Evaluate integral $\int_{-\pi/2}^{\pi/2} \cos \left[\frac{\pi n}{2} +\left(a \sin t+b \cos t \right) \right] dt$, Bessel?

Question

Evaluate definite integral: $$\int_{-\pi/2}^{\pi/2} \cos \left[\frac{\pi n}{2} +\left(a \sin t+b \cos t \right) \right] dt$$

$n$ is an integer. $a,b$ real numbers.

The purpose of the integral - computing matrix elements of an electron Hamiltonian in an elliptic ring in the quantum box basis.

Before you ask me what I've done already, I've got this integral from the original, much more complicated one.

Originally I just gave up and computed it numerically.

But I wonder - is it possible to express the integral in closed form with Bessel functions?

Or maybe some series, still better than numerical integration.

I'm not asking for a full solution, some hint would be fine. Or even just a reassurance that a closed form exists.

Answer 1

Write $a \sin(t) + b \cos(t) = c \cos(t-\delta)$ where $a = c \sin(\delta)$ and $b = c \cos(\delta)$. So now (depending on $n$ mod $4$) we want to look at $\pm \int_{-\pi/2}^{\pi/2} \cos(c \cos(t-\delta))\; dt$ or $\pm \int_{-\pi/2}^{\pi/2} \sin(c \cos(t-\delta))\; dt$.

The integral with $\cos$ turns out nicely, because (due to symmetry) we can take the integral from $-\pi$ to $\pi$ and divide by $2$:

$$ \int_{-\pi/2}^{\pi/2} \cos(c \cos(t-\delta))\; dt = \pi J_0(c)$$

The integral with $\sin$ is not as nice. If we call it $F(c)$, then we have the differential equation

$$ c F''(c) + F'(c) + c F(c) = 2 \cos(\delta) \cos(c \sin(\delta))$$ with initial conditions $$ F(0) = 0,\ F'(0) = 2 \cos(\delta)$$

whose solution, according to Maple, is

$$ F(c) = \pi \cos(\delta) Y_0(c) \int_0^c J_0(z) \cos(z \sin(\delta))\; dz - J_0(c) \int_0^c Y_0(z) \cos(z \sin(\delta))\; dz $$

and I don't think those integrals over $z$ can be done in closed form.

Answer 2

This is not a complete answer. But I think that might be helpful toward the desired solution. Fix $x_n=\frac{\pi n}{2}$. Use the Taylor series $\cos (x_n+h)=\lim_{m\to \infty}\sum^{m}_{k=0} \frac{(-1)^k}{(2k)!}f^{(k)}(x_n)\cdot (x_n+h)^{2k} $ for $h=h(t)=(a\sin t +b\cos t)$ in the interval $ [-|a|-|b|,|a|+|b|]$. Set $ I_n=\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \cos \left( x_n+h(t) \right)\; \mathrm{d}t $. $$ I_n = \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \left[ \lim_{m\to \infty}\sum^{m}_{k=0} \frac{(-1)^k}{(2k)!} \left( \frac{\pi n}{2} +a \sin t+b \cos t \right)^{2k} \right] \mathrm{d}t $$ By Dominated_convergence_theorem \begin{align} I_n=& \lim_{m\to \infty}\sum^{m}_{k=0} \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \frac{(-1)^k}{(2k)!} \left( \frac{\pi n}{2} +a \sin t+b \cos t \right)^{2k} \mathrm{d}t \\ =& \lim_{m\to \infty}\sum^{m}_{k=0} \frac{(-1)^k}{(2k)!} \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \left( \frac{\pi n}{2} +a \sin t+b \cos t \right)^{2k} \mathrm{d}t \end{align} By multinomial theorem \begin{align} I_n= & \lim_{m\to \infty}\sum^{m}_{k=0} \frac{(-1)^k}{(2k)!} \sum_{i_1+i_2+i_3=2k} \frac{(2k)!}{i_1!\cdot i_2!\cdot i_3! } \left( \frac{\pi n}{2} \right)^{i_1} \cdot \left( a \right)^{i_2} \cdot \left( b \right)^{i_3} \cdot \int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \left(\sin t\right)^{i_2} \cdot \left(\cos t\right)^{i_3} \mathrm{d}t \end{align} Now you can search a table with definite integrals ( for exemple Handbook-Mathematical-Formulas-Integrals) for a recursive calculations of $\int_{-\frac{\pi}{2}}^{+\frac{\pi}{2}} \left(\sin t\right)^{i_2} \cdot \left(\cos t\right)^{i_3} \mathrm{d}t$. Let's say $$ \int \sin^{u}(x)\cos^v(x) \mathrm{d} x = \begin{cases} -\frac{sin^{u-1}(x)\cos^{u+1}(x)}{u+v}+\frac{u-1}{u+v}\int \sin^{u-2}(x)\cos^v(x) \mathrm{d} x \\ \frac{sin^{u+1}(x)\cos^{u-1}(x)}{u+v}+\frac{v-1}{u+v}\int \sin^{u}(x)\cos^{v-2}(x) \mathrm{d} x \end{cases} $$ I strongly suspect that this integral can be expressed in terms of Gamma and Beta functions.

Answer 3

Little addition on $\sin$ case:

$$ I=\int^{\pi/2}_{-\pi/2}\sin(c\cos(t-\delta))dt=\int^{\pi}_{0}\sin(c\sin(t-\delta))dt $$

Make substitution: $g=\sin(t-\delta)\quad dt=\frac{dg}{\pm \sqrt{1-g^2}}$

Sign in last expression depends on range. We know that $-\frac{\pi}{2}<\delta<\frac{\pi}{2}$. Researching sign of $\cos$, we get $[-\delta,\frac{\pi}{2}]-$plus sign and $[\frac{\pi}{2},\pi-\delta]-$minus. It leads to partition of the integral:

$$ \left[\int^{1}_{-\sin(\delta)}\frac{\sin(cg)}{\sqrt{1-g^2}}dg \right]-\left[\int^{\sin(\delta)}_{1}\frac{\sin(cg)}{\sqrt{1-g^2}}dg\right] $$

Revert boundaries of 2nd integral and use $\int^{\sin(\delta)}_{-\sin(\delta)}=\int^{-\sin(\delta)}_{\sin(\delta)}=0$ (as far as argument is an odd function). $$ 2\int^{1}_{\max(-\sin(\delta),\sin(\delta))}\to 2\int^{1}_{\sin|\delta|} $$ Then, $$ I=2\int^{1}_{\sin|\delta|}\frac{\sin(cg)}{\sqrt{1-g^2}}dg $$ Where $\sin$ as well as square root can be expanded in Taylor series, then apply formula $\sum^{\infty}_{n=0}a_n\sum^{\infty}_{m=0}b_m=\sum^{\infty}_{n=0}\sum^{n}_{k=0}a_{n-k}b_k$, change order of summation and integration and only thing left is to numerically compute obtained series.