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Difference between revisions of "Henstock-Kurzweil integral"
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Consider the function <math>f:[0,1]\rightarrow\mathh{R}</math>
 
Consider the function <math>f:[0,1]\rightarrow\mathh{R}</math>
 +
 
<math>f\left( \frac{1}{n}\right) =n\forall n\in\mathbb{N}</math>
 
<math>f\left( \frac{1}{n}\right) =n\forall n\in\mathbb{N}</math>
  
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Let <math>\varepsilon>0</math> be given.
 
Let <math>\varepsilon>0</math> be given.
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Consider [[gauge]] <math>\delta:[0,1]\rightarrow\mathbb{R}^+</math>
 
Consider [[gauge]] <math>\delta:[0,1]\rightarrow\mathbb{R}^+</math>
  
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But as <math>\varepsilon>0</math> is arbitrary, we have that <math>f</math> is Generalized Reimann integrable or, <math>\int_0^1 f(x)dx=0</math>
 
But as <math>\varepsilon>0</math> is arbitrary, we have that <math>f</math> is Generalized Reimann integrable or, <math>\int_0^1 f(x)dx=0</math>
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==References==
 
==References==

Revision as of 00:20, 19 February 2008

The Henstock-Kurzweil integral (also known as the Generalized Reimann integral) is one of the most widely applicable generalizations of the Reimann integral, but it also uses a strikingly simple and elegant idea. It was developed independantly by Ralph Henstock and Jaroslav Kurzweil

Definition

Let $f:[a,b]\rightarrow\mathbb{R}$

Let $L\in\mathbb{R}$

We say that $f$ is Generalised Reimann Integrable on $[a,b]$ if and only if, $\forall\epsilon>0$, there exists a gauge $\delta:[a,b]\rightarrow\mathbb{R}^+$ such that,

if $\mathcal{\dot{P}}$ is a $\delta$-fine tagged partition on $[a,b]$, then $|L-S(f,\mathcal{\dot{P}})|<\epsilon$

Here, $S(f,\mathcal{\dot{P}})$ is the Reimann sum of $f$ on $[a,b]$ with respect to $\mathcal{\dot{P}}$


The elegance of this integral lies in in the ability of a gauge to 'measure' a partition more accurately than its norm

Illustration

The utility of the Henstock -Kurzweil integral is demonstrated by this function, which is not Reimann integrable but is Geeralized Reimann Integrable.

Consider the function $f:[0,1]\rightarrow\mathh{R}$ (Error compiling LaTeX. Unknown error_msg)

$f\left( \frac{1}{n}\right) =n\forall n\in\mathbb{N}$

$f(x)=0$ everywhere else.

It can be shown that $f$ is not Reimann integrable on $[0,1]$

Let $\varepsilon>0$ be given.

Consider gauge $\delta:[0,1]\rightarrow\mathbb{R}^+$

$\delta\left( \frac{1}{n}\right) =\frac{\varepsilon}{k2^{k+1}}$

$\delta(x)=1$ everywhere else.

Let $\mathcal{\dot{P}}$ be a $\delta$-fine partition on $[0,1]$

The Reimann sum will have maximum value only when the tags are of the form $t_i=\frac{1}{n}$, $n\in \mathbb{N}$. Also, each tag can be shared by at most two divisions.

$S(f,\mathcal{\dot{P}})\leq\sum_{k=1}^{\infty}\frac{\varepsilon}{2^k}<\varepsilon$

But as $\varepsilon>0$ is arbitrary, we have that $f$ is Generalized Reimann integrable or, $\int_0^1 f(x)dx=0$

References

R.G. Bartle, D.R. Sherbert, Introduction to Real Analysis, John Wiley & sons

See Also

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