To find iteration method to evaluate $ \int_{-\infty}^{+\infty} e^ {-t(t-x)(t-y)(t-z)}\;dt$

Question

$$ f(x,y,z)= \int_{-\infty}^{+\infty} e^ {-t(t-x)(t-y)(t-z)}\;dt$$

$$t=p+x$$ $$ f(x,y,z)= \int_{-\infty}^{+\infty} e^ {-p(p+x)(p-(y-x))(p-(z-x))}\;dp$$

$$ f(-x,y-x,z-x)= f(x,y,z) \tag {1} $$

$$ f(0,a,b)= \int_{-\infty}^{+\infty} e^ {-t^2(t-a)(t-b)}\;dt$$

$$ f(0,a,b)e^ {c}= \int_{-\infty}^{+\infty} e^ {-t^2(t-a)(t-b)+c}\;dt$$

We can find roots of $$t^2(t-a)(t-b)-c=0$$, as $$x_1,x_2,x_3,x_4$$ The roots depend on $a,b,c$ $$ f(0,a,b)e^ {c}= \int_{-\infty}^{+\infty} e^ {-(t-x_1))(t-x_2)(t-x_3)(t-x_4)}\;dt$$

$$t=p+x_1$$

$$ f(0,a,b)e^ {c}= \int_{-\infty}^{+\infty} e^ {-p(p-(x_2-x_1))(p-(x_3-x_1))(p-(x_4-x_1))}\;dp$$

$$ f(0,a,b)e^ {c}= f(x_2-x_1,x_3-x_1,x_4-x_1)=\int_{-\infty}^{+\infty} e^ {-p(p-(x_2-x_1))(p-(x_3-x_1))(p-(x_4-x_1))}\;dp$$

It is enough to solve $$ f(0,a,b)= \int_{-\infty}^{+\infty} e^ {-t^2(t-a)(t-b)}\;dt$$

$$t=p-\frac{(a+b)}{4}$$

$$ f(0,a,b)= \int_{-\infty}^{+\infty} e^ {-p^4+(3\frac{a+b}{3}+ab)p^2-\frac{a+b}{2}(\frac{(a+b)^2}{4}+ab)p+\frac{(a+b)^2}{16}(3\frac{(a+b)^2}{16}+ab)}\;dp$$

$$ f(0,a,b)= e^ {\frac{(a+b)^2}{16}(3\frac{(a+b)^2}{16}+ab)}\int_{-\infty}^{+\infty} e^ {-p^4+(3\frac{a+b}{3}+ab)p^2-\frac{a+b}{2}(\frac{(a+b)^2}{4}+ab)p}\;dt$$

$$m=3\frac{a+b}{3}+ab$$

$$x=-\frac{a+b}{2}(\frac{(a+b)^2}{4}+ab)$$

$$ f(0,a,b)= e^ {\frac{(a+b)^2}{16}(3\frac{(a+b)^2}{16}+ab)}\int_{-\infty}^{+\infty} e^ {-p^4+mp^2+xp}\;dt$$

$$ f(0,a,b)e^ {-\frac{(a+b)^2}{16}(3\frac{(a+b)^2}{16}+ab)}= \int_{-\infty}^{+\infty} e^ {-p^4+mp^2+xp}\;dp$$

$$ y= g_m(x)= \int_{-\infty}^{+\infty} e^ {-p^4+mp^2+xp}\;dp$$ If we define $$y'=\frac{dy}{dx}$$ $$y'''=\frac{d^3y}{dx^3}$$

and $m$ is an constant $$-4y'''+2my'+xy=\int_{-\infty}^{+\infty}(-4p^3+2mp+x) e^ {-p^4+mp^2+xp}\;dp=e^ {-p^4+mp^2+xp}|{_{p=-\infty}^{p=+\infty}}=0$$

I need to solve the third-order linear differential equation $$-4y'''+2my'+xy=0 \tag{2}$$ I do not know the solution of it.

I believe we can also find a way to solve $f(x,y,z)$ via using iteration method .

the solution can be written $$ f(x,y,z)e^{c_1}= \int_{-\infty}^{+\infty} e^ {-t(t-x)(t-y)(t-z)+c_1}\;dt$$

where $c_1$ depends on $x,y,z$

$$ f(x,y,z)e^ {c_1}= \int_{-\infty}^{+\infty} e^ {-(t-x_1))(t-y_1)(t-z_1)(t-b_1)}\;dt$$

$t-->t+x_1$

$$ f(x,y,z)e^ {c_1}= \int_{-\infty}^{+\infty} e^ {-t(t-(y_1-x_1))(t-(z_1-x_1)(t-(b_1-x_1))}\;dt$$
where $c_2$ depends on $(y_1-x_1),(z_1-x_1),(b_1-x_1)$

$$ f(x,y,z)e^ {c_1}e^ {c_2}= \int_{-\infty}^{+\infty} e^ {-(t-x_2))(t-y_2)(t-z_2)(t-b_2)}\;dt$$

And If we iterate in this way

Finally we can get $x_n=y_n=z_n=b_n=k$ where $n-->\infty$

$$ f(x,y,z)e^ {c_1+c_2+c_3+....}=\int_{-\infty}^{+\infty} e^ {-(t-k)^4}\;dt$$

$t-->t+k$

$$ f(x,y,z)e^ {c_1+c_2+c_3+....}=\int_{-\infty}^{+\infty} e^ {-t^4}\;dt=2 Γ(5/4)$$

$$ f(x,y,z)=2 Γ(5/4)e^ {-(c_1+c_2+c_3+....)}$$
My question :

Is it possible to find an iteration algorithm as define above to solve $f(x,y,z)$

How can $c_n$ selected for next iteration that finally we get $x_n=y_n=z_n=b_n=k$ where $n-->\infty$?

If you have other method to solve the $ f(x,y,z)$ please let me know.

Note: This kind of iteration method can be used to solve similar differential equations like equation (2).

Thanks for help and comments

EDIT:

I used the method that @mrc ntn gave clue in his answer below. I would like to find more terms of $f(x,y,z)$

$$g(s)= f(sx,sy,sz)= \int_{-\infty}^{+\infty} e^ {-t(t-sx)(t-sy)(t-sz)}\;dt$$

There is no odd terms in expansion because of $g(-s)=g(s)$

Proof: $$g(-s)=\int_{-\infty}^{+\infty} e^ {-t(t+sx)(t+sy)(t+sz)}\;dt$$

$$g(-s)=\int_{-\infty}^{+\infty} e^ {-t (-1)^3(-t-sx)(-t-sy)(-t-sz)}\;dt$$ $$g(-s)=\int_{-\infty}^{+\infty} e^ {-t (-1)^3(-t-sx)(-t-sy)(-t-sz)}\;dt$$ $t=-u$ $$g(-s)=-\int_{+\infty}^{-\infty} e^ {-u(u-sx)(u-sy)(u-sz)}\;du=\int_{-\infty}^{+\infty} e^ {-u(u-sx)(u-sy)(u-sz)}\;du=g(s)$$ Thus we can write series expansion of $g(s)$ $$g(s)= f(sx,sy,sz)= g(0)+\frac{g''(0)s^2}{2!} ((x+y+z)^2+a_1(xy+yz+xz))+\frac{g^{(4)}(0)s^4}{4!} ((x+y+z)^4+a_2(xy+yz+xz)^2+b_2(xy+yz+xz)(x+y+z)^2+c_2 xyz(x+y+z))+\frac{g^{(6)}(0)s^6}{6!} P_6(x+y+z,xy+yz+xz,xyz) +......$$

if $x=0$ and $y=0$ then

$$g(s)= f(sx,0,0)= \int_{-\infty}^{+\infty} e^ {-t^3(t-sx)}\;dt$$

$$g(1)= f(x,0,0)= \int_{-\infty}^{+\infty} e^ {-t^3(t-x)}\;dt=\int_{-\infty}^{+\infty} e^ {-t(t+x)^3}\;dt$$

$$g(1)= f(x,0,0)= f(-x,-x,-x)$$

$$ f(x,0,0)= \int_{-\infty}^{+\infty} e^ {-t^3(t-x)}\;dt= g(0)+\frac{g''(0)}{2!} x^2+\frac{g^{(4)}(0)}{4!} x^4+\frac{g^{(6)}(0)x^6}{6!}+......$$

$$g^{(n)}(0)=\int_{-\infty}^{+\infty} t^{3n}e^ {-t^4}\;dt$$ $$g(0)=\frac{Γ(1/4)}{2}$$ $$g'(0)=0$$ $$g''(0)=\frac{Γ(7/4)}{2}$$ $$g'''(0)=0$$ $$g^{(4)}(0)=\frac{Γ(13/4)}{2}$$ $$g^{(2n)}(0)=\frac{Γ((1+6n)/4)}{2}$$

$$f(-x,-x,-x)= g(0)+\frac{g''(0)}{2!} (9-3a_1)x^2+\frac{g^{(4)}(0)}{4!} (81+9a_2-27b_2+3c_2)x^4+\frac{g^{(6)}(0)}{6!} P_6(-3x,+3x^2,-x^3) +......$$

$9-3a_1=1$

$a_1=-\frac{8}{3}$

We can find $a_2,b_2,c_2$ via using $ f(x,y,z)= f(-x,y-x,z-x) $

$$(x+y+z)^4+a_2(xy+yz+xz)^2+b_2(xy+yz+xz)(x+y+z)^2+c_2 xyz(x+y+z)=(y+z-3x)^4+a_2(-x(y-x)-x(z-x)+(z-x)(y-x))^2+b_2(-x(y-x)-x(z-x)+(z-x)(y-x))(y+z-3x)^2-c_2x(y-x)(z-x)(y+z-3x)$$

I will write down when I find constants. More terms can be found via this method but maybe someone knows easier way.

EDIT : I have noticed that any term can be calculated via the method below:

$$g(s)= f(sx,sy,sz)= \int_{-\infty}^{+\infty} e^ {-t(t-sx)(t-sy)(t-sz)}\;dt=\int_{-\infty}^{+\infty} e^ {-t^4} e^ {st^3(x+y+z)} e^ {-s^2t^2(xy+xz+yz)} e^ {s^3t(xyz)} \;dt$$

$$g(s)= f(sx,sy,sz)= \int_{-\infty}^{+\infty} e^ {-t^4} (1+st^3(x+y+z)+\frac{s^2t^6(x+y+z)^2}{2!}+....) (1-s^2t^2(xy+xz+yz)+\frac{s^4t^4(xy+xz+yz)^2}{2!}+....) (1+s^3t(xyz)+\frac{s^6t^2(xyz)^2}{2!}+....) \;dt$$

$s^2$ terms can be written for $g(s)$ :

$$ \int_{-\infty}^{+\infty} \frac{s^2(t^3)^2(x+y+z)^2}{2!} e^ {-t^4} \;dt+ \int_{-\infty}^{+\infty} (-s^2)t^2(xy+xz+yz) e^ {-t^4} \;dt=s^2 \frac{(x+y+z)^2}{2!}\int_{-\infty}^{+\infty} t^6 e^ {-t^4} \;dt-s^2(xy+xz+yz) \int_{-\infty}^{+\infty} t^2 e^ {-t^4} \;dt= s^2 \frac{Γ(7/4)(x+y+z)^2}{2.2!} -s^2(xy+xz+yz)\frac{Γ(3/4)}{2}= \frac{s^2}{2!} \frac{Γ(7/4)}{2}[ (x+y+z)^2 -\frac{8}{3}(xy+xz+yz)] $$

$s^4$ terms can be written for $g(s)$ :

$$ \int_{-\infty}^{+\infty} \frac{s^4(t^3)^4(x+y+z)^4}{4!} e^ {-t^4} \;dt+ \int_{-\infty}^{+\infty} \frac{ (s^4)(-t^2)^2(xy+xz+yz)^2}{2!} e^ {-t^4} \;dt+ \int_{-\infty}^{+\infty} \frac{ s^2(t^3)^2(x+y+z)^2(s^2)(-t^2)(xy+xz+yz)}{2!} e^ {-t^4} \;dt+\int_{-\infty}^{+\infty} s^3(t)(xyz) s(t^3)(x+y+z) e^ {-t^4} \;dt=$$

$$ s^4[\frac{Γ(13/4)}{2.4!} (x+y+z)^4 + \frac{ Γ(5/4)}{2.2!}(xy+xz+yz)^2 -\frac{ Γ(9/4)}{2.2!}(x+y+z)^2 (xy+xz+yz) +\frac{ Γ(5/4)}{2}(xyz)(x+y+z)]=\frac{s^4}{4!} \frac{Γ(13/4)}{2} [(x+y+z)^4 + \frac{64}{15}(xy+xz+yz)^2-\frac{16}{3}(x+y+z)^2 (xy+xz+yz)+\frac{128}{15}(xyz)(x+y+z)]$$

Any $s^{2n}$ term can be found via this method but I have not got an answer for my iteration question above. I wonder if we can find a way for iteration solution or not. Thanks a lot for answers and comments

Answer 1

Just a try... you may define $$ g(s) =f(sx,sy,sz) $$ where you already know that $g(0)=2 \Gamma(5/4)= \Gamma(1/4)/2$. You can calculate at least the first terms of the series expansion of $g(s)$. For example, calculating the first one is easy once you know the value for $g(0)$. You should obtain: $$ g(s) = \Gamma(1/4)/2 +O(s^2) $$ because the additional term due to the first order expansion is odd in $t$. A bit more work gives $$ \Gamma(1/4)/2 + s^2 \Gamma(3/4)[3 (x^2+ y^2 + z^2) - 2 (xy+yz+zx)] /16 + O(s^4) $$ Note that you have to expand $g(s)$ to second order in $s$ to obtain the above expression... but the ``rest'' is guaranteed to be $O(s^4)$ and not $O(s^3)$ because the odd terms vanish in the integration over the real line. Maybe using the Faa di Bruno formula applied to the exponential function (https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) it is possible to figure out the general term of the integrand and then figure out the various polynomials in $(x,y,z)$ that appear as coefficients in the power series of $s$. But it seems difficult...

Answer 2

$\color{brown}{\textbf{Transformations.}}$

Let WLOG $\quad 0\le x\le y \le z,$ $$t=u+\dfrac{x+y+z}4.\tag1$$

Then $$256t(t-x)(t-y)(t-z)$$ $$ = (4u+x+y+z)(4u-3x+y+z)(4u+x-3y+z)(4u+x+y-3z)$$ $$=(16u^2+8\sigma_2-3\sigma_1^2)^2 - 8(\sigma_1^3-4\sigma_1\sigma_2+8\sigma_3)(4u+\sigma_1) - 4(\sigma_1^2-4\sigma_2)^2,$$ where $$\sigma_1=x+y+z,\quad\sigma_2=xy+yz+zx,\quad \sigma_3=xyz\tag2$$ (see also Wolfram Alpha test), $$\sigma_1^2-3\sigma_2 = x^2+y^2+z^2 - xy - yz -zx = \frac12((x-y)^2+(y-z)^2+(z-x)^2) \ge 0.$$

Let $$a=\sqrt{\frac{3\sigma_1^2-8\sigma_2}{32}},\quad b = \dfrac1{32}(\sigma_1^3-4\sigma_1\sigma_2+8\sigma_3),\quad c = \dfrac1{64}(\sigma_1^2-4\sigma_2)^2+b\sigma_1,\tag3$$ then $$t(t-x)(t-y)(t-z)=(u^2-2a^2)^2-4bu-c.$$

Therefore, $$I=\int\limits_{-\infty}^\infty e^{-t(t-x)(t-y)(t-z)}\,\mathrm dt =\int\limits_{-\infty}^\infty e^{-(u^2-2a^2)^2+4bu+c}\,\mathrm du = 2a\,e^{c-4a^4}\int\limits_{-\infty}^\infty e^{16a^4(v^2-v^4)} e^{8abv}\,\mathrm dv,$$ $$I= 2a\,e^{c-4a^4}G(16a^4,8ab),\quad\text{where}\quad G(p,q)=\int\limits_{-\infty}^\infty e^{p(v^2-v^4)} \cosh(qv)\,\mathrm dv.\tag4$$

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$\color{brown}{\textbf{Checking.}}$

$$\text{If}\quad \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac12 \\ 1 \\ \frac52 \end{pmatrix}, \quad\text{then}\quad \begin{pmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \end{pmatrix} =\begin{pmatrix} 4 \\ \frac{17}4 \\ \frac54 \end{pmatrix},\quad \begin{pmatrix} a \\ b \\ c \end{pmatrix} =\begin{pmatrix} \frac{\sqrt7}4 \\ \frac3{16} \\ \frac{49}{64} \end{pmatrix},$$ $$I(\,^1/_2,1,\,^5/_2)\approx 5.37992,$$ $$\frac{\sqrt7}2 G\left(\frac{49}{16},\frac{3\sqrt7}8\right) \approx 5.37992.$$

$$\text{If}\quad \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}, \quad\text{then}\quad \begin{pmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \end{pmatrix} =\begin{pmatrix} 8 \\ 17 \\ 10 \end{pmatrix},\quad \begin{pmatrix} a \\ b \\ c \end{pmatrix} =\begin{pmatrix} \frac{\sqrt7}2 \\ \frac32 \\ \frac{49}4 \end{pmatrix},$$ $$I(1,2,5)\approx 1.16955\cdot 10^{10},$$ $$\sqrt7 G\left(49,6\sqrt7\right)\approx 1.16955\cdot 10^{10}.$$

These tests confirm the executed transformations.

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$\color{brown}{\textbf{Calculations.}}$

Using the Maclaurin series in the form of

$$\cosh qv = \sum\limits_{n=0}^\infty \dfrac{(q v)^{2n}}{(2n)!},$$

easily to get $$G(p,q) = \sum\limits_{n=0}^\infty \dfrac{q^{2n}}{(2n)!} g_n(p), \quad\text{where}\quad g_n(p) = \int\limits_{-\infty}^\infty v^{2n} e^{p(v^2-v^4)}\,\mathrm dv.\tag5$$

Functions $g_n(p)$ are known for $n=0,$ $$g^\,_0(p)=\frac\pi{2\sqrt2}e^{\large^p\negthinspace/\negthinspace_8}\!\left( \operatorname{I_{\large -^1\!/\negthinspace_4}}\negthickspace\left(\frac p8\right) +\operatorname{I_{\large ^1\!/\negthinspace_4}}\negthickspace\left(\frac p8\right)\right) \tag{6.0},$$

$n=1,$ $$g^\,_1(p)=\frac\pi{8\sqrt2}e^{\large^p\negthinspace/\negthinspace_8}\!\left( p\operatorname{I_{\large -^1\!/\negthinspace_4}}\negthickspace\left(\frac p8\right) +(p+4)\operatorname{I_{\large ^1\!/\negthinspace_4}}\negthickspace\left(\frac p8\right) +p\operatorname{I_{\large ^3\!/\negthinspace_4}}\negthickspace\left(\frac p8\right) +p\operatorname{I_{\large ^5\!/\negthinspace_4}}\negthickspace\left(\frac p8\right)\right) \tag{6.1},$$

$n=2,$

$n=3,\dots$

where $I_\nu$ is Modified Bessel function of the first kind.

On the other hand, $$2g_{n+1}(p)-4g_{n+2}(p) = \dfrac1p\int\limits_{-\infty}^\infty v^{2n+1} \,\mathrm d e^{p (v^2-v^4)} = \dfrac1p v^{2n+1}e^{v^2-v^4}\Bigg|_{-\infty}^\infty-\dfrac{2n+1}p g_n(p)\\ = -\dfrac{2n+1}p g_n(p),$$ $$g_{n+2}(p) = \frac12 g_{n+1}(p) + \frac{2n+1}{4p}g_n(p).\tag{7.1}$$

Recurrency relations $(7.1)$ allows to calculate the arbitrary quantity of terms of formulas $(5).$

Alternative approach is transforming the series for $G(p,q).$

Really, for arbitrary $p$ \begin{align} &g_{n+2} = \frac12\left(\frac12 g_{n} + \frac{2n-1}{4p}g_{n-1}\right) + \frac{2n+1}{4p}g_n,\\ &g_{n+2}= \frac{p+2n+1}{4p}g_n + \frac{2n-1}{8p} g_{n-1},\tag{7.2}\\ &g_{n+2}= \frac{p+2n+1}{4p}\left(\frac12 g_{n-1} + \frac{2n-3}{4p}g_{n-2}\right) + \frac{2n-1}{8p}g_{n-1},\\ &g_{n+2}= \frac{p+4n}{8p}g_{n-1} + \frac{(2n-3)(p+2n+1)}{16p^2}g_{n-2}, \tag{7.3}\\ &g_{n+2}= \frac{p+4n}{8p}\left(\frac12 g_{n-2} + \frac{2n-5}{4p}g_{n-3}\right) + \frac{(2n-3)(p+2n+1)}{16p^2}g_{n-2},\\ &g_{n+2}= \frac{p^2+3(2n-1)p+(4n^2-4n-3)}{16p^2}g_{n-2} + \frac{(2n-5)(p+4n)}{32p^2}g_{n-3},\dots, \tag{7.4}\\ &g_2 = \frac12 g_1 + \frac1{4p}g_0,\tag{6.2}\\ &g_3 = \frac{p+3}{4p}g_1 + \frac1{8p} g_0,\tag{6.3}\\ &g_4 = \frac{p+8}{8p}g_1 + \frac{p+5}{16p^2}g_0,\tag{6.4}\\ &g_5 = \frac{p^2+15p-6}{16p^2}g_1 + \frac{p+12}{32p^2}g_{0}\dots,\tag{6.5}\\ &G(p,q) \approx \sum\limits_{n=0}^5 \dfrac{q^{2n}}{(2n)!} g_n(p) =g_0 + \dfrac{q^2}{2!}g_1 + \dfrac{q^4}{4!}\left(\frac12 g_1 + \frac1{4p}g_0\right)\\ &+ \dfrac{q^6}{6!}\left(\frac{p+3}{4p}g_1 + \frac1{8p} g_0\right) + \dfrac{q^8}{8!}\left(\frac{p+8}{8p}g_1 + \frac{p+5}{16p^2}g_0\right)\\ &+ \dfrac{q^{10}}{10!}\left(\frac{p^2+15p-6}{16p^2}g_1 + \frac{p+12}{32p^2}g_{0}\right)\\ &=\left(1 + \dfrac{60q^4+q^6}{6!\cdot8p} + \dfrac{q^8(p+5)}{8!\cdot16p^2} + \dfrac{q^{10}(p+12)}{10!\cdot32p^2}\right)g^\,_{0}\\ &+ \left(\dfrac{24q^2+q^4}{48} + \dfrac{q^6(p+3)}{6!\cdot4p}+ \dfrac{q^8(p+8)}{8!\cdot8p} + \dfrac{q^{10}(p^2+15p-6)}{10!\cdot16p^2}\right)g^\,_1. \end{align}

In the case $x=\frac12,\quad y=1,\quad z=\frac52,$ one can get $$\frac{\sqrt7}2 g_0\left(\frac{49}{16}\right)\approx4.4083178,$$ $$\frac{\sqrt7}2 g_1\left(\frac{49}{16}\right)\approx1.8650221,$$ $$I\approx 5.37994$$ (see Wolfram Alpha calculations).

At the same time, the sequence of the terms for $\cosh qv$ increases until $n=\frac q2,$ and good approximation requires the longer series.