Improper mutiple integral

Question

I would like to solve the following integral:

Integrate[y Erf[a Sqrt[x^2 + y^2]]/Sqrt[x^2 + y^2]
            Exp[-b ((x - c)^2 + y^2)], {x, -∞, ∞}, 
         {y, 0, ∞}, Assumptions -> {a > 0, b > 0, c > 0}]

If I add such input in Mathematica, it takes ages to solve it and in the end it could not. The solution is obviously a function of a, b, and c. Any ideas, tricks how to tackle this problem?

Answer 1

Here's a way...

Convert to polar coordinates

Integrate[
 r (r Sin[t]) Erf[a r]/r Exp[-b ((r Cos[t] - c)^2 + (r Sin[t])^2)],
 {r, 0, Infinity}, {t, 0, Pi}, Assumptions -> {a > 0, b > 0, c > 0}]

But that doesn't work, so...

Do the t integral first

tInt = Integrate[r r Sin[t] Erf[a r]/r Exp[-b ((r Cos[t] - c)^2 + (r Sin[t])^2)],
 {t, 0, Pi}]
(*
  (E^(-b (c + r)^2) (-1 + E^(4 b c r)) Erf[a r])/(2 b c)
*)

The r integral won't evaluate still, so...

Try differentiating with respect to the parameter a

This trick I learned in Salomon Bochner's Fourier Integrals. In this case, I should note that tInt equals zero at a -> 0. The value of daInt is the partial derivative with respect to a of the integral we wish to find. Later we will antidifferentiate with respect to a (using a substitution). Normally, by the Fundamental Theorem of Calculus, I would subtract the value at a == 0, but since the value is zero, you won't see that step below.

daInt = Integrate[D[tInt, a], {r, 0, Infinity}, Assumptions -> {a > 0, b > 0, c > 0}]
(*
   E^(-((a^2 b c^2)/(a^2 + b)))/(a^2 + b)^(3/2)
*)

But this won't integrate with respect to a, so...

Try another substitution, u = a^2 + b

The Jacobian factor is 1/(2 Sqrt[u - b]).

aInt = Integrate[1/(2 Sqrt[u - b]) daInt /. a^2 -> u - b,
          u, Assumptions -> {a > 0, b > 0, c > 0}];
Simplify[aInt /. u -> a^2 + b, a > 0 && b > 0 && c > 0]
(* (Sqrt[π] Erf[a Sqrt[b/(a^2 + b)] c])/(2 b^(3/2) c) *)

To me, getting an answer at this step was a miracle....so

Check numerically

With[{a = 2, b = 1, c = 1},
 {NIntegrate[
   y Erf[a Sqrt[x^2 + y^2]] / Sqrt[x^2 + y^2] Exp[-b ((x - c)^2 + y^2)],
   {x, -∞, ∞}, {y, 0, ∞}],
  (Sqrt[π] Erf[a Sqrt[b/(a^2 + b)] c])/(2 b^(3/2) c) // N}
 ]
(* {0.70375, 0.70375} *)

You can verify for other values as well. Seems like I didn't make a mistake. So the answer is $$\frac{\sqrt{\pi }\ \text{erf}\left(a c \sqrt{\frac{b}{a^2+b}}\right)}{2 b^{3/2} c}$$