I am trying to evaluate the following integral:
Here, $n$ is a natural number (so that the corresponding cosines complete a cycle along $[0,1]$) and all other parameters are real numbers (even positive, if that helps).
Of course, I have no particular reason to think that there is a closed form for this, I am just hoping that there is.
I am not sure how to solve this, but one things that immediately came to mind is the integral representation of the $0$th Bessel function:
$$ J_0 (z) = \frac{1}{\pi }\int_0^\pi {\cos (z\sin \theta )\,{\rm d}\theta } = \frac{1}{\pi }\int_0^\pi {\cos (z\cos \theta )\,{\rm d}\theta } $$
This does not exactly fit the integral I want to compute. I can maybe scale/shift things to get rid of $L,r$, but more importantly, I have an additional cosine which multiplies the integrand, so this is not exactly Bessel.
But I thought of maybe somehow identifying this as a Fourier transform (or at least the real part of a Fourier transform), due to the integration against cosine. Again, this does not exactly fit, since I don't have an additional integral over $(-\infty,\infty)$. Also, the variable $x$ here seems to simultaneously "serve" as the integration variable for the Bessel function AND the Fourier transform.
Nevertheless, maybe someone here has an idea on how to continue this approach in order to evaluate the integral?
Thanks in advance.
Too long for a comment. I will write some ideas and hints. Maybe you or someone else will know how to continue.
First of all, the integral can be split onto two integrals (why?). I will write down my ideas about the first part, the other one will be similar indeed.
So
$$\int_0^1 A \sin(L + r \cos(2\pi n x + \phi))\cos(2\pi nx + \phi)\ \text{d}x$$
Via a simple substitution we can write it in a more "simple" way
$$y = 2\pi n x + \phi \qquad \qquad \text{d}y = 2\pi n \text{d}x$$
so that
$$\frac{A}{2\pi n}\int_{\phi}^{2\pi n + \phi} \sin(L + r\cos(y))\cos(y)\ \text{d}y$$
Now we expand the sine function via trigonometric rules, and this gives us two more integrals which show a "simpler" form
$$I_1 = \frac{A \sin(L)}{2\pi n}\int_{\phi}^{2\pi n + \phi} \cos(r\cos(y))\cos(y)\ \text{d}y$$
$$I_2 = \frac{A \cos(L)}{2\pi n }\int_{\phi}^{2\pi n + \phi} \sin(r\cos(y))\cos(y)\ \text{d}y$$
Again, let's focus on the first one. Considering
$$t = r\cos(y) \qquad \qquad \text{d}t = -r\sin(y)\text{d}y = -r\sqrt{1-t^2}\ \text{d}y$$
We read
$$I_1 = -\frac{A \sin(L)}{2\pi n r}\int_{r\cos(\phi)}^{r(\cos(\phi) - \sin(\phi))} \frac{t \cos(t)}{\sqrt{1-t^2}}\ \text{d}t$$
Now, to say: if $r = 1$ and $\phi = 0$ the latter integral is zero. With the same assumptions, we can prove that $I_1$ is zero too, so until now things work.
This being said, the "problems" lie upon all tose variables. We can get rid of $A$ and $L$ with simple substitutions, and we can manage to simplify things thank to $n$ being natural (or integer), but for what concerns $r$ and $\phi$ we hit a wall. Not a dead end, for we can make assumptions, but they are too weak.
Numerical integration works but if we have well defined parameters.
An intersting example: if there are some $(r, \phi)$ values for which $r\cos(\phi) = 0$ and $r(\cos(\phi) - \sin(\phi)) = 1$ then we have something interesting:
$$\int_0^1 \frac{t\cos(t)}{\sqrt{1-t^2}}\text{d}t = \frac{\pi \pmb{H}_{-1}(1)}{2}$$
Where the special function "Struve Function H" appears.
For example, this happens for $r\to -1, x\to 2 \pi c_1+\frac{\pi }{2}\text{ if }c_1\in \mathbb{Z}$ so we can conclude for this special case
$$I_1 = \frac{A \sin(L) \pmb{H}_{-1}(1)}{4n} \qquad \qquad n\in\mathbb{Z}\backslash\{0\}$$
One can have fun in other wasys with other assumptions though.