Re: Antiderivative of e^(x^2)?

From: Justin Davis (jkd3_at_duke.edu)
Date: 03/02/05 

Date: Wed, 02 Mar 2005 14:48:33 GMT

"nimcha" <nimcha@NO.uga.SPAM.edu> wrote in message
news:vdhb211urvrc37uv9hqnk5ricnqv00o5ns@4ax.com...
> Hello,
>
> Just as a mental exercise, I was trying to evaluate the antiderivative
> of e^(x^2). (My Differential Equations textbook claims it cannot be
> done, which piqued my curiosity.) So, is the following correct?
>

It can be done. After all, you're familiar with series, presumably, and
could come up with a series for e^(x^2). You could integrate this term by
term and find a series that would give you the antiderivative. This is
considerably more messy, however, than int(dx/x) = ln(x) + C.

This isn't the sort of thing we have to do for most functions. We know how
to write int(dx/x) in a nice, concise way because we have the ln function;
in fact, some people define ln(x) = int(t = 1 to x, dt/t). You could, if you
wanted to, write pq(x) = int(t = 0 to x, exp(t^2) dt). So when someone asked
you what the indefinite integral of e^(x^2) was, you'd say pq(x) + C.

The difference here is that pq, unlike ln, is not an "elementary" function.
It's not something like a logarithm, sine, cosine, a polynomial, or any
other beastie that has been used enough to have a name attached to it.

Can't be done? Not at all - don't believe them. Can't be done in closed form
(no infinite series) in terms of elementary functions? Seems so, but I don't
think anyone has really proven that.

> e^(x^2) = (e^x)^x = ((e^x)^(x-1))*(e^x)
>
> Substitute u = e^x and du = (e^x)dx:
>
> int(((e^x)^(x-1))*(e^x)dx) = int((u^(x-1))du)
>
> where u^(x-1) = e^((x-1)*ln(u)), so, my question is, is it possible to
> integrate this exponential?
>
> Possible methods I could think of were:
>
> int((e^((x-1)*ln(u)))du) = (e^((x-1)*ln(u)))*(1/(x-1)) + C =
> (u^(x-1))/(x-1) + C = (e^((x^2)-x))/(x-1) + C
>

Remember first that u = e^x = exp(x) and x = ln u, so you can't treat x like
a constant when you are integrating with respect to u.

Secondly, are you claiming that int[exp(c*ln(u))du] = (1/c) exp[c*ln(u)] +
C? You'll note that exp[c*ln(u)] = u exp c, so int[exp(c*ln(u))du] = int (u
exp c du) = (exp c/2) u^2 + C. This is rarely equal to (1/c) exp[c*ln(u)] +
C, so something must be going wrong.

The formula you are probably half-remembering is int(exp(au)du) = (1/a)
exp(au) + C. Unfortunately, you have ln(u) instead of u itself in the
exponent, so you can't use this formula.

> int((u^(x-1))du) = (u^x)/x + C = e^(x^2) + C
>

Same thing here; u is a function of x and x is a function of u, so in an
integration with respect to one or the other, you can't treat the other as a
constant.

> int((e^((x-1)*ln(u)))du) = (e^((x-1)*ln(u)))*1/ln(u) + C =
> (u^(x-1))/ln(u) + C = (e^((x^2)-x))/x + C
>

I'm not sure what's going on here, but it looks like despite writing du you
actually took the integral with respect to x.

> (Sorry the plain-text-only format of Usenet makes these expressions
> kind of awkward. They're prettier on paper...)
>
> Any input appreciated. =)
>
> Thanks,
>
> George
> (to email me, remove the NO SPAM from my email address!)

Relevant Pages

  • Re: Ist order ODE
    ... Lynn Kurtz wrote: ... integrating factor exp), ... side can't be expresses in terms of elementary functions. ... It turns out that my case has a positive coefficient of x^2 in the exponent, ...
    (sci.math)
  • Re: Ist order ODE
    ... Presumably you got that by multiplying the equation through by its ... integrating factor exp), ... side can't be expresses in terms of elementary functions. ... how to complete the square and change variables to express it in terms ...
    (sci.math)
  • Re: Ist order ODE
    ... Presumably you got that by multiplying the equation through by its ... integrating factor exp), ... side can't be expresses in terms of elementary functions. ... Google for that. ...
    (sci.math)