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I am trying to solve the following integral, which may be thought of as an integral of an exponentially-decaying distance from a point $r$ on the $x$-axis to a circumference of a circle of radius $s$.

$$\int_0^{2\pi} \exp\left(-\delta\sqrt{r^2-2\cos(\phi)rs+s^2}\right)\,d\phi$$

Does this even have a closed form?

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No. But is $r \gg s$ then you can Taylor expand the square root. To first order in the small quantity $s/r$, the integral may be expressed as a Bessel function. –  Ron Gordon Dec 20 '12 at 16:22
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I expand upon my comment. I do not believe this has a closed form. But consider the following integral:

$$\int_{0}^{2 \pi} {d \phi} \exp{(- \sqrt{a + b \cos{\phi}})}$$

When $a \gg b$, you can Taylor expand the square root term to get

$$\int_{0}^{2 \pi} {d \phi} \exp{\left [- \sqrt{a} \left (1 + \frac{b}{a} \cos{\phi} \right ) \right ]}$$

which evaluates to

$$ 2 \pi \exp{(- \sqrt{a})} I_0 \left (\frac{b}{\sqrt{a}} \right ) $$

with relative error $O \left ( \frac{b}{a} \right )$.

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