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In my physics problem I've modelled the following integral.

\begin{equation} p(X,Y,x_0,y_0) = \int_{0}^{2\pi} \frac{1}{\sqrt{(X-R\cos(\Psi))^2+(Y-R\sin(\Psi))^2}} \delta(g) d\Psi \end{equation}

Where $g=Y[x_0-R\cos(\Psi)]-X[y_0-R\sin(\Psi)]-R[x_0\sin(\Psi)-y_0\cos(\Psi)]$, and $\delta$ is the dirac delta function. From the research I've done I think the answer is:

\begin{equation} p = \sum_i \frac{f(\Psi_i)}{|g'(\Psi_i)|} \end{equation}

Where $\Psi_i$ are the roots of $g(\Psi)=0$, and $g'= \frac{\partial g}{\partial\Psi}$

Matthew James
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    The integral is of the form

    $$\displaystyle{\int}\dfrac{d\Psi}{\sqrt{A-B\cos\Psi-C\sin\Psi} }.$$

    WolframAlpha's solution includes a hypergeometric function as well as lots of trig. functions. I'd suggest you approximate the integral by using $\sin\Psi \sim \Psi$ and $\cos\Psi\sim 1-\frac{\Psi^2}{2}$ if you want a nicer solution.

     – Joan S. Guillamet F. Nov 13 '23 at 15:02
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    Also if $g=0$, then that means that either there's some $\Psi=\text{ct.}$ that satisfies it or that it is not constant but $X=x_o$ and $Y=y_o$. If it's the first, then the integral is $0$, if it's the latter you get what I mentioned on top. – Joan S. Guillamet F. Nov 13 '23 at 15:31
  • Thanks for your suggestion. I've also played around with WolframAlpha for the form you're talking about. Sadly, the linearization you're looking at is only valid for small values of $\Psi$ when my $\Psi$ will actually vary quite a lot.

    What do you mean by $\Psi = ct$? There are many points that satisfy $g=0$, not just $X=x_0$ and $Y=y_o$.

    Do you think my line integral expression is correct or not?

     – Matthew James Nov 13 '23 at 16:16
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    You can find, isolating or numerically, a $\Psi$ that makes $g=0$ so that means $\Psi$ would be a constant (ct.). The other way would be to choose $X$ and $Y$ so that $g=0$ with $\Psi$ not being constant so $d\Phi\neq 0$ and thus the integral wouldn't be $0$. I don't know what would the limits of integration for the latter be, though. – Joan S. Guillamet F. Nov 13 '23 at 19:08
  • After reflecting on your comment I realized there was a problem in my original formulation. The $e^{kg^2}$ would simply turn the whole integral to $0$ for $k=\infty$. I realized what I actually wanted was the dirac delta. I changed the question. – Matthew James Nov 13 '23 at 21:08
  • Well, yes, you got it then: https://math.stackexchange.com/questions/276583/dirac-delta-function-of-a-function . Aren't there infinitely many roots though? – Joan S. Guillamet F. Nov 13 '23 at 23:11
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    $g'(\Psi_i)$ should have an absolute value by the way. – Joan S. Guillamet F. Nov 13 '23 at 23:13

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\begin{equation} d = \sum_i \frac{f(\Psi_i)}{|g'(\Psi_i)|} \end{equation}

Where $\Psi_i$ are the roots of $g(\Psi)=0$, and $g'= \frac{\partial g}{\partial\Psi}$

Matthew James
  • 115