I have a complicated function as shown

It doesn't have a standard integration as far as I know.
Now, numerical integration of the left hand side function over a set of points gives the right hand side value (sorry about the missing integration sign in the image).
I know the value of parameter $X0$ as constant and $x$ is the independent variable. I wish to extract parameter alpha which I know is a real number.
How shall I do that in Mathematica? I have tried FindRoot method,didn't work.
FindRoot[(NIntegrate[(Abs[
0.5*Sqrt[Pi/(alpha)]*
Exp[-((Pi*x)^2/(alpha))]*(1 -
Erf[(-32.3077)*Sqrt[(alpha)] +
I*Pi*x/Sqrt[(alpha)]])])^2, {x, 0.00081846, 0.000818573,
0.000818686, 0.0008188, 0.000818913, 0.000819026, 0.000819139,
0.000819253, 0.000819366, 0.000819479, 0.000819592, 0.000819706,
0.000819819, 0.000819932, 0.000820045, 0.000820159, 0.000820272,
0.000820385, 0.000820498, 0.000820612, 0.000820725, 0.000820838,
0.000820951, 0.000821065, 0.000821178, 0.000821291, 0.000821404,
0.000821518, 0.000821631, 0.000821744, 0.000821857, 0.000821971,
0.000822084, 0.000822197, 0.00082231, 0.000822424, 0.000822537,
0.00082265, 0.000822763, 0.000822877, 0.00082299, 0.000823103,
0.000823216, 0.00082333, 0.000823443, 0.000823556, 0.000823669,
0.000823783, 0.000823896, 0.000824009, 0.000824122, 0.000824236,
0.000824349, 0.000824462, 0.000824575, 0.000824689, 0.000824802,
0.000824915, 0.000825028}]) == 0.0139828, {alpha, 0.001}]
$X0$ for a particular case is -32.3077, becomes $0$ for particular $x$ and varies upto +30.3699 for different cases.
'alpha' for that particular X0 should come around 0.03 and thus alpha varies with $X0$

NumericQof this Q&A: What are the most common pitfalls awaiting new users? – Jens Sep 03 '15 at 17:05f[alpha_?NumericQ, x_?NumericQ] := Abs[0.5*Sqrt[Pi/alpha]*Exp[-(Pi*x)^2/alpha]*(1 - Erf[-32.3077*Sqrt[alpha] + I*Pi*x/Sqrt[alpha]])]^2followed byFindRoot[NIntegrate[f[alpha, x], ...in my answer, but the error message persisted. Of course, even the code in the question gives the correct answer despite the error message. – bbgodfrey Sep 03 '15 at 17:21FindRoot[eq[alpha],{alpha,0.001}]works if I defineeq[alpha_?NumericQ]:=NIntegrate[Abs[0.5 Sqrt[\[Pi]/alpha] Exp[-((\[Pi] x)^2/alpha)] (1-Erf[-32.3077 Sqrt[alpha]+(I \[Pi] x)/Sqrt[alpha]])]^2,{x,..]-0.0139828... but I upvoted your answer, too. – Jens Sep 03 '15 at 17:25