Why Does the Integral of a Step Function from -∞ to +∞ Equal Infinity?

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SUMMARY

The integral of the step function, represented as \(\int_{-\infty}^{\infty}\theta (x) dx\), equals infinity, as confirmed by the graphical representation of the step function. However, when evaluated mathematically using limits, \(\lim_{a\rightarrow\infty}\int_{-a}^{a}\theta(x) dx\) results in zero due to the properties of the Dirac delta function, which is zero except at \(x=0\). The confusion arises from the misinterpretation of the relationship between the step function \(\theta(x)\) and the Dirac delta function \(\delta(x)\), where \(\delta(x) = \frac{d \theta}{dx}\).

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  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with the Dirac delta function and its properties.
  • Knowledge of distribution theory and its applications in mathematical analysis.
  • Basic concepts of limits and continuity in calculus.
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  • Study the properties of the Dirac delta function in distribution theory.
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coki2000
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Hello,
I have a question about dirac-delta function. Integral of step function from -infinity to +infinity equals to infinity from the graph of step function.

[tex]\int_{-\infty}^{\infty}\theta (x) dx=\infty[/tex]

But calculating mathematically

[tex]\lim_{a\rightarrow\infty}\int_{-a}^{a}\theta(x) dx=\lim_{a\rightarrow\infty}(\delta \left(a \right)-\delta \left(-a\right))=0[/tex]

Because delta function is 0 except x=0. Please explain to me. Thanks
 
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It seems that you're using

[tex]\theta \left( x \right) = \frac{d \delta}{dx} \left( x \right),[/tex]

when actually (in a distributional sense)

[tex]\delta \left( x \right) = \frac{d \theta}{dx} \left( x \right).[/tex]

Also, for [itex]a > 0[/itex],

[tex]\int^a_{-a} \theta \left( x \right) dx = a.[/tex]
 
George Jones said:
It seems that you're using

[tex]\theta \left( x \right) = \frac{d \delta}{dx} \left( x \right),[/tex]

when actually (in a distributional sense)

[tex]\delta \left( x \right) = \frac{d \theta}{dx} \left( x \right).[/tex]

Also, for [itex]a > 0[/itex],

[tex]\int^a_{-a} \theta \left( x \right) dx = a.[/tex]

Oh, yes, thanks for your help.
 

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