By changing variables, the definite integral can be rewritten as a constant times the following definite integral (where a
and d
are not the same as in the original integral):
int = Inactive[Integrate][Erfc[x + d] Exp[-a x^2], {x, -Infinity, Infinity}];
Let's series expand the integrand, and then integrate:
integrand = Erfc[x + d] Exp[-a x^2];
series = Series[integrand, {d, 0, 20}];
terms = Integrate[series[[3]], {x, -Infinity, Infinity}, Assumptions->a>0];
terms //TeXForm
$\left\{\frac{\sqrt{\pi }}{\sqrt{a}},-\frac{2}{\sqrt{a+1}},0,\frac{2 a}{3 (a+1)^{3/2}},0,-\frac{a^2}{5
(a+1)^{5/2}},0,\frac{a^3}{21 (a+1)^{7/2}},0,-\frac{a^4}{108 (a+1)^{9/2}},0,\frac{a^5}{660
(a+1)^{11/2}},0,-\frac{a^6}{4680 (a+1)^{13/2}},0,\frac{a^7}{37800 (a+1)^{15/2}},0,-\frac{a^8}{342720
(a+1)^{17/2}},0,\frac{a^9}{3447360 (a+1)^{19/2}},0\right\}$
Dropping the constant term, and noticing that the series alternates, we can try using FindGeneratingFunction
as follows:
gf = FindGeneratingFunction[terms[[2 ;; -1 ;; 2]] /. a->λ/(1-λ) //Simplify, d];
Sqrt[d] gf /. {λ -> a/(1+a), d->d^2} //Simplify
-((Sqrt[π] Erf[Sqrt[a/(1 + a)] Sqrt[d^2]])/Sqrt[a])
Hence, the definite integral might be:
dint[a_, d_] = Sqrt[π/a](1 - Erf[Sqrt[a/(1+a)] d])
Sqrt[1/a] Sqrt[π] (1 - Erf[Sqrt[a/(1 + a)] d])
For comparison, here's the numerical integration version:
nint[a_, d_] := NIntegrate[Erfc[x + d] Exp[-a x^2], {x, -Infinity, Infinity}]
Spot checks:
dint[Pi, 1.1]
nint[Pi, 1.1]
dint[2, -1.]
nint[2, -1.]
0.175458
0.175458
2.19554
2.19554
So, it looks like FindGeneratingFunction
successfully converted the series into a function. I will let you perform the shift + rescaling necessary to convert the above integral into the form you desire.
Integrate[ Erfc[c*x + d]*Exp[-a*x^2]*Exp[b*x], {x, -Infinity, Infinity}, Assumptions -> c > 0 && a > 0 && d == 0 && b == 0]
, which yieldsSqrt[π]/Sqrt[a]
. $\endgroup$ – bbgodfrey Apr 12 '18 at 1:32