4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I tried a few other substitutions and no result ! $\endgroup$ – Claude Leibovici Jun 1 at 5:42
  • $\begingroup$ 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. $\endgroup$ – Laxmi Narayan Bhandari Jun 1 at 8:54
  • 1
    $\begingroup$ @LaxmiNarayanBhandari please post it, it would be nice to see! $\endgroup$ – Robbie Rosati Jun 1 at 14:32
  • 1
    $\begingroup$ 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. $\endgroup$ – skbmoore Jun 1 at 14:55
  • $\begingroup$ @skbmoore I'm not sure if it helps, but $x$ and $y$ are about equal, positive, and $\lesssim \mathcal{O}(1)$. $\endgroup$ – Robbie Rosati Jun 1 at 15:04
3
$\begingroup$

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 $x=y=1$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.