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Hello everyone

Today in my QM class, a discussion arose on the definition of the delta function using the Heaviside step function [itex]\Theta(x)[/itex] (= 0 for x < 0 and 1 for x > 0). Specifically,

[tex]\Theta(x) = \int_{-\infty}^{x}\delta(t) dt[/tex]

which of course gives

[tex]\frac{d\Theta(x)}{dx} = \delta(x)[/tex]

Some books (esp those on Communications and Signal analysis) define [itex]\Theta(0) = 1/2[/itex]. However, if I set [itex]x = 0[/itex] in the above integral, I get

[tex]\int_{-\infty}^{0}\delta(t) dt = \Theta(0) = \frac{1}{2}[/tex]

To me, this is an ambiguous result, because even though this would follow if [itex]\delta(x)[/itex] were a "normal function" by virtue of its evenness, the point [itex]x = 0[/itex] is a singularity point of the integrand and besides, the normal Riemann integral would implicitly assume an open interval formed by the limits of integration: [itex](-\infty,0)[/itex] and not a closed interval [itex](-\infty,0][/itex].

Now, I have the following question:

Is the expression [itex]\int_{-\infty}^{0}\delta(t) dt = \frac{1}{2}[/itex] correct?

If I construct a sequence of well behaved functions (rectangular, gaussian, or something else) [itex]\{\delta_{n}(x)\}[/itex] which converge to [itex]\delta(x)[/itex], if the elements of this sequence are even then indeed

[tex]\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}[/tex]

But can one infer

[tex]\int_{-\infty}^{0}\delta(t) dt = \lim_{n \rightarrow \infty}\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}[/tex]

from this always? I think this should depend on the definition of the sequence, and that such a result in general does not make sense as there is an ambiguity when one writes 0 as the upper limit: does it mean 0- or 0+? (For 0-, the integral is zero and for 0+, the integral is 1).

Can a rigorous justification and answer be given for this?

Thanks in advance.

Today in my QM class, a discussion arose on the definition of the delta function using the Heaviside step function [itex]\Theta(x)[/itex] (= 0 for x < 0 and 1 for x > 0). Specifically,

[tex]\Theta(x) = \int_{-\infty}^{x}\delta(t) dt[/tex]

which of course gives

[tex]\frac{d\Theta(x)}{dx} = \delta(x)[/tex]

Some books (esp those on Communications and Signal analysis) define [itex]\Theta(0) = 1/2[/itex]. However, if I set [itex]x = 0[/itex] in the above integral, I get

[tex]\int_{-\infty}^{0}\delta(t) dt = \Theta(0) = \frac{1}{2}[/tex]

To me, this is an ambiguous result, because even though this would follow if [itex]\delta(x)[/itex] were a "normal function" by virtue of its evenness, the point [itex]x = 0[/itex] is a singularity point of the integrand and besides, the normal Riemann integral would implicitly assume an open interval formed by the limits of integration: [itex](-\infty,0)[/itex] and not a closed interval [itex](-\infty,0][/itex].

Now, I have the following question:

Is the expression [itex]\int_{-\infty}^{0}\delta(t) dt = \frac{1}{2}[/itex] correct?

If I construct a sequence of well behaved functions (rectangular, gaussian, or something else) [itex]\{\delta_{n}(x)\}[/itex] which converge to [itex]\delta(x)[/itex], if the elements of this sequence are even then indeed

[tex]\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}[/tex]

But can one infer

[tex]\int_{-\infty}^{0}\delta(t) dt = \lim_{n \rightarrow \infty}\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}[/tex]

from this always? I think this should depend on the definition of the sequence, and that such a result in general does not make sense as there is an ambiguity when one writes 0 as the upper limit: does it mean 0- or 0+? (For 0-, the integral is zero and for 0+, the integral is 1).

Can a rigorous justification and answer be given for this?

Thanks in advance.

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