Consider the integral:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\delta|^2\frac{e^{\kappa \cos (2\psi)}}{\pi I_0(\kappa)}\cos^2(2\psi)d\psi$$
Where $I_0(\kappa)$ is the modified Bessel function of the first kind of order $0$
What's the solution in terms of $\delta$ and $I_n(\kappa)$
From mathematica, I get:
$$\frac{|\delta|^2(I_1(\kappa )+\kappa I_2(\kappa ))}{\kappa I_0(\kappa )}$$
But the article that I'm studying, suggests it to be:
$$\frac{|\delta|^2(I_0(\kappa)+I_2(\kappa))}{2I_0(\kappa)}$$
Which one is right?
Also about the following integral:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\delta|^2\frac{e^{\kappa \cos (2\psi)}}{\pi I_0(\kappa)}\sin^2(2\psi)d\psi$$
I get this solution from mathematica:
$$\frac{|\delta|^2 I_1(\kappa)}{\kappa I_0(\kappa)}$$
but the paper has computed it as:
$$\frac{|\delta|^2(I_0(\kappa)-I_2(\kappa))}{2I_0(\kappa)}$$
Which one is right?
Both of them, since:
$$\kappa I_0(\kappa)+\kappa I_2(\kappa) = 2 I_1(\kappa)+2\kappa I_2(\kappa) \tag{1}$$ is a consequence of: $$ e^{z\cos\theta}=I_0(z)+2\sum_{n\geq 1}I_n(z)\cos(n\theta)\tag{2} $$ by differentiating twice both terms of $(2)$ with respect to $\theta$.
As an alternative, the identity: $$ I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{3}$$ comes from applying integration by parts, plus the cosine sum formula, to the integral: $$ I_n(\alpha)=\frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx.\tag{4}$$