# Closed form solution for this integral?

I'm hitting a road block in finding an expression (closed form preferably) for the following integral:

$$\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx$$

where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, while $c$ is a positive or negative odd multiple of $0.5$; $u$ is a (positive) parameter.

Things I have considered or tried:

• look up in tables (Gradshsteyn and Ryzhik): there are very few explicit results for integrals involving $\exp(-a x^3)$ (or for the other factors after transforming via $y=x^3$). Also, tabulated results involving $\exp(-a x^p)$ for more general $p$ do not include the other factors $x^b (1-x/u)^c$. One exception is (3.478.3): $$\int^{u}_0 x^b (1-ux)^c \exp(-a x^3) dx,$$ but the limits of integration do not match with my case;

• there is a closed form solution (3.478.1) for the simpler integral $$\int^{+\infty}_0 x^{d-1} \exp(-a x^3) dx = \frac{a^{-d/3}}{3} \Gamma(d/3).$$ (NB: there is also an expression for the indefinite integral.) A binomial expansion of $[1-(x/u)]^n$ for integer $n$ would produce a solution in series form. However, in my case, the exponents $b$ and $c$ are strictly half-integer. For the same reason, integration by parts does not lead to a simpler integral without the factor $[1-(x/u)]^c$;

• Wolfram Math online did not produce a result;

• the integral is an intermediate step in a longer analysis, so numerical solution (with given values for the parameter) is not practical.

Grateful for any pointers or solution.

• There is a solution which involves hypergeometric function : it is a monster ! – Claude Leibovici Sep 1 '18 at 13:30
• I don't mind monsters :) Please list the solution if you can.Thanks. – Lucozade Sep 1 '18 at 13:37
• Maybe for half-integer values of b and c, the hypergeometrics may simplify and the monster becomes a puppy? Even if a solution for only integer values of b and c is sufficiently simple and amenable, it might be possible to find or approximate the solution for half-integer b and c by interpolation, if smooth enough? – Lucozade Sep 1 '18 at 15:03
• Do you enjoy the puppy ? – Claude Leibovici Sep 2 '18 at 6:45
• Let's agree on calling it a midget :). Few of us would have the stamina to get to this by hand IMO. Thanks to technology for allowing us to get on with the more interesting aspects of maths. – Lucozade Sep 2 '18 at 10:57

Long Comment:

Since it was explicitly asked for, Mathematica 11.3 when provided with

$$\text{Integrate}\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>1\land a>0\land u>0\right]$$

gives up the monster

$$\text{ConditionalExpression}\left[\frac{1}{6} \left(2 a^{\frac{1}{3} (-b-c)} \left(\frac{\left(-\frac{1}{u}\right)^c \Gamma \left(\frac{1}{3} (b+c+1)\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},-\frac{c}{3};\frac{1}{3},\frac{2}{3},-\frac{b}{3}-\frac{c}{3}+\frac{2}{3};-a u^3\right)}{\sqrt[3]{a}}-c u \left(-\frac{1}{u}\right)^c \Gamma \left(\frac{b+c}{3}\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},1-\frac{c}{3};\frac{2}{3},\frac{4}{3},-\frac{b}{3}-\frac{c}{3}+1;-a u^3\right)+\frac{\pi \sqrt[3]{a} (-1)^c 3^{\frac{3}{2}-c} u^{2-c} \Gamma (c+1) \Gamma \left(\frac{1}{3} (b+c-1)\right) \, _3F_3\left(\frac{2}{3}-\frac{c}{3},1-\frac{c}{3},\frac{4}{3}-\frac{c}{3};\frac{4}{3},\frac{5}{3},-\frac{b}{3}-\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\Gamma \left(\frac{c-1}{3}\right) \Gamma \left(\frac{c}{3}\right) \Gamma \left(\frac{c+1}{3}\right)}\right)+\frac{3^{-b-c-\frac{1}{2}} u^{b+1} \Gamma (c+1) \left(\frac{4 \pi ^2 \Gamma (b+1)}{\Gamma \left(\frac{1}{3} (b+c+2)\right) \Gamma \left(\frac{1}{3} (b+c+3)\right) \Gamma \left(\frac{1}{3} (b+c+4)\right)}+\frac{(-1)^c \Gamma \left(-\frac{b}{3}-\frac{c}{3}\right) \Gamma \left(\frac{1}{3} (-b-c-1)\right) \Gamma \left(\frac{1}{3} (-b-c+1)\right)}{\Gamma (-b)}\right) \, _3F_3\left(\frac{b}{3}+\frac{1}{3},\frac{b}{3}+\frac{2}{3},\frac{b}{3}+1;\frac{b}{3}+\frac{c}{3}+\frac{2}{3},\frac{b}{3}+\frac{c}{3}+1,\frac{b}{3}+\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\pi }\right),\Re(c)>-1\right]$$,

including the condition that the only half integer negative value $c$ can take is $-\frac{1}{2}$ given the above assumptions.

Further Thoughts...

By differentiating $e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c$ w.r.t $x$ and then integrating between $0$ and $\infty$ we obtain the integral identity

$$\frac{c}{u} \int_0^\infty e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^{c-1} \, dx =b \int_0^\infty e^{-a x^3} x^{b-1} \left(1-\frac{x}{u}\right)^c \, dx-3 a \int_0^\infty e^{-a x^3} x^{b+2} \left(1-\frac{x}{u}\right)^c \, dx$$

since $\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c \right]_0^\infty=0$

Mathematica again gives the condition that $c>-1$ with the following assumptions

$$\text{Integrate}\left[e^{-a x^3} x^{b} \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>0\land a>0\land u>0\right]$$.

which implies $c$ must be positive for your integral to exist.

• Thanks, also for the additional thought, but I cannot quite follow your final conclusion. The constant $c$ is in fact $n-(3/2)$, with $n$ a strictly positive integer. From my (physical) application from which this originates, $n$ is a degrees-of-freedom in a probability distribution that should be able to be $n=1$ as well (and indeed $c=-1/2$ is then the only negative value possible). Anyway, my case of interest is $n=2$. But can you clarify further how you concluded that $c$ cannot be negative? – Lucozade Sep 2 '18 at 10:50
• According to Mathematica 11.3 given the other assumptions then $c>-1$ must hold for your inegral to be defined. All I am saying in the case of $c=-1/2$ the right hand side of the above identity is defined and finite and the left hand side of the identity is undefined. This means the identity does not hold in the case of $c=-1/2$, which I am interpreting this as a contradiction in our assumptions that disallows $c=-1/2$ and other negative values of $c$ between $0$ and $-1$. $c$ must always be positive to avoid this contradiction. – James Arathoon Sep 2 '18 at 13:27

An integral of a product of two Meijer G-functions of rational powers of the argument gives a G-function: $$\int_0^\infty x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx = \\ \int_0^u x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx + (-1)^c \int_u^\infty x^b \left( \frac x u - 1 \right)^c e^{-a x^3} dx = \\ \Gamma(c + 1) \int_0^\infty x^b G_{1, 1}^{1, 0} \left( \frac x u \middle| {c + 1 \atop 0} \right) G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx + \\ (-1)^c \Gamma(c + 1) \int_0^\infty x^b G_{1, 1}^{0, 1} \left( \frac x u \middle| {c + 1 \atop 0} \right) G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx = \\ 3^{-c - 1} \Gamma(c + 1) u^{b + 1} G_{3, 4}^{1, 3} \left(a u^3 \middle| { \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop 0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right) + \\ 3^{-c - 1} (-1)^c \Gamma(c + 1) u^{b + 1} G_{3, 4}^{4, 0} \left(a u^3 \middle| { \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop 0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right).$$ When $b$ and $c$ are half-integers, one of the three numbers $(-b - c - 1)/3, (-b - c)/3, (-b - c + 1)/3$ is an integer. Therefore, the ratio of the gamma functions in the $G_{3, 4}^{4, 0}$ term always has an infinite sequence of double poles inside the integration contour, and the G-function cannot be converted to a sum of hypergeometric functions. The sum will be valid only as a limit, which will give derivatives of ${_pF_q}$ wrt parameters.

• Great insight, and also useful to see the split as $\int^u_0 + \int^\infty_u$. In fact, I've just realized I'm also going to need the integral for $\int^\infty_v$ with $v \not = u$. Is there an expression for that result (in terms of Meijer G or otherwise) as well? – Lucozade Sep 2 '18 at 15:14
• Restricting the domain to $[v, \infty)$ is the same as multiplying the integrand by $H(x - v)$. Now you have a product of three G-functions. It's likely that you'll need more general functions for a closed form expression. – Maxim Sep 2 '18 at 18:12