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pedroobv
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[SOLVED] Dirac delta function and Heaviside step function
In Levine's Quantum Chemistry textbook the Heaviside step function is defined as:
Dirac delta function is:
Now, the integral:
Is evaluated using integration by parts considering
[tex]u=f(x), du=f'(x)[/tex]
[tex]dv=\delta (x-a)dx, v=H(x-a)[/tex]
We have then:
Since [tex]H(x-a)[/tex] vanishes for [tex]x<a[/tex], the integral becomes:
This is the point where my question arrives. [tex]H(x-a)[/tex] is considered to have a value of unity for all the integral and that's why it is pulled out of the integral as a constant, however the lower bound of the integral is [tex]a[/tex] and in this point [tex]H(x-a)=1/2[/tex]. Could you please tell me if the following explanation is correct?
I think that because in all the integral, except in [tex]a[/tex], [tex]H(x-a)=1[/tex] and since the upper bound is infinity the value of the integral at the point [tex]a[/tex] can be ignored.
If I'm wrong, any suggestion for correcting my explanation will be appreciated.
In Levine's Quantum Chemistry textbook the Heaviside step function is defined as:
[tex]H(x-a)=1,x>a[/tex]
[tex]H(x-a)=0,x<a[/tex]
[tex]H(x-a)=\frac{1}{2},x=a[/tex]
[tex]H(x-a)=0,x<a[/tex]
[tex]H(x-a)=\frac{1}{2},x=a[/tex]
Dirac delta function is:
[tex]\delta (x-a)=dH(x-a) / dx[/tex]
Now, the integral:
[tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx[/tex]
Is evaluated using integration by parts considering
[tex]u=f(x), du=f'(x)[/tex]
[tex]dv=\delta (x-a)dx, v=H(x-a)[/tex]
We have then:
[tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(x)H(x-a)|^{\infty}_{-\infty}-\int ^{\infty}_{-\infty}H(x-a)f'(x)dx[/tex]
[tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(\infty)-\int ^{\infty}_{-\infty}H(x-a)f'(x)dx[/tex]
[tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(\infty)-\int ^{\infty}_{-\infty}H(x-a)f'(x)dx[/tex]
Since [tex]H(x-a)[/tex] vanishes for [tex]x<a[/tex], the integral becomes:
[tex]\int ^{\infty}_{-\infty}f(x)\delta (x-a)dx=f(\infty)-\int ^{\infty}_{a}H(x-a)f'(x)dx=f(\infty)-\int ^{\infty}_{a}f'(x)dx[/tex]
This is the point where my question arrives. [tex]H(x-a)[/tex] is considered to have a value of unity for all the integral and that's why it is pulled out of the integral as a constant, however the lower bound of the integral is [tex]a[/tex] and in this point [tex]H(x-a)=1/2[/tex]. Could you please tell me if the following explanation is correct?
I think that because in all the integral, except in [tex]a[/tex], [tex]H(x-a)=1[/tex] and since the upper bound is infinity the value of the integral at the point [tex]a[/tex] can be ignored.
If I'm wrong, any suggestion for correcting my explanation will be appreciated.