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

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! ^_^
