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Definite integral of gaussian times sqare root of x gives wrong result

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.
POSTED BY: Wolfgang Brehm
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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))
POSTED BY: Wolfgang Brehm
Jim Baldwin
Jim Baldwin, Retired
Posted 11 years ago

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.)
POSTED BY: Jim Baldwin
POSTED BY: Ilian Gachevski
Jim Baldwin
Jim Baldwin, Retired
Posted 11 years ago

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.
POSTED BY: Jim Baldwin
POSTED BY: Daniel Lichtblau

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
POSTED BY: Ilian Gachevski

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
POSTED BY: Ilian Gachevski

I get the same thing - the numerical integration is fine, but the symbolic integration seems to be broken.
POSTED BY: Wolfgang Brehm
Jim Baldwin
Jim Baldwin, Retired
Posted 11 years ago

It is curious. If I enter the indefinite integral into Wolfram Alpha

Integrate[Sqrt[x] Exp[-(x - 1)^2]]

I get the following output:

Integral

I am hoping for an enlightening answer from Daniel Lichtblau.
POSTED BY: Jim Baldwin

Thanks for your reply,

I know that you can compute the numeric integral which is real and positive. But this is sadly not satisfactory:

  1. It highlights even more that there is an error
  2. I wold like to know the symbolic integral
POSTED BY: Wolfgang Brehm
Jim Baldwin
Jim Baldwin, Retired
Posted 11 years ago

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.
POSTED BY: Jim Baldwin
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