- #1
JorgeM
- 30
- 6
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?
Thanks for your advise.
Jorge M
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?
Thanks for your advise.
Jorge M
Last edited: