What is the value of this integral?

In summary: I don't see how the author could have converged to the 0.438392 answer for the principal value integral without actually doing the calculation.The author has written down a (numerically) exact value of 0.438392.Does anybody know how he got that? As far as I know, there is no representation of that integral in terms of special functions. Mathematica doesn't give me anything (numerical or otherwise), and Gradshteyn and Ryzhik's Table of Integrals doesn't seem to have it.It looks to me like there's quite a nasty singularity at y = 0.25 (decimal numbers :() so I don't even see how this conver
  • #1
rsq_a
107
1
There is a paper that compares numerical methods of calculating certain principal value integrals. One of them is,

[tex]\displaystyle P \int_0^1 \frac{e^{-y^2}}{y-0.25} dy[/tex]

The author has written down a (numerically) exact value of 0.438392.

Does anybody know how he got that? As far as I know, there is no representation of that integral in terms of special functions. Mathematica doesn't give me anything (numerical or otherwise), and Gradshteyn and Ryzhik's Table of Integrals doesn't seem to have it.
 
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  • #2
It looks to me like there's quite a nasty singularity at y = 0.25 (decimal numbers :() so I don't even see how this converges.
 
  • #3
CompuChip said:
It looks to me like there's quite a nasty singularity at y = 0.25 (decimal numbers :() so I don't even see how this converges.

It's a principal value integral. I wrote it in the intro. But I'll clarify the latex.
 
  • #4
Whoops, I completely missed that. If I push Mathematica to its precision limits, I get approximately the same answer but I don't get the decimals right. I don't see any way of evaluating this analytically either...
 
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  • #5
CompuChip said:
Whoops, I completely missed that. If I push Mathematica to it's precision limits, I get approximately the same answer but I don't get the decimals right. I don't see any way of evaluating this analytically either...

Yes, Maple can do it, too.

I doubt there is an analytical answer. What I assumed the author had was an expression in terms of special functions -- which could be 'exactly' calculated via tables -- or at least some numerical method not limited by the singularity.

Odd though. The other functions he looks at are,

[tex]\int_0^1 \frac{y^6}{y-0.25} dy[/tex]
[tex]\int_0^1 \frac{\cos{y}}{y-0.25} dy[/tex]
[tex]\int_0^1 \frac{exp^{-y}}{y-0.25} dy[/tex]

And each of these have exact answers (in terms of special functions in the second and third cases).
 

FAQ: What is the value of this integral?

What is an integral and why is it important?

An integral is a mathematical concept that represents the area under a curve. It is important because it allows us to find the total value of a continuous function and is used in various fields such as physics, engineering, and economics to solve real-world problems.

How do you calculate the value of an integral?

The value of an integral can be calculated by using various methods, such as the fundamental theorem of calculus, integration by substitution, or integration by parts. Each method involves breaking down the integral into smaller, more manageable parts and using known rules and formulas to solve for the final value.

What are the applications of integrals?

Integrals have a wide range of applications in various fields. They are used in physics to calculate work, in engineering to determine the area under a stress-strain curve, in economics to calculate total profit, and in statistics to find probabilities and expected values.

What are the different types of integrals?

There are two main types of integrals: definite and indefinite. A definite integral has specific bounds or limits, and the result is a single value. An indefinite integral has no bounds and represents a family of functions, with a constant added at the end.

Why is it important to understand the value of an integral?

Understanding the value of an integral is crucial for solving real-world problems and making accurate predictions. It also helps us understand the behavior of a function and how it changes over a given interval. Additionally, integrals are the foundation for more advanced mathematical concepts such as differential equations and multivariable calculus.

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