction, multiplication, division, and exponentiation.

2. How is a closed-form solution different from a numerical solution?

A closed-form solution gives the exact value of an integral, while a numerical solution approximates the value using numerical methods. A closed-form solution is generally preferred as it provides a more accurate and precise answer, whereas a numerical solution may have some error due to the approximation process.

3. What types of integrals can have closed-form solutions?

Simple integrals, such as polynomials, trigonometric functions, and exponential functions, can have closed-form solutions. However, more complex integrals, such as those involving special functions or non-elementary functions, may not have closed-form solutions and require numerical methods for evaluation.

4. How do you find a closed-form solution for an integral?

Finding a closed-form solution for an integral involves applying various integration techniques, such as substitution, integration by parts, and trigonometric identities. It also requires knowledge of basic algebra and calculus principles. Some integrals may be more challenging and require advanced techniques, such as contour integration or the use of special functions.

5. Why are closed-form solutions important in mathematics and science?

Closed-form solutions are important because they provide exact and precise solutions to mathematical problems, which are essential in understanding and analyzing various phenomena in science and engineering. They also allow for easier and faster calculations compared to numerical methods, making them useful in practical applications.

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