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Recently, I encountered the following integral. Given that it is relatively of a simple form, I suspect it is written in terms of known special functions. I know for $\phi = \pi/2$, it corresponds to Craig formula.

$$ I(a, \phi) = \int_{0}^{\phi} \exp \left[ -\frac{a^2}{(\sin{\theta})^2} \right] d\theta $$

Can anyone enlighten me regarding this integral?

Sungmin
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    Simple functions abound whose integrals have no expression in terms of special functions. I cannot say for sure that this is one of them, but having a simple integrand is not a strong indicator that the integral has a closed form. – Paul Sinclair Dec 14 '16 at 04:51
  • @PaulSinclair I understand your comment. However, at the same time, it is the reason why various special functions are defined. They appear in many different contexts while not allowing to be written as a closed form. I just suspected that this integral is still simple enough to be touched by someone long time ago. Anyway, I happened to find that $I$ can be written in terms of the error function and Owen's T function. In case you are interested, please check my answer below. – Sungmin Dec 15 '16 at 11:00

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It is embarrassing to write my own answer to my question, but I do so for future readers seeking for a solution of similar integrals.

The integral $I(a,\theta)$, defined in the question turns out to be written in terms of error functions and Owen's T function, (i.e., it implies that the integral has something to do with bivariate normal distribution). The answer is $$ I(a,\theta) = 2 \pi T\left(\sqrt{2} a \cot (\theta ),\tan (\theta )\right)+\frac{1}{2} \pi \text{erf}(a) (\text{erf}(a \cot (\theta ))-1), $$ where $T$ is Owen's T function and $a$ is assumed to be positive.

Edit:

Let me explain how it was derived. This identity is evaluated first by calculating $\frac{\partial I}{\partial a}$ which is $$ \int -2 a e^{-a^2 \csc ^2(\theta )} \csc ^2(\theta ) \, d\theta = \sqrt{\pi } e^{-a^2} \text{erf}(a \cot (\theta )), $$ by making the change of variable $\cot \theta = y$.

Next, $I$ is recovered by integrating the RHS of the above equation with regard to $a$ using the fact that this integral amounts to calculating the CDF of skew-normal distribution](https://en.wikipedia.org/wiki/Skew_normal_distribution).

Sungmin
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  • You might want to at least include the integral definition of the Owen function for completeness. – J. M. ain't a mathematician Dec 15 '16 at 11:27
  • @J.M. thanks, I tried to include relevant information that is needed to derive the equation. Instead of writing down the definition of Owen's T function, a link to the corresponding Wikipedia page is added. – Sungmin Dec 15 '16 at 12:24
  • Since your expression above depends on $a$ while the original integral does not, this cannot be correct as written. Perhaps $I$ is the value for a particular value of $a$? – Paul Sinclair Dec 15 '16 at 19:30
  • @PaulSinclair Oh, I made a mistake. Initially the integral was defined as a function of $r$ instead of $a$, but I used different symbol in the answer. I changed the notation in the question now. – Sungmin Dec 15 '16 at 21:29