Gaussian integral with a sine in the exponential

Question

This Gaussian integral came up while working on a likelihood analysis for pulsar timing arrays:

$$ \int_{-\pi}^{\pi} \exp\left( -(x-y \sin{\gamma} )^2 \right) \mathrm{d}\gamma $$

I've tried everything I can think of, but I can't get an analytic solution.

If you expand the square, the cross term can be written in terms of a cosh, but I don't know where to go from there. That gives the integrand as $\exp{(-x^2)} \exp{(-y^2 \sin^2\gamma)} \cosh{(2 x y \sin{\gamma})}$.

The substitution $\beta\equiv\sin{\gamma}$ leads to $$ \oint_0^0 \frac{\exp\left( -(x-y \beta )^2 \right)}{\sqrt{1-\beta^2}} \mathrm{d}\beta $$

I also tried $\arctan \alpha \equiv \gamma$, which gives

$$ \int \frac{\exp\left[-\left( x- \frac{\alpha y}{\sqrt{1+\alpha^2}} \right)^2\right]}{1+\alpha^2} \mathrm{d}\alpha $$

Neither Mathematica nor Rubi are able to evaluate any of these integrals. I was hoping the residue theorem might be applicable, or one of these substitutions might get into a form that Mathematica knows. Any help will be greatly appreciated and certainly land you in the acknowledgements of the paper my collaborators and I are working on:)

Any approximate results (saddle point approximation?) would also be appreciated.

Answer 1

Quite a simple approach with series, and certainly not an analytic solution, but this is all I have come up with:

$$ I(x,y):=\int_{-\pi}^\pi \exp{\big(-(x-y\sin {t})\big)}dt $$ $$ =\int_{-\pi}^\pi \exp{\big(-x^2-y^2 \sin^2{t}+2xy \sin t )\big)}dt $$ $$ =\exp(-x^2) \int_{-\pi}^\pi \sum_{r=0}^\infty \frac{(-y^2 \sin^2{t})^r}{r!}\sum_{s=0}^\infty \frac{(2xy \, \sin t)^s}{s!} dt $$ $$ =\exp(-x^2) \sum_{r=0}^\infty \sum_{s=0}^\infty \frac{(-y^2 )^r}{r!} \frac{(2xy )^s}{s!} \int_{-\pi}^\pi \sin^{2r+s} t \, dt $$ The integral is zero if $s$ is odd and otherwise we use: $$\int_{-\pi}^\pi \sin^{2n} t \, dt = 2 \pi \frac{(2n)!}{2^{2n}(n!)^2}$$ So, we take $s$ to be even and write $s=2p$: $$ I =\exp(-x^2) \sum_{r=0}^\infty \sum_{p=0}^\infty \frac{(-y^2 )^r}{r!} \frac{(2xy )^{2p}}{(2p)!} 2 \pi \frac{(2(r+p))!}{2^{2(r+p)}((r+p)!)^2} $$ $$ =2\pi \exp(-x^2) \sum_{r=0}^\infty \sum_{p=0}^\infty \frac{(-y^2 )^r}{r!} \frac{(xy)^{2p}}{(2p)!} \frac{(2(r+p))!}{2^{2r}((r+p)!)^2} $$ This is obviously not a closed form, but the terms get small fairly quickly even if $x=y=1$.

Answer 2

With $x \sim y$ and neither very large, the best approximation I can think of is a series expansion in the difference $x=y+\epsilon.$ You get an expansion in the generalized hypergeometric ${}_2F_2,$ a less-frequently encountered function in mathematical physics. The derivation is as follows.

Use symmetry to argue that $$ I(x,y):=\int_{-\pi}^\pi \exp{\big(-(x-y\sin^2{t})\big)}dt = \int_{-\pi}^\pi \exp{\big(-(x-y\cos^2{t})\big)}dt $$ Then set $x=y+\epsilon,$ use the half-angle trig ID and expand to order $\epsilon$ $$ I(x,y) \sim \int_{-\pi}^\pi \exp{\big(-(4y^2 \sin^4{(t/2)}+ 4y\epsilon \sin^2{(t/2)})\big)}dt \sim$$ $$\sim \int_{-\pi}^\pi \exp{\big(-4y^2 \sin^4{(t/2)}\big)} \Big(1+4y(y-x)\sin^2{(t/2)} \Big) dt $$ $$\sim 2\pi\Big({}_2F_2(1/4,3/4;1/2,1,-b) + 2y(y-x){}_2F_2(3/4,5/4;3/2,1,-b) \Big) $$ where the integrals have been done in Mathematica and $b=4y^2.$ This should work O.K. as long as $|y-x|<< y.$ For a numerical example, I took $x=0.88$ and $y=0.90.$ The value of the integral is 3.18050 and the approximation 3.18086. More terms in $\epsilon$ are easily derived.

I found several other expressions for the integral, but they don't seem to lead to known closed-forms. If, however, $x\sim1$ and $y\sim1,$ (an additional constraint not mentioned in the comments) there might be something else that can be said about an approximation to the function.