Fdtxjr

I'd like to calculate the following integral:

$$int^infty_0 mathrmerfleft(fracalphasqrt1+x - fracsqrt1+xbetaright) expleft(-fracxgammaright), dx,$$

where $beta > 0$, $gamma > 0$ and $alpha in mathbbR$.

I've tried a few approaches, but with no success.

The form is similar to Equation 12 on page 177 of Erdelyi's Tables of Integral Transforms (Vol. 1):

$$int^infty_0 mathrmerfleft(fracalphasqrtt - fracsqrttbetaright) expleft(-fractgammaright), dt$$

but the change of variables requires a change in limits.

Any advice would be greatly appreciated!

If you let: $$t = sqrt1 + x,$$

and change limits appropriately, Mathematica evaluates it as:

$$gamma texterfleft(alpha -frac1beta right) ++fracalpha sqrtbeta ^2+gamma
.$$

Not as neat as I'd like it, but there you go.

I'd start with an integration by parts, which should give you
$$ texterf(alpha - 1/beta) gamma - fracgammabeta sqrtpi
exp(2alpha/beta - 1/beta^2) int_0^infty expleft(-(1/beta^2+1/gamma) x - alpha^2/(1+x) right) fracalpha beta + 1 + x(1+x)^3/2 dx$$
Now according to Maple, for $A > 0$
$$ int_0^infty exp(-A x - alpha^2/(1+x)) dfracdx(1+x)^1/2 =
1/2,frac rm e^A-2,alpha,sqrt A left(
rm erfleft(-sqrt A+alpharight)+1+ left( 1-
rm erfleft(sqrt A+alpharight) right) rm e^4,
alpha,sqrt A right) sqrt pi sqrt A$$
However, I wasn't able to get a closed form for the integral with $(1+x)^3/2$ instead of $(1+x)^1/2$.

Let us define:
begineqnarray
J(alpha)&:=& intlimits_0^infty erfleft( fracalphasqrt1+x - fracsqrt1+xbeta right) e^-fracxgamma dx\
&underbrace=_t=sqrt1+x& intlimits_1^infty erfleft( fracalphat - fractbeta right) e^-fract^2-1gamma 2 t dt
endeqnarray
Now we differentiate with respect to $alpha$. We have:
begineqnarray
&&fracpartialpartial alpha J(alpha)=\
&& intlimits_1^infty frac4sqrtpi e^-(fracalphat - fractbeta)^2 e^-fract^2-1gamma dt =\
&& frac4sqrtpi e^2 fracalphabeta+frac1gamma intlimits_1^infty e^-left( fracalpha^2t^2 + (frac1beta^2+frac1gamma) t^2right) dt underbrace=_beginarrayrrr A&:=&alpha^2\B&:=&1/beta^2+1/gammaendarray\
&&
frac4sqrtpi e^2 fracalphabeta+frac1gamma cdot
left.fracsqrtpi left(e^-2 sqrtA sqrtB left(1-texterfleft(fracsqrtAt-sqrtB tright)right)+e^2 sqrtA sqrtB left(texterfleft(fracsqrtAt+sqrtB tright)-1right)right)4
sqrtB right|_1^infty =\
%
&&= frace^frac1gsqrtfrac1b^2+frac1g left(e^a left(-Delta_-right) texterfleft(a-sqrtfrac1b^2+frac1gright)-e^a left(Delta_+right)
texterfleft(a+sqrtfrac1b^2+frac1gright)+e^a left(-Delta_-right)+e^a left(Delta_+right)right)
endeqnarray
where $delta:=sqrtfrac1b^2+frac1g$ and $Delta_pm:=2 delta pm frac2b
$.

Now all we need to do is to integrate over $alpha$. A glimpse at the bottom line above suffices to realize that the anti-derivative with respect to $alpha$ can be easily found using integration by parts. In addition it is also easy to see that $J(infty) = gamma$. Therefore the final result reads:
begineqnarray
gamma- J(alpha) =
%
gammacdot (1-erf(alpha-frac1beta)) +
fracgamma e^frac1gamma e^-alpha Delta_-2 delta
left(frac1beta - fracDelta_-2 e^4 alpha delta +deltaright) +
fracgamma e^frac1gamma4 delta
left(
Delta_+ e^-alpha Delta_- erf(a-delta) + e^alpha Delta_+ Delta_- erf(a+delta)
right)
endeqnarray

In[441]:= b, g = RandomReal[0, 2, 2, WorkingPrecision -> 50];
dd = Sqrt[1/b^2 + 1/g];
Dm, Dp = 2 -1/b + dd, 1/b + dd;
a = Range[0, 2, 0.001];
f = Interpolation[
 Transpose[a, 
 NIntegrate[
 Erf[#/Sqrt[1 + x] - Sqrt[1 + x]/b] Exp[-x/g], x, 0, 
 Infinity] & /@ a]]



In[446]:= a = RandomReal[0, 2, WorkingPrecision -> 50];
f'[a]
E^(1/g)/Sqrt[
 1/b^2 + 1/
 g] (E^(a (2 /b - 2 Sqrt[1/b^2 + 1/g])) + E^(
 a (2 /b + 2 Sqrt[1/b^2 + 1/g])) + 
 E^(a (2 /b - 2 Sqrt[1/b^2 + 1/g])) Erf[a - Sqrt[1/b^2 + 1/g]] - 
 E^(a (2 /b + 2 Sqrt[1/b^2 + 1/g])) Erf[a + Sqrt[1/b^2 + 1/g]])

g - f[a]


 g (1 - Erf[a - 1/b]) + (g E^(1/g) E^(-a Dm))/(
 2 dd) (1/b + E^(4 a Sqrt[1/b^2 + 1/g]) (-Dm/2) + dd) + (g E^(1/g))/(
 4 dd) ((Dp) E^(-a Dm) Erf[a - dd] + E^(a Dp) (Dm) Erf[a + dd])


Out[447]= 0.648692

Out[448]= 0.648691540832040074541903643440002542062490436

Out[449]= 0.681556

Out[450]= 0.681555757815934967025234698639289684849178686