Herman Rubin wrote:
> 
> In article <1107559162.508814.123220@f14g2000cwb.googlegroups.com>,
> Vladimir Bondarenko <vb@cybertester.com> wrote:
> >isr...@math.ubc.ca (Robert Israel) writes on Feb 4, 2:39 pm
> 
> >With a serene look, I continue, What then about the following
> >integral
> 
> >int(erf(z)*exp(z^2-z^4), z= 0..infinity);
> 
> >?
> 
> >If you would not be able to get to the answer here, Is there a
> >person who can find the exact value of this integral?
> 
> I have not summed the series of expression, but it can be
> done in the following manner:
> 
> The not sufficiently well known expansion of exp(z^2)*erf(z)
> is
>         C*(z + 2*z^3/3 + 2^2*z^5/(3*5) + 2^3*z^7/(3*5*7) + ... )
> 
> This being a series of positive terms, it can be integrated
> term by term.  It is certainly expressible in terms of
> hypergeometric functions.

Yes, that's true:

# rewrite the question:
'2/sqrt(Pi)*Int(erf(z)/D(erf)(z)*exp(-z^4), z)';
  expand(%): combine(%,exp);

# prove identity
Sum((z*2^(1/2))^(2*i-1)/(2*i)!*i!*2^i,i = 1 .. infinity)/sqrt(2)='erf(z)/D(erf)(z)';
  value(%):; simplify(%) assuming 0<x: is(%);

# apply it and integrate term by term
2/sqrt(Pi)*Sum(Int(exp(-z^4)*(z*2^(1/2))^(2*i-1),z=0..infinity)/(2*i)!*i!*2^i,
  i = 1 .. infinity)/sqrt(2);
  value(%): simplify(%): expand(%);
  evalf(%,14);

  1/2+1/3*1/Pi^(1/2)*hypergeom([1, 1],[5/4, 7/4],1/4)+
  1/15*hypergeom([1, 3/2],[7/4, 9/4],1/4)

  .78489072897265

# check it numerical
Int(erf(z)*exp(z^2-z^4), z= 0..infinity): evalf(%,14);

  .78489072897265

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