Convergence of improper integrals

In summary: This is why for these cases we use the Cauchy Principal Value to define the integral. In the case of the example you gave, the integral is not bounded and therefore diverges, but the Cauchy Principal Value is well defined and is equal to 0.
  • #1
lwebb
2
0
What is the difference between

[itex]\int_{-\infty}^{\infty} \frac{x}{1+x^2}dx[/itex]

and

[itex]\lim_{R\rightarrow \infty}\int_{-R}^{R} \frac{x}{1+x^2}dx[/itex] ?

And why does the first expression diverge, whilst the second converges and is equal to zero?
 
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  • #2
lwebb said:
What is the difference between

[itex]\int_{-\infty}^{\infty} \frac{x}{1+x^2}dx[/itex]

and

[itex]\lim_{R\rightarrow \infty}\int_{-R}^{R} \frac{x}{1+x^2}dx[/itex] ?

And why does the first expression diverge, whilst the second converges and is equal to zero?

Welcome to Physics Forums!

The first is an improper integral and is defined as:

[itex]\int_{-\infty}^{\infty} \frac{x}{1+x^2}dx = \int_{-\infty}^{0} \frac{x}{1+x^2}dx + \int_{0}^{\infty} \frac{x}{1+x^2}dx [/itex]

In other words, for the improper integral to be finite both positive and negative parts must be finite.

The second is called the "Cauchy Principal Value". You can google for this. Odd functions like yours or like ##sin(x)## will have CPV = 0, even though the improper integral is not finite.
 
  • #3
I apologise in advance if this is a stupid question, but could one say that the two expressions are different because the infinities in the first expression may or may not be of the same magnitude, and so the area under curve is therefore undefined. Whereas the infinities in the second expression must be of the same magnitude, and so the areas either side of the y-axis must therefore be equal, leading them to cancel out as in an odd function?
 
  • #4
lwebb said:
I apologise in advance if this is a stupid question, but could one say that the two expressions are different because the infinities in the first expression may or may not be of the same magnitude, and so the area under curve is therefore undefined. Whereas the infinities in the second expression must be of the same magnitude, and so the areas either side of the y-axis must therefore be equal, leading them to cancel out as in an odd function?

They don't need to be infinities. Take the example of ##sin(x)## or ##cos(x)##: the improper integral is not well defined because the integrals are oscillating functions and never settle down to a finite value, although the integrals themselves are bounded.
 

FAQ: Convergence of improper integrals

What is the definition of convergence for improper integrals?

The improper integral of a function f(x) over an interval [a, b] is said to converge if the limit of the integral from a to b as the upper limit approaches infinity or the lower limit approaches negative infinity, exists and is finite.

What is the difference between absolute and conditional convergence of improper integrals?

Absolute convergence of an improper integral means that the integral converges regardless of the order in which the integrals are taken. Conditional convergence means that the integral only converges if the integrals are taken in a specific order.

What is the comparison test for determining the convergence of improper integrals?

The comparison test states that if the integral of a function f(x) over an interval [a, b] is less than or equal to the integral of a function g(x) over the same interval, and the integral of g(x) converges, then the integral of f(x) also converges.

How does the limit comparison test work for determining the convergence of improper integrals?

The limit comparison test states that if the limit of the ratio of f(x) and g(x) as x approaches infinity is a finite number, then the improper integral of f(x) and g(x) either both converge or both diverge.

What is the Cauchy criterion for determining the convergence of improper integrals?

The Cauchy criterion states that if for any positive number ε, there exists a number M such that for any t > b > a, the integral of f(x) over the interval [a, b] is less than ε, then the improper integral of f(x) converges.

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