

Integral Solution
Name: Danilo V.
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A
Question:
Which is the analytical solution to this integral?
Int=Integrate[Erf[x]^n Exp[x^2+ax],{x,Infinity,Infinity}],
where n>0 (integer), and a>0 (real).
Note here I use Mathematica notation. The integral is over x from
minus infinity to plus infinity, and erf[x] is the error function.
I have tried to find the answer using substitution, by parts, etc
(with the old pencil and paper method).
I also tried looking in Tables of Integrals, using Mathematica,
Maple, etc. Perhaps there is no answer.
Any hint would be very much appreciated.
P.S. Is the method of residues useful for an integral like this?
Replies:
Danilo,
I do not know right off whether the integral exists, but it is possible that
the integral is infinite. A way to approach it is as follows. First, graph
the function you are integrating. Mathematica may allow this to be done
quickly. After verifying that the function has nothing nasty in it, use
Mathematica to do finite integrals, perhaps (10,10), then (1000,1000),
then
(100000 , 100000). Verify that the integral does not shoot up as the limit
gets too large. If the error function is approximated by a polynomial or a
series of functions, the approximation may become inaccurate when x is too
big.
Dr. Ken Mellendorf
Illinois Central College
Hello,
You can definitely integrate this equation numerically using mathematica
or a number of other techniques. If you are looking for an analytical
solution, one has to try all sort of possible substitutions and there is
no guarantee that one exists. Approximate solutions may exist, and they
can be obtained by series expansion of the terms in the intergrand.
Also, you may want to look at this text: Table of Integrals, Series, and
Products (by I.S. Gradshteyn; Academic Press). This is the best source I
know for locating analytical solutions to integrals. I hope this is
helpful.
AK
Some time ago this question was posted. While I was not able to find an
analytical solutions,
I was able to integrate it numerically. Some of the results are:
n=1, a=1 0.87148956
n=2, a=2 0.95388311
n=3, a=1 0.59679878
n=5, a=1 0.46616033
n=10,a=1 0.35786485
n=1, a=2 3.28921783
n=2, a=2 2.96518548
n=3, a=2 2.54325302
n=4, a=2 2.37223565
n=5,a=2 2.14963856
n=10, a=2 1.65148716
n=10, a=10 1.27624646E+11
It seems to be blowing up for large values of 'n' and 'a', but rather
slowly.
Vince Calder
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Update: June 2012

