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I am trying to evaluate the following expression numerically $$\frac{d^2}{dt^2}e^{-2t^2}\int_0^\infty\frac{\xi/\sqrt{2}}{\xi^{3/2}}e^{(-\xi^2/2-2\xi t))}$$

My code is as follows

f[t_]:=Exp[-2*t^2]*NIntegrate[Erf[\[Xi]/Sqrt[2]]/\[Xi]^(3/2)*Exp[-(\[Xi]^2/2)-2*\[Xi]*t],{\[Xi],0,Infinity}]
Der[t_] := f''[t]

But when I evaluate, say by entering

Der[5]

I am getting an error that the integrand has evaluated to non-numerical values for all sampling points in the region. Now I know that this is a common error, and other threads have revealed that this is happening because Mathematica is not recognizing something in my expression (likely $\xi$) as a numerical variable. However, I am not sure how I would go about fixing this. I have tried putting in ?NumericQ beside all instances of $\xi$ but that didn't seem to work.

How would I go about numerically evaluating this expression (and ideally, plotting it as a function of t) without the NIntegrate::inumr error?

Thank you in advance!

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  • $\begingroup$ Why not simplify the integrand by dividing out $\xi$ and bringing out constants? And you need a $d \xi$ within the integral. $\endgroup$ Feb 9 '17 at 20:59
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Needs["NumericalCalculus`"]

f[t_?NumericQ] := 
 Exp[-2*t^2]*
  NIntegrate[
   Erf[ξ/Sqrt[2]]/ξ^(3/2)*
    Exp[-(ξ^2/2) - 2*ξ*t], {ξ, 0, Infinity}]

Use ND

Der[t_?NumericQ] := ND[f[x], {x, 2}, t]

Der[5]

(*  3.41091*10^-20  *)
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  • $\begingroup$ @PratikSamant - I just copied and pasted and evaluated without any problems. You apparently have problems with prior definitions. Either clear all variables or start with a fresh kernel. $\endgroup$
    – Bob Hanlon
    Feb 13 '17 at 16:51
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Notice: my formula contains a sign error (-t instaed of +t). But the derivation is still valid. Only the final numeric value must be taken at t = -5.

The integral can be written as

f[t_] := 
 NIntegrate[\[Xi]^(-3/2 )
    Erf[\[Xi]/Sqrt[2]] Exp[-(1/2) (\[Xi] - 2 t)^2], {\[Xi], 
   0, \[Infinity]}]

Let us form the second derivative with respect to t under the integral:

D[Exp[-1/2 (\[Xi] - 2 t)^2], {t, 2}]

(* Out[22]= -4 E^(-(1/2) (-2 t + \[Xi])^2) + 
 4 E^(-(1/2) (-2 t + \[Xi])^2) (-2 t + \[Xi])^2 *)

and call f''[t] = f2[t] which is

f2[t_] := 
 NIntegrate[\[Xi]^(-3/2 )
    Erf[\[Xi]/Sqrt[2]] 4 E^(-(1/
     2) (-2 t + \[Xi])^2) (-1 + (-2 t + \[Xi])^2), {\[Xi], 
   0, \[Infinity]}]

The value requested is then

f2[5]

(* Out[24]= 0.0129202 *)

This is very different from Bob Hanlon's result which is, in fact

f2[-5]

(* Out[30]= 3.43309*10^-20 *)
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