Is there an analytical expression for the following integral? As a solution for the definite integral or the indefinite integral form?
$$\int _{\cos\left(\frac{2\pi}{L}\right)} ^{1} \frac{\cos\left(A y-B\sqrt{1-y^2}\right)}{\sqrt{1-y^2}} dy,$$
where $A\in \mathbb{R}$, $B\in \mathbb{R}$ and $L>0$, $L \in \mathbb{Z}$.
Yes, I got one solution but it isn't too good. It's quite messy with double sums over incomplete beta function. Let $$I(t)=\int_{\cos \frac{2\pi}{L}}^{1}\frac{\cos (Ay-Bt\sqrt{1-y^2}}{\sqrt{1-y^2}}dy$$ Let, $A(l)=\cos \frac{2\pi}{L}$. Differentiating both sides of $I(t)$ leads to, $$I'(t)=B\int_{A(l)}^{1}\sin (Ay-Bt\sqrt{1-y^2})dy$$ By series expansion of sine function, $$\sin x=\sum_{k\ge 0}\frac{(-1)^{k}x^{2k-1}}{(2k+1)!}$$ $$I'(t)=\sum_{k\ge 0}\frac{(-1)^{k}}{(2k+1)!}\int_{A(l)}^{1}(Ay-Bt\sqrt{1-y^2})^{2k+1}dy$$ Now use the binomial theorem to expand the stuff inside the expression. $$I'(t)=\sum_{k\ge 0}\frac{(-1)^{k}}{(2k+1)!}\sum_{u=0}^{2k+1}\binom{2k+1}{u}\int_{A(l)}^{1}(Ay)^{2k+1-u} (Bt\sqrt{1-y^2})^{u}dy$$ Bring the constants $A,B$ outside the integrand. $$I'(t)=\sum_{k\ge 0}(-1)^{k}\frac{(-1)^{k}}{(2k+1)!}\sum_{u=0}^{2k+1}\binom{2k+1}{u}A^{2k+1-u}B^{u}\int_{A(l)}^{1} y^{2k+1-u} (1-y^2)^{\frac{u}{2}}dy$$ Substitute $x=y^2 \implies dy=\frac{dx}{2\sqrt x}$ $$I'(t)=2\sum_{k\ge 0}(-1)^{k}\frac{(-1)^{k}}{(2k+1)!}\sum_{u=0}^{2k+1}\binom{2k+1}{u}A^{2k+1-u}B^{u}\int_{B(l)}^{1} x^{k-u/2} (1-x)^{\frac{u}{2}} dx$$ Where $B(l)=A^2(l)$. Now let's focus on $A(l)=\cos \frac{2\pi}{L}$. Since $L \gt 0, L\in \mathbb{Z}$, it have only one zero at $L=4$. So let me simplify the case to $L>4$,As for all $L>4$ the value of $A(l)>0$ and so do $B(I)$. $$I'(t)=2\sum_{k\ge 0}(-1)^{k}\frac{(-1)^{k}}{(2k+1)!}\sum_{u=0}^{2k+1}\binom{2k+1}{u}A^{2k+1-u}B^{u}\int_{B(l)}^{1} x^{k-u/2} (1-x)^{\frac{u}{2}} dx$$ I can split the integral as follows, $$\int_{B(l)}^{1} x^{k-u/2} (1-x)^{\frac{u}{2}} dx=\int_{0}^{1} x^{k-u/2} (1-x)^{\frac{u}{2}} dx - \int_{0}^{B(l)} x^{k-u/2} (1-x)^{\frac{u}{2}} dx $$ The first integral will be, $B(k-\frac{u}{2}+1,\frac{u}{2}+1)$. It's the beta function. Similarly the second integral would be incomplete beta function. It would be $B(B(l);k-\frac{u}{2}+1,\frac{u}{2}+1)$ Now we got the closed form for its derivative. Integrate on both sides, will give you the closed form for $I(t)$ and finally set $t=1$