Let $$ I(n,\eta) = \int_{0}^{\eta} \cos nt \, \cos t \, \sqrt{\cos^2 t - \cos^2 \eta}\; dt $$

where it is known that $0 < \eta \leq \frac \pi 2$.

Is it possible to evaluate $S$, the infinite sum of (even indexed) integrals of the form $I(2k,\eta)$, in closed form?

$$ S(\eta) = \sum_{k=0}^{\infty} \; \frac{2k}{(2k)^2-1} \int_{0}^{\eta} \cos 2kt \, \cos t \, \sqrt{\cos^2 t - \cos^2 \eta} \; dt $$

This integral contains terms similar to that in this question and arises in a similar context.


This is not a complete solution, but a start.

You just need to use the same idea as in the answer to the question you linked.

The sum has a closed form (check here) and can be moved under the integral:

$$ S_0 = \sum_{k=0}^{\infty} \; \frac{2k}{(2k)^2-1} \cos 2kt=\frac{1}{4} \cos t \log \frac{1+\cos t}{1-\cos t} -\frac{1}{2} $$

$$ S(\eta) = \frac{1}{4} \int_{0}^{\eta} \cos^2 t \log \frac{1+\cos t}{1-\cos t} \, \sqrt{\cos^2 t - \cos^2 \eta} \; dt -\frac{1}{2} \int_{0}^{\eta} \cos t \, \sqrt{\cos^2 t - \cos^2 \eta} \; dt $$

The second integral is:

$$\int_{0}^{\eta} \cos t \, \sqrt{\cos^2 t - \cos^2 \eta} \; dt=\int_{0}^{\sin \eta}\sqrt{\sin^2 \eta - \sin^2 t} \; d \sin t=\frac{\pi \sin^2 \eta}{4}$$

The first integral is tricky, I'm not sure how to go about it yet. When I get an idea, I'll update the answer.


A little addition: Let's transform the logarithm the following way:

$$\log \frac{1+\cos t}{1-\cos t}=\log \frac{(1+\cos t)^2}{1-\cos^2 t}=2\log \frac{1+\cos t}{\sin t}$$

And try to solve the related integral:

$$I_0=\int_{0}^{\eta} \cos^2 t \log \sin t \, \sqrt{\cos^2 t - \cos^2 \eta} \; dt=\int_{0}^{\sin \eta} \cos t \log \sin t \, \sqrt{\sin^2 \eta - \sin^2 t} \; d \sin t$$

Make a change of variable:

$$\frac{\sin t}{\sin \eta}=z,~~~~,\sin \eta = p$$

$$I_0=p^2 \int_{0}^{1} \sqrt{1-z^2} \sqrt{1-p^2z^2} (\log p +\log z) dz$$

The first part of this integral has a closed form in terms of complete elliptic integrals, for a solution see this answer:

$$p^2 \log p \int_{0}^{1} \sqrt{1-z^2} \sqrt{1-p^2z^2}dz=\frac{\log p}{3} \left((1+p^2)E(p)-(1-p^2)K(p) \right)$$

The second part I'm not sure how to solve, but with the help of Mathemarica I was able to get a 'closed' form:

$$\int_{0}^{1} \sqrt{1-z^2} \sqrt{1-p^2z^2} \log z~ dz=\frac{\pi}{8} \left(\text{Hypergeometric2F1Regularized}^{(0,0,1,0)}\left(-\frac{1}{2},\frac{1}{2},2,p^2\right)+\text{Hypergeometric2F1Regularized}^{(0,1,0,0)}\left(-\frac{1}{2}, \frac{1}{2},2,p^2\right)+\psi ^{(0)}\left(\frac{1}{2}\right) \, _2F_1\left(-\frac{1}{2},\frac{1}{2};2;p^2\right)\right)$$

It was done by evaluating:

$$\int_{0}^{1}\sqrt{1-z^2} \sqrt{1-p^2z^2} ~~z^a~ dz$$

Then taking $a$ derivative and setting $a=0$.

This 'closed' form seems to be correct, I checked using the 'easy' values $p=0$ and $p=1$, it gives the same values as direct integration.

  • $\begingroup$ Thanks - I have used ordinary Gauss Hypergeometric functions before, and am aware of numerous identities to simplify them, but these Regularized 2F1 functions seem quite a bit more general. $\endgroup$ – user_of_math Feb 28 '16 at 19:51

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