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$$\frac{1}{\sqrt{(x-a)^{2} + (y-b)^2 + (z-c)^2}} = \frac{1}{2\pi} \int_{0}^{2\pi}\frac{du}{(z-c) + i(x-a)\cos(u) + i(y-b)\sin(u)}$$

I am trying to understand the first part of E.T Whittaker's "On the Partial Differential Equations of Mathematical Physics". Embarrassingly, I can't get past this identity, which is stated without proof. I assume that this problem involves some basic contour integration but this is not my strong-suit. Some help would be much appreciated

Gary
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Ajay Srinivasan
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    Hint: first, convince yourself $\int_0^{2\pi}\frac{du}{A+B\cos(u-\phi)}$ doesn't depend on $\phi$. – J.G. Feb 12 '24 at 21:29
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    Potential hint: $A \cos(u) + B\sin(u) = \sqrt{A^2 + B^2} \sin(u + \arctan{\frac{A}{B}})$ – Dan Feb 13 '24 at 01:00
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    @Dan By $\arctan\frac{A}{B}$ I think you mean $\operatorname{atan2}(A,,B)$. – J.G. Feb 13 '24 at 07:47

1 Answers1

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Thanks J.G for the very useful hint! I think I have it:

first, let $A=z-c$ and $B = \sqrt{(x-a)^2 + (y-b)^2}$, an define an angle $\phi$ such that:

$$ \tan \phi = \frac{y-b}{x-a} $$

then the integrand becomes (through some algebra and use of the angle-difference trig identities):

$$\frac{1}{A + iB(\cos \phi \cos u + \sin \phi \sin u)} = \frac{1}{A + iB\cos(u-\phi)}$$

now by J.G's hint, a simple change of variables ($ u^{'} = u-\phi$) will show that our integral (without the $2\pi^{-1}$ factor) is equivalent to:

$$\int_{0}^{2\pi}\frac{du^{'}}{A + B^{'}\cos(u^{'})} = \frac{1}{B^{'}}\int_{0}^{2\pi}\frac{du^{'}}{A^{'} + \cos(u^{'})}$$

where $B^{'} = iB$ and $A^{'} = \frac{A}{B^{'}}$. From here I made use of the result I obtained from this post:

$$\int_{0}^{2\pi} \frac{du}{A + \cos u} = \frac{2\pi}{\sqrt{A^2 - 1}}$$

from which (by some more algebra) we can obtain the desired result:

$$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{du}{(z-c) + i(x-a)\cos u + i(y-b)\sin u} = \frac{1}{\sqrt{A^2 - (iB)^2}} = \frac{1}{\sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}}$$

Ajay Srinivasan
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