The first one, Integral(-inf to +inf) e^(x^2)dx, is not in Thomas because it does not converge. But \int_{-infty}^\infty e^{-x^2}dx, as dextercioby suggests, certainly is in Thomas, and the second is a variation. You may be trying to find an anti-derivative formula and not finding that- neither integrand has an "elementary" anti-derivative. Both are used extensively in probability (e^{-x^2} is the "bell shaped curve") and so in quantum mechanics.
Here is a simple way to get the first integral:
Let I= \int_{-\infty}^\infty e^{-x^2}dx. Since the integrand is symmetric about x=0, we also have I/2= \int_0^\infty e^{-x^2}dx.
And, we can write I/2= \int_0^\infty e^{-y^2}dy. Multiplying those together, I^2/4= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right). By Fubini's theorem, we can write that product of integrals as a double integral:
I^2/4= \int_{x= 0}^{\infty}\int_{y=0}^\infty e^{-(x^2+ y^2}dydx
Now, change to polar coordinates: x^2+ y^2= r^2cos^2(\theta)+ r^2sin^2(\theta)= r^2 and dydx= r drd\theta. The area of integration, with both x going from 0 to infinity is the first quadrant. In polar coordinates, r goes from 0 to infinity while \theta goes from 0 to \pi/2. The integral becomes<br />
I^2/4= \int_{\theta= 0}^{\pi/2}\int_{r=0}^\infty e^{-r^2} rdrd\theta= \frac{\pi}{2}\int_0^\infty e^{-r^2} rdr<br />
<br />
That extra 'r' in the integrand now allows us to make the change of variable u= r^2 so du= 2r dr and the integral becomes<br />
I^2/4= \frac{\pi}{4}\int_0^\infty e^{-u}du<br />
which is easy.<br />
<br />
It was necessary, to make that change to polar coordinates, that the integral be over the entire first quadrant. Again, that function, e^{-x^2}, has no elementary anti-derivative. In fact, its anti-derivative is typically written "erf(x)", the "error function", and it values are got by a numerical integration.
