General closed form of $\int_{0}^{\theta_0} \frac{1}{\sqrt{\cos \theta-\cos \theta_0}} d\theta$

Question

I once asked a question about how to integrate the reciprocal of the square root of cosine.

Is there a general closed form for the integral $$\int_{0}^{\theta_0} \frac{1}{\sqrt{\cos \theta-\cos \theta_0}} d\theta$$

in terms of the gamma function?

Thank you.

Answer 1

The general form is not in gamma function but something else.

Let $s = \sin\frac{\theta}{2}$ and $s_0 = \sin\frac{\theta_0}{2}$ and introduce $s = s_0 x$, we have

$$\begin{align} \int_0^{\theta_0} \frac{d\theta}{\sqrt{\cos\theta - \cos\theta_0}} & = \int_0^{s_0} \frac{1}{\sqrt{2(s_0^2 - s^2)}}\frac{2ds}{\sqrt{1-s^2}} = \sqrt{2}\int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-s_0^2 x^2)}}\\ &= \sqrt{2}K\left(\sin\frac{\theta_0}{2}\right) \end{align} $$

where $\displaystyle\;K(k) = \int_0^{\pi/2} \frac{dt}{\sqrt{1-k^2\sin(t)^2}}\;$ is the complete elliptic integral of the first kind.

Answer 2

Mathematica produces:

$$\frac{2F(\theta_0/2, \csc^2(\theta_0/2))}{\sqrt{1 - \cos(\theta_0)}}$$

Where $$F(\phi,m)=\int_0^\phi (1-m \sin^2\theta)^{-1/2}d\theta$$

Answer 3

The simplest result for $$I=\int_{0}^{\theta_0} \dfrac{d\theta}{\sqrt{\cos (\theta)-\cos (\theta_0)}} =\frac{2 F\left(\frac{\theta_0}{2}|\csc ^2\left(\frac{\theta_0}{2}\right)\right)}{\sqrt{1-\cos (\theta_0)}}$$ where $F$ is the elliptic integral of the first kind.

Now, how to express this result on the basis of the $\Gamma$ function ?