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I try to integrate over an error function and evaluate the following integral

Integrate[ 
 Erf[(cep (re + rp) + 2 \[Alpha]2)/(Sqrt[2] Sqrt[cep])], {rp, 
  0, \[Infinity]}, 
 Assumptions -> 
  re \[Element] Reals && rp \[Element] Reals && cp \[Element] Reals &&
    re > 0]

but it gives nothing. why??!

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  • 1
    $\begingroup$ Try: Integrate[Erf[x], {x, 0, Infinity}] To see why this doesn't converge, plot it. $\endgroup$
    – bill s
    Apr 17, 2021 at 10:32
  • $\begingroup$ Thanks but it does not give "Integral of Erf[x] does not converge on {0,[Infinity]}" in my case, it just returns the initial expression! $\endgroup$
    – Wisdom
    Apr 17, 2021 at 10:35
  • $\begingroup$ Try: NIntegrate[Erf[x], {x, 0, Infinity}] $\endgroup$ Apr 17, 2021 at 10:36
  • $\begingroup$ @MariuszIwaniuk Thanks, but I need an analytical answer! $\endgroup$
    – Wisdom
    Apr 17, 2021 at 10:37
  • $\begingroup$ $\text{erf}(x)\approx 1$ and "does not give not coverage........".Try: Integrate[1, {x, 0, Infinity}] ? Analytical answer is: $$\infty$$ $\endgroup$ Apr 17, 2021 at 10:40

1 Answer 1

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 f = Integrate[Erf[(cep (re + rp) + 2 \[Alpha]2)/(Sqrt[2] Sqrt[cep])], rp]
 
  (*(E^(-((cep (re + rp) + 2 \[Alpha]2)^2/(2 cep))) Sqrt[
 2/\[Pi]])/Sqrt[cep] + (re + rp + (2 \[Alpha]2)/cep) Erf[(
 cep (re + rp) + 2 \[Alpha]2)/(Sqrt[2] Sqrt[cep])]*)

And using fundamental theorem of calculus:

 Limit[f, rp -> Infinity, Assumptions -> {cep > 0, re > 0, \[Alpha]2 > 0}] - 
 Limit[f, rp -> 0, Assumptions -> {cep > 0, re > 0, \[Alpha]2 > 0}]

 (*\[Infinity]*)

On MMA 12.2.0 works fine.

enter image description here enter image description here

To work quit kernel and start again.

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  • $\begingroup$ So if I obtain an analytical solution I should change the upper limit to a finite number? $\endgroup$
    – Wisdom
    Apr 17, 2021 at 10:59
  • $\begingroup$ @Wisdom $\infty$ is infinty number.You can try: Integrate[ Erf[(cep (re + rp) + 2 \[Alpha]2)/(Sqrt[2] Sqrt[cep])], {rp, 0, INF}, Assumptions -> {cep > 0, re > 0, \[Alpha]2 > 0, INF > 0}] where INF is a finite number. $\endgroup$ Apr 17, 2021 at 11:02

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