Message Boards

WOLFRAM COMMUNITY

13616 Views

11 Replies

7 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Mathematics Wolfram|Alpha Algebra Symbolic Computations

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

Wolfgang Brehm

Wolfgang Brehm, Universität Konstanz

Posted 9 years ago

POSTED BY: Wolfgang Brehm

11 Replies

Sort By:

Wolfgang Brehm

Wolfgang Brehm, Universität Konstanz

Posted 9 years ago

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/(2s))+BesselI(-1/4,m*2/(2s)) ) * sqrt(m) * exp( - m*2/(2s))

POSTED BY: Wolfgang Brehm

Jim Baldwin

Jim Baldwin, Retired

Posted 9 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

Ilian Gachevski

Ilian Gachevski, Wolfram Research

Posted 9 years ago

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.

POSTED BY: Ilian Gachevski

Jim Baldwin

Jim Baldwin, Retired

Posted 9 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.

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

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 9 years ago

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.

POSTED BY: Daniel Lichtblau

Ilian Gachevski

Ilian Gachevski, Wolfram Research

Posted 9 years ago

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

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

Ilian Gachevski

Ilian Gachevski, Wolfram Research

Posted 9 years ago

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

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

Wolfgang Brehm

Wolfgang Brehm, Universität Konstanz

Posted 9 years ago

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 9 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: I am hoping for an enlightening answer from Daniel Lichtblau.