$\frac{1}{2 \pi a} \int_{-\pi a}^{\pi a} \exp \left(-\frac{1}{2} \left( \frac{2 a \sin^{-1} \left( \frac{x}{2 a} \right)}{\sigma} \right)^2 \right)dx$

Question

I have a normal distribution whose variable depends on an $\arcsin$ function as follows:

${\displaystyle\frac{1}{2 \pi a} \int_{-\pi a}^{\pi a} \exp \left(-\frac{1}{2} \left( \frac{2 a \sin^{-1} \left( \frac{x}{2 a} \right)}{\sigma} \right)^2 \right) \textrm{d}x}$

Where $a$ is a parameter.

I'm trying to use a substitution method for $u = 2 a \sin^{-1} \left( \frac{x}{2 a} \right)$ but I can't find a solution. I'm wondering if this is even solvable.

Basically the integral is performed over a circumference and the probability depends on the variable $x$ is the the length of the chord. What I'm trying to evaluate is the average probability density along the circle. Probably it would be possible to reformulate the integral in a much more convenient way, but I cannot understand how.

Thanks in advance to anyone who might give it a try!

Answer 1

Mathematica gives the following result: $$\frac{1}{2 \pi a}\int_{-\pi a}^{\pi a} e^{-\frac{1}{2} \left(\frac{2 a \sin ^{-1}\left(\frac{x}{2 a}\right)}{\sigma }\right)^2} \, dx=\frac{i \sigma e^{-\frac{\sigma ^2}{8 a^2}} \left(\text{erfi}\left(\frac{\sigma ^2-4 i a^2 \csc ^{-1}\left(\frac{2}{\pi }\right)}{2 \sqrt{2} a \sigma }\right)-\text{erfi}\left(\frac{\sigma ^2+4 i a^2 \csc ^{-1}\left(\frac{2}{\pi }\right)}{2 \sqrt{2} a \sigma }\right)\right)}{2 \sqrt{2 \pi } a}$$

Answer 2

For a step-by-step solution of the integral:

$$ I := \frac{1}{2\,\pi\,a}\int_{-\pi\,a}^{\pi\,a} \exp\left(-\frac{2\,a^2}{\sigma^2}\,\arcsin^2\left(\frac{x}{2\,a}\right)\right)\text{d}x $$

first of all, by symmetry:

$$ I = \frac{1}{\pi\,a}\int_0^{\pi\,a} \exp\left(-\frac{2\,a^2}{\sigma^2}\,\arcsin^2\left(\frac{x}{2\,a}\right)\right)\text{d}x $$

therefore, substituting $u = \frac{x}{2\,a}$:

$$ I = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} \exp\left(-\frac{2\,a^2}{\sigma^2}\,\arcsin^2\left(u\right)\right)\text{d}u $$

while, substituting $v = \arcsin u$:

$$ I = \frac{2}{\pi}\int_0^{v^*} \exp\left(-\frac{2\,a^2}{\sigma^2}\,v^2\right)\cos v\,\text{d}v $$

with $v^* \equiv \arcsin\left(\frac{\pi}{2}\right)$, i.e.

$$ I = \frac{1}{\pi}\int_0^{v^*} \exp\left(-\frac{2\,a^2}{\sigma^2}\,v^2\right)\left(e^{-\text{i}\,v} + e^{\text{i}\,v}\right)\text{d}v $$

from which:

$$ I = \frac{1}{\pi}\left[\int_0^{v^*} \exp\left(-\frac{2\,a^2}{\sigma^2}\,v^2 - \text{i}\,v\right)\text{d}v + \int_0^{v^*} \exp\left(-\frac{2\,a^2}{\sigma^2}\,v^2 + \text{i}\,v\right)\text{d}v\right] $$

i.e.

$$ \small I = \frac{1}{\pi}\,e^{-\frac{\sigma^2}{8\,a^2}}\left[\int_0^{v^*} \exp\left(-\left(\frac{\sqrt{2}\,a}{\sigma}\,v + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}\right)^2\right)\text{d}v + \int_0^{v^*} \exp\left(-\left(\frac{\sqrt{2}\,a}{\sigma}\,v - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}\right)^2\right)\text{d}v\right]. $$

This done, substituting respectively $w = \frac{\sqrt{2}\,a}{\sigma}\,v + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}$ and $z = \frac{\sqrt{2}\,a}{\sigma}\,v - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}$:

$$ I = \frac{\sigma}{\sqrt{2}\,\pi\,a}\,e^{-\frac{\sigma^2}{8\,a^2}}\left(\int_{\frac{\sigma}{2\sqrt{2}\,a}\,\text{i}}^{\frac{\sqrt{2}\,a}{\sigma}\,v^* + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}} e^{-w^2}\,\text{d}w + \int_{-\frac{\sigma}{2\sqrt{2}\,a}\,\text{i}}^{\frac{\sqrt{2}\,a}{\sigma}\,v^* - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}} e^{-z^2}\,\text{d}z\right) $$

i.e.

$$ I = \frac{\sigma}{2\sqrt{2\,\pi}\,a}\,e^{-\frac{\sigma^2}{8\,a^2}}\left(\int_{\frac{\sigma}{2\sqrt{2}\,a}\,\text{i}}^{\frac{\sqrt{2}\,a}{\sigma}\,v^* + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}} \frac{2}{\sqrt{\pi}}\,e^{-w^2}\,\text{d}w + \int_{-\frac{\sigma}{2\sqrt{2}\,a}\,\text{i}}^{\frac{\sqrt{2}\,a}{\sigma}\,v^* - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}} \frac{2}{\sqrt{\pi}}\,e^{-z^2}\,\text{d}z\right) $$

from which what desired:

$$ I = \frac{\sigma}{2\sqrt{2\,\pi}\,a}\,e^{-\frac{\sigma^2}{8\,a^2}}\left[\text{erf}\left(\frac{\sqrt{2}\,a}{\sigma}\,v^* + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}\right) + \text{erf}\left(\frac{\sqrt{2}\,a}{\sigma}\,v^* - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i}\right)\right]. $$

The result obtained is equivalent to that of Steven Clark obtained with Mathematica.

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If it's considered appropriate to divide the real part from the imaginary part, we have:

$$ \begin{aligned} & \frac{\sqrt{2}\,a}{\sigma}\,v^* + \frac{\sigma}{2\sqrt{2}\,a}\,\text{i} = \frac{\pi\,a}{\sqrt{2}\,\sigma} + \frac{\sigma^2-4\,a^2\,\log\left(\frac{\pi}{2} + \sqrt{\frac{\pi^2}{4}-1}\,\right)}{2\,\sqrt{2}\,a\,\sigma}\,\text{i} \equiv \alpha + \beta\,\text{i} \\ & \frac{\sqrt{2}\,a}{\sigma}\,v^* - \frac{\sigma}{2\sqrt{2}\,a}\,\text{i} = \frac{\pi\,a}{\sqrt{2}\,\sigma} - \frac{\sigma^2+4\,a^2\,\log\left(\frac{\pi}{2} + \sqrt{\frac{\pi^2}{4}-1}\,\right)}{2\,\sqrt{2}\,a\,\sigma}\,\text{i} \equiv \alpha + \gamma\,\text{i} \\ \\ \end{aligned} $$

from which it follows that:

$$ \begin{aligned} & \text{Re}(I) = \frac{\sigma}{2\sqrt{2\,\pi}\,a}\,e^{-\frac{\sigma^2}{8\,a^2}}\left[\frac{\text{erf}(\alpha + \beta\,\text{i}) + \text{erf}(\alpha - \beta\,\text{i})}{2} + \frac{\text{erf}(\alpha + \gamma\,\text{i}) + \text{erf}(\alpha - \gamma\,\text{i})}{2}\right] \\ & \text{Im}(I) = \frac{\sigma}{2\sqrt{2\,\pi}\,a}\,e^{-\frac{\sigma^2}{8\,a^2}}\left[\frac{\text{erf}(\alpha + \beta\,\text{i}) - \text{erf}(\alpha - \beta\,\text{i})}{2\,\text{i}} + \frac{\text{erf}(\alpha + \gamma\,\text{i}) - \text{erf}(\alpha - \gamma\,\text{i})}{2\,\text{i}}\right] \\ \end{aligned} $$

as proof here. In particular, the 3D-plot of $\text{Re}(I)$ when $\text{Im}(I)=0$ turns out to be the following:

where I limited myself to graphing in the intervals $-50 \le a \le 50$ and $-20 \le \sigma \le 20$.

I hope it's clear enough, good luck! ^_^