Integration of sin(2kz)

  • Thread starter amjad-sh
  • Start date
200
6
Problem 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.
 

lurflurf

Homework Helper
2,416
120
The integral does not converge in the traditional sense, but for particular applications assign a useful value using limits.

for example

$$\int_{-\infty}^{z} \sin(2 k t) \, dt \\
=\lim_{s\rightarrow 0^+}\int_{-\infty}^{z} \sin(2 k t) e^{s t}\, dt$$
 
349
189
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.
 
200
6
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?
 
200
6
However, I know such beasts come up in physics and physicists do something not quite rigorous to define them.
HAHAHAHAHA :thumbup:
 

vanhees71

Science Advisor
Insights Author
Gold Member
11,603
4,228
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 :-).
 
349
189
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.
 

Want to reply to this thread?

"Integration of sin(2kz)" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top