3
$\begingroup$

Given two polynomials $a(t)$ and $b(t)$, consider the integral

$$C(z)=\int_0^1 \frac{dt}{\sqrt{(z-b(t))^2+a(t)}}.$$

For $z$ near $\infty$, $C(z)$ is a well-defined analytic function which can be extended analytically to the multi-valued function belonging to the so-called Nilsson class. In particular, by some general results there exists a linear non-homogeneous differential operator in $z$ with polynomial coefficients of order $2g$ which should annihilate $C(z)$. Here $g$ is the genus of a certain hyperelliptic curve related to $b(t)$ and $a(t)$.

I have real difficulties to find such an operator explicitly even when $a(t)$ and $b(t)$ are of low degree. Even when both polynomials are linear, it seems to be quite non-trivial. Maybe somebody knows of a method and/or software to do this? I would be happy to be able to handle some cases when $a(t)$ and $b(t)$ are quadratic (which should be much easier, than the general case).

Sincerely, B.Shapiro, Stockholm University

$\endgroup$
  • $\begingroup$ Oops, I forgot the square root in the denominator of the integral! It should be \int_0^1 dt/Sqrt ((z-b(t))^2+a(t)) of course. $\endgroup$ – user53092 Mar 3 '17 at 13:29
4
$\begingroup$

In general, there's nothing special about $t=0$ and $t=1$, so to get a closed form you're going to need an antiderivative. An antiderivative of the $-1/2$ power of a quadratic in $t$ is well-known:

$$ \int \frac{dt}{\sqrt{A t^2 + B t + C}} = \frac{1}{\sqrt{A}} \ln \left(2 \sqrt{A} t + 2 \sqrt{A t^2 + B t + C} + B/\sqrt{A}\right) $$

In Maple, once you have the integral you might use holexprtodiffeq in the gfun package to find a linear differential equation it satisfies.

For example, this tells me that $$\int_0^1 \frac{dt}{\sqrt{(z-t)^2 + t + 1}} = \ln(3+2\sqrt{z^2-2z+3}-2z) - \ln(1 + 2 \sqrt{z^2+1}-2z)$$ satisfies the differential equation $$ \left( 16\,{z}^{7}+4\,{z}^{6}+4\,{z}^{5}-592\,{z}^{4}+352\,{z}^{3}- 254\,{z}^{2}-128\,z+138 \right) {\frac {\rm d}{{\rm d}z}}f \left( z \right) + \left( 32\,{z}^{8}-8\,{z}^{7}-17\,{z}^{6}-501\,{z}^{5}+400 \,{z}^{4}-790\,{z}^{3}-265\,{z}^{2}-177\,z+126 \right) {\frac {{\rm d} ^{2}}{{\rm d}{z}^{2}}}f \left( z \right) + \left( 8\,{z}^{9}-6\,{z}^{8 }+11\,{z}^{7}-53\,{z}^{6}+89\,{z}^{5}-155\,{z}^{4}-103\,{z}^{3}-135\,{ z}^{2}-189\,z-27 \right) {\frac {{\rm d}^{3}}{{\rm d}{z}^{3}}}f \left( z \right) =0$$

For a quartic, you have an incomplete elliptic integral. In Maple's notation,

$$\eqalign{&\int \!{\frac {1}{\sqrt { \left( t-r_{{1}} \right) \left( t-r_{{2}} \right) \left( t-r_{{3}} \right) \left( t-r_{{4}} \right) }}} \,{\rm d}t\cr & \ \ = {\frac {\pm 2}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{1}}-r_{{3} }}}{\it EllipticF} \left( {\frac {\sqrt {r_{{1}}-r_{{3}}}\sqrt {t-r_{{ 4}}}}{\sqrt {r_{{1}}-r_{{4}}}\sqrt {t-r_{{3}}}}},{\frac {\sqrt {-r_{{3 }}+r_{{2}}}\sqrt {r_{{1}}-r_{{4}}}}{\sqrt {r_{{2}}-r_{{4}}}\sqrt {r_{{ 1}}-r_{{3}}}}} \right) }} $$

In your application, $r_1 \ldots r_4$ will be the roots of a quartic depending on the parameter$z$, and though these can be represented explicitly in terms of radicals the result is generally not going to be pretty. It's probably best to leave these in implicit form.

Unfortunately, it looks like holexprtodiffeq will not handle expressions involving EllipticF. In principle, I guess the differential operator should be determined by the partial differential equations satisfied by EllipticF. $$\eqalign{ {\frac {\partial ^{2}}{\partial {k}^{2}}}f \left( x,k \right) &+{\frac { \left( 3\,{k}^{2}-1 \right) }{{k}^{3}-k}} {\frac {\partial }{\partial k}}f \left( x,k \right) +{\frac { \left( -{x}^{3}+x \right) }{ (k^2-1)(k^2 x^2-1)}} {\frac {\partial }{\partial x}}f \left( x,k \right)+{\frac {f \left( x ,k \right) }{{k}^{2}-1}}=0 \cr {\frac {\partial ^{2}}{\partial x \partial k}}f \left( x,k \right) & +{\frac { \left( {\frac {\partial }{ \partial x}}f \left( x,k \right) \right) k{x}^{2}}{{k}^{2}{x}^{2}-1}} =0\cr {\frac {\partial ^{2}}{\partial {x}^{2}}}f \left( x,k \right) &-{ \frac { \left( -2\,{k}^{2}{x}^{3}+ \left( {k}^{2}+1 \right) x \right) {\frac {\partial }{\partial x}}f \left( x,k \right) }{1+{k}^{2}{x}^{4} + \left( -{k}^{2}-1 \right) {x}^{2}}}=0} $$

$\endgroup$
  • $\begingroup$ Many thanks, Robert. It is a pity that holexpprttodiffeq is rather weak, but it might reflect the difficulty of finding an algorithm solving my problem. Sincerely, B.Shapiro $\endgroup$ – user53092 Mar 6 '17 at 18:37
2
$\begingroup$

If you have access to Mathematica, you may also want to try my package HolonomicFunctions on your problem. Using the method of creative telescoping, it operates directly on the integral, so there is no need to find an antiderivative or a closed form. The package can find a homogeneous or an inhomogeneous linear ODE.

When the polynomials $a(t)=t+1$ and $b(t)=t$ are linear, as in the example above, the input

Annihilator[Integrate[1/Sqrt[(z - t)^2 + t + 1], {t, 0, 1}], Der[z]]

yields the same differential operator of order 3.

For quadratic polynomials (I chose "randomly" $a(t)=2 t^2+3t+4$ and $b(t)=t^2+t+1$) let me demonstrate how to obtain an inhomogeneous differential equation. The following command

Annihilator[
  Integrate[1/Sqrt[(z - (t^2 + t + 1))^2 + (2 t^2 + 3 t + 4)], {t,0,1}], 
  Der[z], Inhomogeneous -> True]

yields the second-order differential operator $$ P=(2z+3)(8z+5)(1024z^4-1280z^3+3712z^2+240z+1825)\partial_z^2+ (32768z^5+83968z^4-25600z^3+64768z^2+52960z-58450)\partial_z+ 4096z^4+23040z^3-128z^2-6048z+9320 $$ and the inhomogeneous part $$ h(z)=\frac{2 \left(-3328 z^6+17376 z^5-23208 z^4+564178 z^3-158634 z^2-5076 z+747765\right)}{\left((z-3)^2+9\right)^{3/2}}-\frac{2 \left(-1280 z^6+8928 z^5-6824 z^4+88994 z^3-1430 z^2-4050 z+105250\right)}{\left((z-1)^2+4\right)^{3/2}} $$ such that the integral satisfies $P(C(z))+h(z)=0$.

$\endgroup$

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.