I cannot do it either, although I tried. I went round and around only.
Try "erf" or error function. Search the internet.
Hey,
Im having trouble integrating this function: 1/1+Exp(-x^2)
None of the conventional methods I have learned will work on it. Substitution does not work, by parts does not work etc. And worst of all, I have yet to see any text give an example of how to inegrate it. All test only stick to giving examples where Exp is raised to a linear form of x, no nonlinear forms of x.
I have seen many examples were Exp is raised to a linear form of x, such as 1/1+Exp(-x) or 1/1+Exp(2x). Of course I have no trouble integrating these. But the same method I used to integrate these functions does not work on the function above. Could someone please tell me what the anti-derivative is for the function above, and the method that would be used to get it. Please post it step by step.
Thanx in advance
integrals.wolfram.com couldn't do it either ...
You can't calculate it becoase there is no solution in terms of an elementary functions (at least not in finite number of terms of elementary functions). You need to expand the function in a row or something like that.
I'm wondering whether you meant something like 1/(1+e^(-x^2))...in that case, the antiderivative is what we like to call a "non-elementary function"; however, it may still be possible to evaluate certain definite integrals of this function by a method like the calculus of residues...but I'm not sure that that's what you are looking for here.
A famous example of an integral that can be calculated without possessing an elementary antiderivative is Integral[e^(-x^2),x,-infinity,infinity]; it is possible to use an argument involving multiplying it by itself (with a different variable of integration) and then transforming the resulting double integral into polar coordinates, and then evaluating that integral the usual way, and taking the square root:
Let I=Integral[e^(-x^2),x,-infinity,infinity]; then also, I=Integral[e^(-y^2),y,-infinity,infinity].
Therefore, I^2=Integral[e^(-x^2),x,-infinity,infinity]*Integral[e^(-y^2),y,-infinity,infinity]=Integral[Integral[e^(-(x^2+y^2)),x,-infinity,infinity],y,-infinity,infinity]=Integral[Integral[r*e^(-r^2),r,0,infinity],theta,-Pi,Pi]=2*Pi*Integral[r*e^(-r^2),r,0,infinity]=2*Pi*(-e^(-infinity)/2+e^0/2)=Pi, so I=Sqrt(Pi).
Something called the calculus of residues in complex analysis can allow you to evaluate integrals is also useful; basically, the line integral of a function in the clockwise direction around any loop in the complex plane not intersecting any of the nonremovable singularities is equal to 2*Pi*i*sum(Res(w)) over all w in the interior of the loop at which a singularity occurs and Res(w) is the residue of w, the coefficient of the 1/(z-w) ("simple pole") term in the Laurent series about w. Laurent series are similar to Taylor series except that they also have negative-power terms; the lowest exponent that appears is called the order of the singularity, and if that exponent is nonnegative, it is called a removable singularity; if it is -1, it is a simple pole; if -2, a double pole; if -n, a pole of order n, or a multiple pole if n>1; and if there is no lowest exponent in the series, it is called an essential singularity.
Anyway, it is possible to evaluate many real integrals as complex line integrals in the "upper half plane" around loops that surround all singularities in the upper half plane, usually a semicircle of radius R centered at the origin, and letting R->infinity.
Interestingly, the method typically used to evaluate the integral of e^(-x^2) over the real line has no interesting generalizations; see Robert J. MacG. Dawson, On a "Singular" Integration Technique of Poisson, The American Mathematical Monthly, March 2005, p. 270.