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$$\int_{0}^{2\pi}\exp\left(-iA(x\cos\varphi+y\sin\varphi)\right)\cos(l\varphi)\,d\varphi$$

I'm trying to evaluate the integral for an interference problem in Physics. When $y=0$, this reduces to the Bessel Function of the first kind, and when $l=1$, I can differentiate under the integral w.r.t. $x$ and evaluate the integral (which gives a first order Bessel function of the first kind). However, I'm looking for a more general answer, where $l$ is any integer, and $A$ is an arbitrary constant.

Here's a link to a similar question posted 2 years ago. How to solve integral $\int_0^{2\pi} e^{i(a\cos\phi + b\sin\phi)} \cos\phi\ d\phi$

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  • $\begingroup$ Why don't you simply define a new coordinate system by rotation in which $(x,y)$ lies strictly on one axis. $\endgroup$ – Hamed Aug 23 '18 at 16:05
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Let $\rho = \sqrt {x^2 + y^2}, \phi = \arctan(x, y)$. Then $$\int_0^{2 \pi} e^{-i A (x \cos t + y \sin t)} \cos l t \,dt = \\ \int_0^{2 \pi} e^{-i A \rho \cos(t - \phi)} \cos l t \,dt = \cos l \phi \int_0^{2 \pi} e^{-i A \rho \cos t} \cos l t \,dt = \\ 2 \pi (-i)^l \cos l \phi \,J_l(A \rho).$$

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  • $\begingroup$ Hey! Thanks for your answer. Before posting this question, I was able to get to $\int_{0}^{2\pi}e^{-iA\rho\cos(t-\phi)}\cos(lt)\,dt$, but it's the next step where you factored out $\cos(l\phi)$ that I couldn't get to, and still don't understand. Can you please show the working? $\endgroup$ – Pooja Jayachandran Aug 23 '18 at 22:12
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    $\begingroup$ $\int_{-\phi}^{2 \pi - \phi} e^{-i A \rho \cos t} \cos l (t + \phi) dt = \int_0^{2 \pi} e^{-i A \rho \cos t} \cos l (t + \phi) dt$ because the integrand is periodic with period $2 \pi$. After expanding $\cos l (t + \phi)$, the integral of $e^{-i A \rho \cos t} \sin l t$ is zero. Similar to this question. $\endgroup$ – Maxim Aug 23 '18 at 23:18
  • $\begingroup$ That makes sense, thank you! $\endgroup$ – Pooja Jayachandran Aug 24 '18 at 10:42
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Let's recall an integral form of Bessel function of first kind $J_\ell(y)$ for $\ell$ integer: $$ J_\ell(y)=\frac{1}{2\pi}\int_0^{2\pi} e^{i(y\sin \varphi-\ell\varphi)}d\varphi $$ which is a real number.

Now for your question, define the unit vector $\vec{n}(\varphi)=(\cos\varphi, \sin\varphi)$ and $\vec{r}=(x,y)$. Note that the exponent in the integrand is now $$ -i(x\cos \varphi+y\sin\varphi)=-i\vec{n}\cdot \vec{r} $$ Define the following integrals Similarly, define two more integrals $$\begin{aligned} I_\ell(\vec{r})&=\int_0^{2\pi} e^{-i\vec{n}(\varphi)\cdot\vec{r}}\cos (\ell \varphi) d\phi\\ I'_\ell(\vec{r})&=\int_0^{2\pi} e^{-i\vec{n}(\varphi)\cdot \vec{r}}\sin \ell \varphi d\varphi\\ \widetilde{I_\ell}(\vec{r})&= \int_0^{2\pi} e^{-i[\vec{n}(\varphi)\cdot \vec{r}-\ell\varphi]} d\varphi = I_\ell(\vec{r})+iI'_\ell(\vec{r}) \end{aligned} $$ Note that $I_\ell$ and $I'_\ell$ are both real if $\ell$ is even, and purely imaginary otherwise (use the fact that $\vec{n}\to -\vec{n}$ means $\varphi\to \pi+\varphi$). This means $I_\ell$ is the real part of $\widetilde{I}$ if $\ell$ is even, and it is equal to $i\mathfrak{I}(\widetilde{I_\ell})$ if $\ell$ is odd.

Moving on... Let $R_{\theta}$ be a rotation matrix (around $z$-axis, i.e. a 2D rotation) by an amount $\theta$. Note that in general $\vec{u}\cdot R_\theta \vec{v}=R_{-\theta}\vec{u}\cdot \vec{v}$. Moreover, $R_\theta \vec{n}(\varphi)=\vec{n}(\varphi+\theta)$. Using these facts, you find (I'll leave the details to you) $$ \widetilde{I_\ell}(R_\theta\vec{r})= e^{-i\theta \ell} \widetilde{I_\ell}(\vec{r}) $$ At the same time, let $\theta(\vec{r})$ be such that $R_\theta \vec{r}=(0,Y)$ for $Y(\vec{r})=\sqrt{x^2+y^2}$ (no $x$-component). If you need a formula $\theta(\vec{r})=\pi/2-\arctan y/x$. Then $$ \widetilde{I_\ell}(\vec{r})=e^{i\theta(\vec{r})\ell}\int_0^{2\pi} e^{-i(Y(\vec{r})\sin \varphi-\ell\varphi)}d\varphi= 2\pi e^{i\theta(\vec{r})\ell} J_\ell(Y(\vec{r})) $$ Therefore $$ \boxed{ I_\ell(\vec{r})=\begin{cases} 2\pi \cos [\theta(\vec{r})\ell] J_\ell(Y(\vec{r})) & \ell\text{ even}\\ 2\pi i \sin [\theta(\vec{r})\ell] J_\ell(Y(\vec{r})) & \ell\text{ odd} \end{cases} } $$

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  • $\begingroup$ Note that the integral is not zero for $\ell = 0, y = 0$. The factor depending on $\phi = \arctan(x, y)$ is $\cos \ell \phi$. $\endgroup$ – Maxim Aug 23 '18 at 20:45
  • $\begingroup$ Thanks, it was a propagating typo. $\endgroup$ – Hamed Aug 23 '18 at 20:58
  • $\begingroup$ The sign for odd $\ell$ is reversed, $I_1((x, 0)) = -2 \pi i J_1(x)$. $\endgroup$ – Maxim Aug 23 '18 at 21:18

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