# Integration of sin(2kz)

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.

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#### lurflurf

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

for example

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

#### RPinPA

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.

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

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

However, I know such beasts come up in physics and physicists do something not quite rigorous to define them.
HAHAHAHAHA

#### vanhees71

Gold Member
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 :-).

#### RPinPA

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.

#### lurflurf

Homework Helper
$\Longrightarrow \int_{-\infty}^{z} sin(2kt) \, dt=-\dfrac{1}{2k}cos(2kz)$

yes looks good

"Integration of sin(2kz)"

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