Integration of sin(2kz)

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In summary: This method is known as the "principal value" of an integral, and is a common way to assign a value to divergent integrals in physics.
  • #1
amjad-sh
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Homework Statement
In fact, I am working on an condensed matter physics paper, where I reached to an integral of this form: $$\int_{-\infty}^{z} sin(2kz) \, dz$$.
offcourse, I have to solve this integral.
Relevant Equations
There is no relevant equations.
##\int_{-\infty}^{z}sin(2kz)\,dz=\dfrac{-1}{2k} \Big [cos(2kz) \Big ]_{-\infty}^z=-\dfrac{cos(2kz)}{2k}+\dfrac{cos(-\infty)}{2k}##.
I ended up here, and I don't know how to proceed.
One recommended me to use contour integration, but I have no idea about it.
 
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  • #2
The integral does not converge in the traditional sense, but for particular applications assign a useful value using limits.

for example

$$\int_{-\infty}^{z} \sin(2 k t) \, dt \\
=\lim_{s\rightarrow 0^+}\int_{-\infty}^{z} \sin(2 k t) e^{s t}\, dt$$
 
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  • #3
It's not well-defined. An integral like that is defined as a limit, ##\lim_{a \rightarrow -\infty} \int_a^z \sin(2kz) dz## and due to the oscillatory nature of the sine, that limit will oscillate over a range of values and never converge.

You absolutely don't just "plug in infinity". ##\infty## is not a real number and it doesn't have a cosine.

However, I know such beasts come up in physics and physicists do something not quite rigorous to define them. I suspect the different limit which @lurflurf has suggested is probably the right one.
 
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  • #4
lurflurf said:
The integral does not converge in the traditional sense, but for particular applications assign a useful value using limits.

for example

$$\int_{-\infty}^{z} \sin(2 k t) \, dt \\
=\lim_{s\rightarrow 0^+}\int_{-\infty}^{z} \sin(2 k t) e^{s t}\, dt$$
##\int_{-\infty}^{z} sin(2kt) \, dt =\lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z} sin(2kt)e^{st} \, dt }##

Now , by using integration by parts:
let ##du=sin(2kt) \rightarrow u=\dfrac{-1}{2k}cos(2kt)##
## \quad v=e^{st} \rightarrow dv=se^{st}##

##\Longrightarrow \lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z} sin(2kt)e^{st} \, dt }=\lim_{ s \rightarrow 0^+}{\Big [-\dfrac{1}{2k}cos(2kt)e^{st} \Big ]_{-\infty}^{z}}+\lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z}\dfrac{s}{2k}cos(2kz)e^{st} \, dt} ##
##= \lim_{ s \rightarrow 0+}{\dfrac{1}{2k}cos(2kz)e^{sz}} +\lim_{ s \rightarrow 0^+}{\dfrac{1}{2k}cos(2k\infty)e^{-\infty}}+\lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z}\dfrac{s}{2k}cos(2kz)e^{st} \, dt} ##
##=\lim_{ s \rightarrow 0^+}{\dfrac{1}{2k}cos(2kz)e^{sz}}+\lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z}\dfrac{s}{2k}cos(2kz)e^{st} \, dt}##Now,##\lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z}\dfrac{s}{2k}cos(2kz)e^{st} \, dt}=\lim_{ s \rightarrow 0^+}{\dfrac{s}{2k} \Big [ \dfrac{1}{2k}sin(2kt)e^{st} \Big ]_{-\infty}^{z}}+\lim_{ s \rightarrow 0^+} {-\dfrac{s^2}{(2k)^2}\int_{-\infty}^{z} sin(2kz)e^{st} \, dt}##
##=\lim_{ s \rightarrow 0^+}{\dfrac{s}{(2k)^2}sin(2kz)e^{sz}}+\lim_{ s \rightarrow 0^+} {-\dfrac{s^2}{(2k)^2}\int_{-\infty}^{z} sin(2kz)e^{st} \, dt}=\lim_{ s \rightarrow 0^+} {-\dfrac{s^2}{(2k)^2}\int_{-\infty}^{z} sin(2kz)e^{st} \, dt}##

##\Longrightarrow \lim_{ s \rightarrow 0^+}{\int_{-\infty}^{z} sin(2kt)e^{st} \, dt \Big(1+\dfrac{s^2}{(2k)^2} \Big)}=-\lim_{ s \rightarrow 0^+}{\dfrac{1}{2k}cos(2kz)e^{sz}}##
##\Longrightarrow \int_{-\infty}^{z} sin(2kt) \, dt=-\dfrac{1}{2k}cos(2kz)##

Is my answer right?
 
  • #5
RPinPA said:
However, I know such beasts come up in physics and physicists do something not quite rigorous to define them.
HAHAHAHAHA :thumbup:
 
  • #6
Well, well, well. Physicist know at such a point that they deal with distributions rather than usual functions. They are sloppy up to this point, getting an ill-defined and thus diverging integral. Then they remember (hopefully) their pedantic quantum-mechanics professor who told them in their QM 1 lecture that plane waves are to read as "generalized functions" or "distributions", and then they regularize their nonsensical integral from the sloppy derivation using an idea as in #2 :-).
 
  • #7
I was trained in physics first, then mathematics. So I'm aware of the non-rigorous methods of physics and why they work out. For instance, treating differentials like dx as a finite thing for awhile before you decide it's now infinitesimal. There is something rigorous underneath, but the rigor is inconvenient.
 
  • #8
amjad-sh said:
##\Longrightarrow \int_{-\infty}^{z} sin(2kt) \, dt=-\dfrac{1}{2k}cos(2kz)##

Is my answer right?

yes looks good
 
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1. What is the formula for integrating sin(2kz)?

The formula for integrating sin(2kz) is ∫sin(2kz) dz = -1/2k cos(2kz) + C, where C is the constant of integration.

2. How do you solve an integral with sin(2kz)?

To solve an integral with sin(2kz), you can use the formula ∫sin(2kz) dz = -1/2k cos(2kz) + C and then plug in the limits of integration and simplify the expression.

3. What are the steps for integrating sin(2kz)?

The steps for integrating sin(2kz) are:
1. Use the formula ∫sin(2kz) dz = -1/2k cos(2kz) + C
2. Plug in the limits of integration
3. Simplify the expression
4. Add the constant of integration (C)
5. Write the final answer.

4. Is there a specific method for integrating sin(2kz)?

Yes, the specific method for integrating sin(2kz) is using the formula ∫sin(2kz) dz = -1/2k cos(2kz) + C and then simplifying the expression by plugging in the limits of integration and adding the constant of integration.

5. Can you use the power rule for integrating sin(2kz)?

No, you cannot use the power rule for integrating sin(2kz) because the power rule only applies to functions in the form of x^n, and sin(2kz) is not in this form. You must use the specific formula for integrating sin(2kz) to solve the integral.

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