Integration question: integral ( x^2 e^(x^2) ) dx

[jwpaine]

Today in a lab we went through a series of integrals, and we had to do each one and then identify the one which "we have not learned a technique for yet".

Number 5 was (if I can remember correctly): \(\displaystyle \int x^{2}e^{x^2}\)
I thought, "hey, maybe I can do integration by change of variables followed by integration by parts..."

I began by attempting a change of variables: \(\displaystyle u = x^2, \frac{d}{u}dx = 2x, \frac{1}{2}du = xdx\)

Now, since I had an x in du, than \(\displaystyle \sqrt{u}du = x^2\)

so I got \(\displaystyle \frac{1}{2}\int \sqrt{u} \cdot e^{u}du\)

Now, at this point, if I did integration by parts, I would not be able to get \(\displaystyle \frac{d^{n}}{du}\sqrt{u}\) to equal one, for any integration by parts, n times.

How else could I find an antiderivitive for this indefinite integral? Was I on the right track? Does this require integration by trig?

 

[galactus]

Re: Integration question

Actually, JW, that integral is not easily done by elementary means. It can be done, but not with methods you've been exposed to. Are you sure you have the correct one?.

If it were \(\displaystyle \int{x^{2}e^{x}}dx\), then that's another matter. That can be easily done with parts.

Take note that the gamma function is: \(\displaystyle {\Gamma}(3)=\int_{0}^{\infty}x^{2}e^{-x}dx\)

Your integral works out to something like:

\(\displaystyle \frac{x}{2}e^{x^{2}}+\frac{1}{4}i\sqrt{\pi}*erf(ix)}\)

where 'erf' is the error function.

I told you it was ugly as is.

 

[stapel]

Number 5 was (if I can remember correctly): \(\displaystyle \int x^{2}e^{x^2}\)

The results from The Integrator agree with the previous reply:

. . . . .\(\displaystyle \frac{1}{4}\, \left(2e^{(x^2)}\, x\, -\, \sqrt{\pi}\, \mbox{erfi} (x)\right)\)

...where "erfi(x)" is the "imaginary error function". :shock:

Eliz.