# Gaussian integral with a sine in the exponential

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.

• I tried a few other substitutions and no result ! – Claude Leibovici Jun 1 at 5:42
• Can I post a partial solution? I think it would be really helpful. It is a bit complicated, and I fear if there exists an analytic solution. – Laxmi Narayan Bhandari Jun 1 at 8:54
• @LaxmiNarayanBhandari please post it, it would be nice to see! – Robbie Rosati Jun 1 at 14:32
• An asymptotic formula is likely. Please state if $x \to \infty,$ $y \to \infty,$, both $x,y \to \infty$ but the ratio $y/x < 1,$ etc. Often asymptotic formulas are more useful than exact solutions, especially if the exact formula is in terms of higher-order hypergeometric functions, which then need subsequent analysis for large parameters. – skbmoore Jun 1 at 14:55
• @skbmoore I'm not sure if it helps, but $x$ and $y$ are about equal, positive, and $\lesssim \mathcal{O}(1)$. – Robbie Rosati Jun 1 at 15:04

$$I(x,y):=\int_{-\pi}^\pi \exp{\big(-(x-y\sin {t})\big)}dt$$ $$=\int_{-\pi}^\pi \exp{\big(-x^2-y^2 \sin^2{t}+2xy \sin t )\big)}dt$$ $$=\exp(-x^2) \int_{-\pi}^\pi \sum_{r=0}^\infty \frac{(-y^2 \sin^2{t})^r}{r!}\sum_{s=0}^\infty \frac{(2xy \, \sin t)^s}{s!} dt$$ $$=\exp(-x^2) \sum_{r=0}^\infty \sum_{s=0}^\infty \frac{(-y^2 )^r}{r!} \frac{(2xy )^s}{s!} \int_{-\pi}^\pi \sin^{2r+s} t \, dt$$ 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 $$x=y=1$$.
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.