Evaluating this generalised integral

Question

I have the following integral

$$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\exp \left(a u^2+b v^2+c u v\right) \; dvdu,$$

which returns the following solution:

$$\frac{2 \pi }{\sqrt{4 a b-c^2}}.$$

I would like to generalise the integral to define the following function:

$$f[x, y] := \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\exp \left(a u^2+b v^2+c u v\right) u^x v^y\; dvdu,$$

which Mathematica generates a solution to in terms of Hypergeometric functions but crucially it is a function of x and y. I would like to use this function for a range of different x and y, i.e.

Table[f[x,y], {x,0,10}, {y,0,10}] 

As a cross check, I would like to get a consistent result when $x=y=0$. The issue is that when I try to evaluate f[0,0], I get Indeterminate due to a 0 ComplexInfinity. I.e. this doesn't reproduce the result at the top. A same issue occurs for f[0,2].

Is there a fix to this? Can I define cases where it breaks?

Answer 1

modified answer (11.04.2021)

(Thanks @DanielLichtblau for his helpful comment)

Here my ideas to make it work:

Mathematica evaluates the first integral to

Integrate[Exp[a u^2 + b v^2 +c u v], {v, -∞,∞}, {u, -∞, ∞},Assumptions -> {Element[{a, b, c}, Reals] ]
(*ConditionalExpression[(2 \[Pi])/(Sqrt[-b]Sqrt[-4 a + c^2/b]),
4 a b^2 < b c^2] *)

result is correct but the condition is wrong!

stepwise integration

Assuming a!=0 the integral

Integrate[Exp[a (u + c/(2 a) v)^2 ]Exp[a (b/a - (c/(2 a))^2) v ^2 ] , {u, -\[Infinity], \[Infinity]}, {v, -\[Infinity], \[Infinity]}] is transformed to

Integrate[Integrate[Exp[a (u + c/(2 a)v)^2], {u, -\[Infinity], \[Infinity]} ] Exp[a (b/a - (c/(2 a))^2) v^2], {v, -\[Infinity], \[Infinity]}]`
(* ConditionalExpression[(2 \[Pi])/(Sqrt[-a]Sqrt[-4 b + c^2/a]), Re[a] < 0 && Re[c^2/a] > 4 Re[b]]*)

with correct conditions a<0 && c^2< 4 a b

Take these condition for the general case

int = Function[{x, y}, 
Evaluate@Integrate[Exp[a u^2 + b v^2 + c u v] u^x v^y, {v, -∞, ∞}, {u, -∞, ∞}, 
Assumptions -> {Element[{a, b, c}, Reals], a<0,c^2< 4 a b}]]

These evaluate the mentioned cases as expected

Limit[int[x, y], {x -> 0, y -> 0}]
(*ConditionalExpression[(2 \[Pi])/(Sqrt[-a] Sqrt[-b] Sqrt[4 - c^2/(a b)]), c <0]*)     

int[0,2]
(*ConditionalExpression[(a \[Pi])/(2 (-a)^(3/2) Sqrt[-b] b (1 - c^2/(4 ab))^(3/2)), c < 0]*)