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Name: Danilo V.
Status: student	
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Date: N/A 

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?

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


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.


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

Vince Calder

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