I encounter the following integral when trying to find the inverse Fourier transform of the characteristic function of a certain sum of random variables. Here, $0\le\lambda\le1$, $p\ge0$, $q\ge0$ are real, and $n$ is an integer. I want to compute the following integral:

$$\int_0^{2 \pi} e^{p \cos (\lambda \tau) + q \cos ((1 - \lambda) \tau)} \cos (n \tau) \frac{d \tau}{2 \pi}$$

This is a generalization of a Bessel integral, in that for $q = 0$ and $\lambda=1$, I know that:

$$\int_0^{2 \pi} e^{p \cos (\tau)}\cos (n \tau) \frac{d \tau}{2 \pi} =I_n(p)$$

where $I_n(p)$ is the modified Bessel function of the first kind.


This is not a complete answer, but it's a bit long to fit in a comment, so I'm posting it here in case it's useful to someone.

Let's denote:

$$ I(p, q, \lambda; n) := \int_0^{2 \pi} e^{p \cos (\lambda \tau) + q \cos ((1 - \lambda) \tau)} \cos (n \tau) \frac{\mathrm d \tau}{2 \pi} $$

There's a symmetry in this integral of the form:

$$ I(p, q, \lambda; n) = I(q, p, 1 - \lambda; n) $$

Also note that the integrand is even, i.e. $f(-\tau) = f(\tau)$.

Now, as mentioned in the OP, we have:

$$ I(p, 0, 1; n) = I(0, p, 0; n) = I_n(p) $$

Actually, a slight generalization for $q \neq 0$ gives us:

$$ I(p, q, 1; n) = I(q, p, 0; n) = e^q I_n (p) $$

Due to the symmetry property above, taking $\lambda = 1/2$ we actually obtain a rather simple result:

\begin{align*} I(p, q, \frac{1}{2}; n) =& I(q, p, \frac{1}{2}; n) = \frac{1}{2 \pi} \int_0^{2\pi} \mathrm d\tau\, e^{(p + q)\cos(\tau / 2)} \cos (n\, \tau)\\ =& \frac{1}{\pi} \int_0^{\pi} \mathrm dy\, e^{(p + q)\cos y} \cos (2\, n\, y) = I_{2n}(p + q) \end{align*}

A special case (probably of limited interest though) is $q = 0$, and $\lambda = 1 / m$, where $m \in \mathbb N$. Of course, if $m = 1$ or $m = 2$, we obtain the previous results, namely $I_n(p)$ and $I_{2n}(p)$, so we need to take $m \geq 3$ to obtain anything new. The integral can then be converted into the following form:

$$ I(p, 0, 1/m; n) = \frac{m}{2 \pi} \int_0^{2 \pi / m} \mathrm dx\, e^{p \cos x} \cos (m\, n\, x) $$

Unfortunately, a general analytic solution doesn't seem likely. One value which seems to be solvable though is $m = 4$ (originally found by experimenting in Mathematica); in this case, we write the integral as:

$$ I(p, 0, 1/4; n) = \frac{2}{\pi} \int_0^{\pi/2} \mathrm dx\, e^{p \cos x} \cos (4\, n\, x) $$

We can then use the $n$ angle expansion, so our integral turns into:

$$ I(p, 0, 1/4; n) = \frac{2}{\pi} \sum\limits_{k\;\mathrm{even}} (-1)^\frac{k}{2} \begin{pmatrix}4 n\\ k\end{pmatrix} \int_0^{\pi/2} \mathrm dx\, e^{p \cos x}\, \cos^{4n - k} x\, \sin^k x $$

Let's focus on the integral in the above; using the fact that $4 n - k := 2 \ell$ is even, we can rewrite $\cos^{2\ell} x = (1 - \sin^2 x)^\ell$, expand this using the binomial theorem, and additionally expand $e^u = \cosh u + \sinh u$, so that our final result is a linear combination of integrals of the following form:

$$ \mathcal{I}(\alpha, p) = \int_0^{\pi/2} \mathrm dx\, \sinh(p \cos x)\, \sin^\alpha x\\ \mathcal{J}(\alpha, p) = \int_0^\pi \mathrm dx\, \cosh(p \cos x)\, \sin^\alpha x $$

for some integer values of $\alpha$. Note that the second integral has shifted limits, $[0,\pi]$, instead of $[0,\pi/2]$; this is because the integrand is actually symmetric around $\pi/2$, so we can write $\int_0^{\pi/2}(\cdots) = \frac{1}{2} \int_0^\pi (\cdots)$ (the easiest way to see this is to shift the origin to $\pi/2$, i.e. use the substitution $x \rightarrow x - \pi / 2$; then it's simple to demonstrate that $f(-x) = f(x)$, where $f$ denotes the integrand).

These kinds of integrals are known in the literature, see for instance Gradshteyn and Ryzhik, 7th ed., formulas 3.997.1 and 3.997.2:

$$ \mathcal{I}(\alpha, p) = \frac{\sqrt \pi}{2} \left(\frac{2}{p}\right)^\alpha \Gamma\left(\frac{\alpha + 1}{2}\right) \mathbf{L}_\frac{\alpha}{2}(p)\\ \mathcal{J}(\alpha, p) = \sqrt \pi \left(\frac{2}{p}\right)^\alpha \Gamma\left(\frac{\alpha + 1}{2}\right) I_\frac{\alpha}{2}(p) $$

where $I_\frac{\alpha}{2}$ is the modified Bessel function of the first kind, and $\mathbf{L}_\frac{\alpha}{2}$ is the modified Struve function, see for instance DLMF, chapter 11.

The final result can be shown to be:

\begin{align*} I(p, 0, 1/4; n) =& \frac{2}{\pi} \sum\limits_{k\;\mathrm{even}} \sum\limits_{\ell=0}^{2n - k/2} \begin{pmatrix} 4n \\ k \end{pmatrix} \begin{pmatrix} 2n - k / 2 \\ \ell \end{pmatrix} (-1)^{2 n - \ell} \left[ \mathcal{I}(4n - 2\ell, p) + \frac{1}{2} \mathcal{J}(4n - 2\ell, p) \right] \end{align*}

Evidently, using the symmetry property, this automatically gives us $I(0,p,3/4;n)$ as well.

However, I haven't found anything which works for general $m$; while the procedure itself can more or less be repeated for any even $m$, the integrals $\mathcal{I}$ and $\mathcal{J}$ have different integration limits, which do not have any obvious representations in terms of special functions like the above.


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