# Is this Dirac delta function integral correct?

• I
I have to integrate this expression so I started to solve the delta part from the fact that when n=0 it results equals to 1.
And the graph is continuous in segments I thought as the sumation of integers
$$\int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx$$

From the fact that actually
$$δ(sin(x))= \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$
An in the integral

$$\int_{-(n+1/2)π}^{(n+1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ)$$

For n=0
$$\int_{-(1/2)π}^{(1/2)π} \sum_{zeros}^. \frac{1}{|Cos(nπ)|} δ(x-nπ) = 1$$

But taking a look on the graph of 1/|Cos(x)| and the fact that it is continuous in segments
for n=0 ,1 segment
n=1, 3 segments
n=2, 5 segments
and each one of these is just like the first whose integral is equal to one.
So the value it just a sumation that depends for the n value.
And the result is
$$\int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx= \sum_{N=0}^n (2N+1)$$

Has anyone solved it before? Is this a correct way of doing it?
Jorge M

Last edited:
• Delta2

Mark44
Mentor
$$\int x^2e^x \, dx$$
What does this have to do with the problem in post #1?

• JorgeM
What does this have to do with the problem in post #1?
Nothing really, I figured out how to use Latex, so I started to try to write equations and symbols.
However, I have finally written the equations as good as possible in order to make people easy to understand.
But I do not know if what I have made is well-solved.
Thanks.

jasonRF
Gold Member
$$\int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx= \sum_{N=0}^n (2N+1)$$

Has anyone solved it before? Is this a correct way of doing it?
Jorge M
Not quite, although most of your steps are correct. I get,
$$\begin{eqnarray*} \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx & = & \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{m=-\infty}^{\infty} \delta(x-m\pi) \, dx \\ & = & \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{m=-n}^n \delta(x-m\pi) \, dx \\ & = & \sum_{m=-n}^n 1\\ & = & 2n + 1. \end{eqnarray*}$$

jason

• JorgeM and Delta2
Not quite, although most of your steps are correct. I get,
$$\begin{eqnarray*} \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx & = & \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{m=-\infty}^{\infty} \delta(x-m\pi) \, dx \\ & = & \int_{-(n+1/2)π}^{(n+1/2)π} \sum_{m=-n}^n \delta(x-m\pi) \, dx \\ & = & \sum_{m=-n}^n 1\\ & = & 2n + 1. \end{eqnarray*}$$

jason
Actually, that is the correct result.