For any continuous function $h:[-1,1] \to \mathbb R$, define $I_h:[-1,1] \to \mathbb R$ by $$ I_h(t) := \int_{-\pi}^\pi h(\cos\theta)h(t\cos\theta + (1-t^2)^{1/2}\sin\theta)d\theta = \int_{-\pi}^\pi h(\cos\theta)h(\sin(\theta + \alpha))d\theta, $$ where $\alpha=\alpha(t) := \arcsin(t)$.

For example, in the case $h(t) \equiv t^n$, the integral $I_h(t)$ has been computed here https://math.stackexchange.com/a/4242413/168758. Furthermore, the result in this case can be written compactly as $I_h(t) = 2^{1-n}\pi(-i\cos\alpha)^nP_n(i\tan\alpha)$, where $P_n$ is the $n$th Legendre polynomial.

Question. For $h(t) \equiv e^{ct}$ with $c>0$, is there a closed form solution in terms of special polynomials / functions (Legendre, Hermite, Bessel, etc.) ?

I'm really only interested in the values of $I_h(0)$, $I_h(1)$, and $I_h'(0)$.

Example. If $c=1$, then $I_h(0) = 2\pi J_2(0)$, where $J_2$ is the modified Bessel function of the second kind.


1 Answer 1


The integral is easily computable in terms of a Bessel function representation. For $h(t)=e^{ct}$ the integral in question reads

$$I_h(t)=\int_{-\pi}^{\pi}dx e^{c(1+t)\cos x+c\sqrt{1-t^2}\sin x}=\int_{-\pi}^{\pi}dx e^{c\sqrt{2(1+t)}\cos(x-\gamma)}$$

where the angle $\gamma$ is defined by $\cos\gamma=\sqrt{\frac{1+t}{2}}~,~ \sin\gamma=\sqrt{\frac{1-t}{2}}$. Because of the periodicity of the integrand and the fact that we are integrating over the exact period of the integrand we are allowed to shift away the angle $\gamma$. The resulting integral can be computed in terms of a modified Bessel function as follows

$$I_h(t)=\int_{-\pi}^{\pi}e^{c\sqrt{2(1+t)}\cos y}dy= 2\pi I_0\left(c\sqrt{2(1+t)}\right)$$

  • 1
    $\begingroup$ Thanks. Special functions are indeed special. I need to look more into them! $\endgroup$
    – dohmatob
    Commented Sep 7, 2021 at 21:47

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