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\section{\huge The Analytic Form of the HESSI Point-Spread Function}


The point-spread function is the response of a single point source in
a map.  The present analysis is for the fundamental, but higher
harmonics and the effects of shadowing or 
grid nonuniformities could readily be included.

The instantaneous probability $\Pi_{\alpha}(r,\theta)$ of photon passage at
grid angle $\alpha$ is proportional to $cos(2\pi (r/p) 
cos(\alpha-\theta)-\gamma(\alpha)) $, where $\gamma$ is the collimator phase
at the spin axis.  Multiplying this by the response
$\Pi_{\alpha}(r_0,\theta_0)$ to a point source at ($r_0,\alpha_0$) gives an
instananeous PSF, which when integrated over the roll angles in one
half rotation gives the full PSF:

$$ P =  \int_{\delta}^{\pi+\delta} \Pi_{\alpha}(r,\theta) 
\Pi_{\alpha}((r_0,\theta_0) d\alpha $$

$$ =  \int_{\delta}^{\pi+\delta} 
 cos(2\pi r\ cos(\alpha-\theta)-\gamma(\alpha))  
 \ cos(2\pi r_0\ cos(\alpha-\theta_0)-\gamma(\alpha))\ d\alpha $$

We make the assumption that $\gamma$ is independent of roll angle, at least during one half rotation.
Changing variables, $\beta=\alpha-\delta$ and simplifying the formulas with
$k=2\pi r/p$ and $k_0=2\pi r_0/p$,


$$ P = \int_0^{\pi} cos(k\ cos(\beta-\theta-\gamma)) cos(k_0\ cos(\beta-\theta_0-\gamma))d\beta  $$

Using the cosine sum rule permits us to show the $\gamma$ dependence
explicitly:

$$ P = {\textstyle{1\over2}} P_1(k,k_0,\theta,\theta_0) +
       {\textstyle{1\over2}} cos 2\gamma\ P_2(k,k_0,\theta,\theta_0) +
       {\textstyle{1\over2}} sin 2\gamma\ P_3(k,k_0,\theta,\theta_0) $$

%\int_0^{\pi} cos(k\ cos(\beta-\theta)) cos(k_0\ cos(\beta-\theta_0))d\beta  $$

where the functions $P_1, P_2$ and $P_3$ are:

% $$ +sin^2\gamma \int_0^{\pi} sin(k\ cos(\beta-\theta)) sin(k_0\ cos(\beta-\theta_0))d\beta  $$

% $$ sin\gamma cos\gamma  \int_0^{\pi} \lbrace cos(k\ cos(\beta-\theta)) sin(k_0\ cos(\beta-\theta_0)) + sin(k\ cos(\beta-\theta)) cos(k_0\ cos(\beta-\theta_0)) \rbrace d\beta $$


$$ P_1 = \int_0^{\pi} cos(k\ cos(\beta-\theta)-k_0\ cos(\beta-\theta_0)) d\beta, $$

$$ P_2 = \int_0^{\pi} cos(k\ cos(\beta-\theta)+k_0\ cos(\beta-\theta_0)) d\beta, $$

and
$$ P_3 = \int_0^{\pi} sin(k\ cos(\beta-\theta)+k_0\ cos(\beta-\theta_0)) d\beta $$

All of these integrals can be reduced to the form $\int_0^{\pi} F(zcos(\beta-\psi)) d\beta$, where $F$ is $cos$ or $sin$.  Let $z_1$ and  $z_2$ be defined by:


$$ z_{1,2}=\lbrace k_0^2+k^2 \pm 2kk_0 cos(\theta-\theta_0)\rbrace ^{1/2} $$

and $\psi_{1,2}$ are defined by:

$$sin(\psi_{1,2})=(k sin(\theta) \pm k_0 sin(\theta_0))/z_{1,2} $$

Then we have 

$$P_1=\int_0^{\pi} cos(z_1 cos(\beta-\psi_1)) d\beta$$
$$P_2=\int_0^{\pi} cos(z_2 cos(\beta-\psi_2)) d\beta$$
$$P_3=\int_0^{\pi} sin(z_1 cos(\beta-\psi_1)) d\beta$$

It is easy to show that $P_1$ is independent of $\psi_1$ and $P_2$ is
independent of $\psi_2$  (This follows from the transformation $\beta'=\beta
-\psi_1$, which allows us to evaluate $\partial P_1/\partial \psi_1 =
[cos(z_1 cos(\beta')]^{\pi-\psi_1}_{-\psi_1} = 0$. Similarly for $P_2$.)

Both $P_1$ and $P_2$ may be written in terms of Bessel functions:

$$P_1= J_0(z_1) \qquad {\rm and} \qquad P_2= J_0(z_2) $$

but $P_3$ does not reduce to a Bessel function, and it depends explicitly on
$\psi_1$.

%The functions $z_{1,2}$ are dependent only on the difference 
% $\theta-\theta_0$,
% but $\psi_1$ and $\psi_2$ depend on more than the difference.

Because of the mirror asymmetry of the sine, $P_3$ is zero when $\psi_1=0$, 
and when $\psi_1=\pi/2$,

$$P_3(z_1,\psi_1=\pi/2) = \pi  {\bf H_0}(z_1) $$

where  ${\bf H_0}(z)$ is the Struve function,
${1 \over \pi} \int_0^{\pi/2} sin(z cos\beta) d\beta. $

A little manipulation of integrals shows that $P_3$ may be written as
an ``incomplete'' Struve function:

$$ P_3= \int_0^{\psi_1} sin(z cos\beta) d\beta. $$

Since $P_3$ is obviously an odd periodic function of $\psi_1$, it may
be expressed as a Fourier series with the leading terms:

$$ P_3 = [a_1(z) sin\psi + a_3(z) sin(3\psi) + \cdots] $$

The Fourier coefficients vanish for even n, and for odd n they are:

$$ a_n = \int_0^{2\pi} \int_0^{2\pi} sin[z_1 cos(\beta-\psi_1) sin(n\psi_1) 
d\beta d\psi_1 $$

This may be reduced to a single integral over $\psi$ by changing variables
to $\beta'=\beta-\psi_1$, and integrating over $\beta'$.  

$$a_n = {\textstyle{2 \over n}} \int_0^{2\pi} sin(z\ cos\psi) cos(n\psi) d\psi $$

Using the integral definition of the Bessel function,

$$ J_n(z)={i^{-n} \over 2\pi} \int_0^{2\pi} e^{izcos\psi} cos{n\psi} d\psi, $$

the Fourier coefficients may be written:

$$ a_n={1 \over ni} \int_0^{2\pi} \lbrace
e^{izcos\psi}-e^{-izcos\psi} \rbrace cos n\psi \ d\psi $$

$$ ={\textstyle{{2\pi i^n} \over ni}} \lbrace J_n(z)-J_n(-z)\rbrace = {\textstyle{2\pi i^{n-1}} \over n} J_n(z)  \lbrace 1-(-1)^n \rbrace$$

So the leading terms in the Fourier expansion of $P_3$ are:

$$ P_3= 4\pi \lbrace  J_1(z_1) sin\psi_1 - (1/3) J_3(z_1) sin(3\psi_1) + 
\cdots  \rbrace $$

When $\psi=\pi/2$ this must reduce to $\pi {\bf H_0}(z)$, and it does,
because of the expansion of ${\bf H_0}(z)$ in terms of Bessel
functions:

$$ {\bf H_0}(z)={4 \over \pi} \sum_{k=0}^{\infty} {J_{2k+1}(z) \over {2k+1}} $$

So finally, the HESSI PSF may be written:

$$P= {\textstyle{1\over2}} [J_0(z_1) +cos2\gamma J_0(z_2)]+ 2\pi sin2\gamma \lbrace
 J_1(z_1) sin\psi_1 - (1/3) J_3(z_1) sin3\psi_1 + \cdots  \rbrace $$

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