Definite integrals of the type gaussian distribution times sqare root of x from 0 to inf give imaginary results whereas the functions are real valued in this interval.
int [sqrt(x)*exp(-(x-1)**2),x,0,inf] ? -0.07..i
int [x**(-1/2)*exp(-(x-1)**2),x,0,inf] ? 0.4..i
By the way I would love to know the correct symbolic integral but not only does wolfram alpha struggle but maxima and me as well.
Edit: I was not specific in that I would like to know the symbolic integral.
1 Like
jbaldwin
(Jim Baldwin)
March 13, 2015, 2:01pm
2
I don’t know the syntax for Wolfram Alpha but in Mathematica:
In[1]:= NIntegrate[Sqrt[x] Exp[-(x - 1)^2], {x, 0, Infinity}]
NIntegrate[x^(-1/2)*Exp[-(x - 1)^2], {x, 0, Infinity}]
Out[1]= 1.64362
Out[2]= 1.97373
Added comment: I just now entered your equation into Wolfram Alpha and it does recognize your formula as you intended. Sorry, don’t know what to say.
Thanks for your reply,
I know that you can compute the numeric integral which is real and positive. But this is sadly not satisfactory:
It highlights even more that there is an error
I wold like to know the symbolic integral
jbaldwin
(Jim Baldwin)
March 13, 2015, 2:33pm
4
It is curious. If I enter the indefinite integral into Wolfram Alpha
Integrate[Sqrt[x] Exp[-(x - 1)^2]]
I get the following output:
I am hoping for an enlightening answer from Daniel Lichtblau.
I get the same thing - the numerical integration is fine, but the symbolic integration seems to be broken.
iliang
(Ilian Gachevski)
March 13, 2015, 3:02pm
6
Why would you expect the indefinite integral to have a closed form?
In[1]:= int = Integrate[Sqrt[x] Exp[-(x - 1)^2], {x, 0, Infinity}]
Out[1]= (Gamma[3/4] Hypergeometric1F1[3/4, 1/2, 1] + 2 Gamma[5/4] Hypergeometric1F1[5/4, 3/2, 1])/(2 E)
In[2]:= N[int, 50]
Out[2]= 1.6436165407874952730553651165549411806023882736885
1 Like
iliang
(Ilian Gachevski)
March 13, 2015, 3:08pm
7
Oh, I see now. The incorrect result in Wolfram|Alpha is computed (by Integrate) in Mathematica version 9, while the correct approximate answers seem to use NIntegrate. The issue has been resolved in later versions of Mathematica, and hopefully Alpha will catch up soon.
2 Likes
jbaldwin
(Jim Baldwin)
March 13, 2015, 3:09pm
8
I had no such expectations. I was just hopeful/naïve that requesting the indefinite integral might provide more information. (And it did … in a way.)
jbaldwin
(Jim Baldwin)
March 13, 2015, 3:14pm
9
But I don’t think it’s completely fixed in Mathematica 10.0.2 (Windows):
In[33]:= \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(
\*FractionBox[\(1\), \(
\*SqrtBox[\(x\)] Exp[\((x - 1)\)^2]\)] \[DifferentialD]x\)\)
Out[33]= (I BesselK[1/4, 1/2])/Sqrt[2 E]
This does not give a real result.
danl
(Daniel Lichtblau)
March 13, 2015, 3:33pm
10
Some day…
In[1]:= i1 = Integrate[1/(Sqrt[x]*Exp[(x - 1)^2]), {x, 0, Infinity}]
N[i1]
Out[1]= (\[Pi] (BesselI[-(1/4), 1/2] + BesselI[1/4, 1/2]))/(2 Sqrt[E])
Out[2]= 1.97373215009
So it seems to be repaired in what will become the next released version.
1 Like
iliang
(Ilian Gachevski)
March 13, 2015, 3:33pm
11
The second example has also been fixed in the current development version and release candidates.
In[1]:= int2 = Integrate[(1/Sqrt[x]) Exp[-(x - 1)^2], {x, 0, Infinity}]
Out[1]= (\[Pi] (BesselI[-(1/4), 1/2] + BesselI[1/4, 1/2]))/(2 Sqrt[E])
In[2]:= N[int2, 50]
Out[2]= 1.9737321500898237789841487512373971273398502283395
2 Likes
OK thank you all - the conclusion is: The bug is known and fixed, just not in the version we were using.
For future reference the solution is:
int [ exp(-(x-m)**2/s)/sqrt(x) , {x,0,inf}]
= pi/2 * ( BesselI(1/4,m**2/(2*s))+BesselI(-1/4,m**2/(2*s)) ) * sqrt(m) * exp( - m**2/(2*s))