# why integrating over error function gives nothing?

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??!

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

 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.

To work quit kernel and start again.

• So if I obtain an analytical solution I should change the upper limit to a finite number? – Wisdom Apr 17 at 10:59
• @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. – Mariusz Iwaniuk Apr 17 at 11:02