Is this Dirac delta function integral correct?

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  • Thread starter JorgeM
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  • #1
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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
 
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Answers and Replies

  • #3
30
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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.
 
  • #4
jasonRF
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$$ \int_{-(n+1/2)π}^{(n+1/2)π} δ(sin(x)) \, dx= \sum_{N=0}^n (2N+1)$$

Has anyone solved it before? Is this a correct way of doing it?
Thanks for your advise.
Jorge M
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
 
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  • #5
30
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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.
Thanks a lot for your reply
 

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