I What happens in this step function?

[garylau]

when the x>0,then the theta is equal to 1
So the theta =0 when x=0

in the second term of the integral,
it starts to integrate the function from "0 "to infinity(see the yellow loop inside)Since the" 0 "should not be included ,otherwise theta(0)=0 and (0*df/dt)=0

but why the 0 is still included in the integral to integrate the integrand!??

Sorry
i got confused
can anyone help me

thank you!??

 

[garylau]

An open interval does not include its endpoints, and is indicated with parentheses. For example, (0,1) means greater than 0 and less than 1. A closed intervalincludes its endpoints, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.

however
in the definition of the intregral ,the lower and upper limit should be closed interval rather than open interval

So did they do something wrong?

however,the step function does not include x=0(which is the end point) (or x<0) So it is open interval rather than closed interval
So it didn't fit the definition of an integral ?

 

[lurflurf]

The value of theta(0) is not important it is often taken to be 0, 1/2, or 1; but it does not matter.
The value of a function at one point does not effect an integral.

 

[Svein]

Since the" 0 "should not be included ,otherwise theta(0)=0 and (0*df/dt)=0

but why the 0 is still included in the integral to integrate the integrand!??

• \int_{a}^{\infty}f(x)dx is defined as \lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx
• In the same manner, if \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx exists, \int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx.
• Informally: Changing the value of the integrand at a single point does not affect the value of the integral.

 

[garylau]

The value of theta(0) is not important it is often taken to be 0, 1/2, or 1; but it does not matter.
The value of a function at one point does not effect an integral.

it may be affect so much if f(0) is very large

therefore the difference between f(0)*1 and f(0)*0 will be large?

 

[garylau]

• \int_{a}^{\infty}f(x)dx is defined as \lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx
• In the same manner, if \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx exists, \int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx.
• Informally: Changing the value of the integrand at a single point does not affect the value of the integral.

• \int_{a}^{\infty}f(x)dx is defined as \lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx
• In the same manner, if \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx exists, \int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx.
• Informally: Changing the value of the integrand at a single point does not affect the value of the integral.

But why the definition from the website that i quote is quiet different

The limit is belong to closed interval rather than open interval except the (+ / )infinite large's one

Where is your definition coming from(the one which imclude limit u-->0?
the improper integral is for infinite limit only but not for u=0 ?

 

[Svein]

First: See https://en.wikipedia.org/wiki/Riemann_integral.
Second: Take the integral \int_{\epsilon}^{1}\theta(x)dx. Since θ(x) = 1 for x>0, the value is 1 - ε. Therefore (since ε is arbitrary) the integral is 1.

Where is your definition coming from(the one which imclude limit u-->0?
the improper integral is for infinite limit only but not for u=0 ?

An integral is defined as some limit (see the link above). The only problem with Riemann integrals are that there may not be a limit. In this case there is a limit.

That aside, I definitely do not like the paragraph b) in your quote. It assumes that ∞ - ∞ = 0, which is not necessarily true. In fact, ∞ - ∞ has no definite value.