 So we've mentioned how Gaussian integrals show up a lot in physical chemistry problems. And even if we know that the integral of a Gaussian from zero to infinity, so half of the area underneath of a Gaussian, if I have a Gaussian curve, the area under the right half of that curve or the left half of that curve is this quantity we're calling I naught. Even if we know that, that's not enough to solve all the problems where these Gaussian integrals show up, because another very common form that shows up is not the integral of a Gaussian, but the integral of a Gaussian multiplied by x to some power. So maybe x or x squared or x cubed in front of a Gaussian. So when there's x to the zero or there's no polynomial term in front of the Gaussian, we call that I naught. If we call it I sub n, if there's an x to the n in front of it, we can ask ourselves what's the value of these additional forms of these Gaussian style integrals. So those, it turns out, once we have this one are fairly easy to obtain by sort of normal calculus techniques. So let's first ask ourselves what about this integral of I1, if we'll just stick an x in front of the Gaussian, and I guess I should say over here this I sub n is specifically for the area under that half of the curve, the positive values of x. So graphically, if I naught is the area under half of a Gaussian, this is the area under half of a curve that looks like x e to the minus x squared. So that graph still goes to zero at large and small values of x like a Gaussian does, but also goes to zero at the origin because of the x, x is equal to zero at the origin as well. So we're asking what is the area under this section of the curve. So that's the question we're asking, what is this I sub 1? This integral is actually relatively simple, either by u substitution, noticing that x is the derivative of x squared or related to the derivative of x squared, or just by directly writing down, let's see, so I've got an alpha here. So integral of x e to the minus alpha x squared is minus 1 over 2 alpha e to the minus alpha x squared, which we can confirm if we take the derivative of this Gaussian, the derivative of the exponent is minus 2 alpha x, so the minus and the 2 and the alpha cancel, leaving behind just the x. So this is the integral of x e to the minus alpha x squared. We have to evaluate that from zero to infinity, which gives us minus 1 over 2 alpha e to the minus infinity, which is zero, minus e to the zero. So that negative sign cancels this negative sign, and the result is just 1 over 2 alpha. So I sub 1 was not terribly difficult to calculate, much easier than I sub zero. If we do one additional one, turns out to be a little bit different. The next one in the series is I2, so if I stick n equals 2 into this definition, I'm asking for the integral of x squared e to the minus alpha x squared, and now we're in at least a little bit of trouble because the x squared, we can't use u substitution to call the x squared the derivative or related to the derivative of the x squared, but we can use the other maybe second most common calculus trick, one of the two you're likely to have retained from a calculus class, which is u substitution, not u substitution, but integration by parts. So if we say, I kind of wish the coefficient out front were an x rather than x squared because that one is relatively easy to do. So if I say, let's let dv be equal to x e to the minus alpha x squared because I know how to take the derivative, or I'm sorry, the integral of that quantity, and what's left over is an extra factor of x. So the product of these two things, x times x times e to the minus alpha x squared dx is the integral we're interested in, then du is dx, and if I integrate the one I know how to integrate, x e to the minus alpha x squared, that's exactly the same as it was above. That integral is minus one over two alpha times a Gaussian. Then those are the parts we need in order to do integration by parts. So this integral is going to be u times v minus the integral of v du. So I just need to insert the definitions I've chosen in choosing my us and my dv's. And we find, so u times v, x times minus one over two alpha and a Gaussian, so I've got minus one over two alpha x e to the minus alpha x squared, evaluated between zero and infinity. And then the minus integral of v du, v has a negative sign in it, so that minus becomes a plus. The one over two alpha I can keep outside the integral, but I want the integral of v times du, so that's the integral of the Gaussian, again from zero to infinity. And now we can do both of these pieces. Regardless of whether I plug in an infinity, which dies, becomes zero because of the Gaussian portion. Either the minus infinity squared is zero, or whether I plug in a zero when x equals zero, the polynomial portion, the x, causes this thing to disappear. So this first term, both of the terms are zero, and that difference goes away entirely. For the second term, I have one over two alpha times the integral of a Gaussian from zero to infinity, but we know a name for that. The integral of a Gaussian from zero to infinity is exactly the i naught that we started with. So it turns out this i sub two, the second, not the zero of the first, but the second one of these integrals is related to the zeroth one, the basic one for a plain Gaussian, just multiplied by an extra factor of one over two alpha. So if we care about the numerical value, that's one over two alpha times this existing value, so now I have a one over four, I still have a square root of pi on top, and I have the alpha to the three halves on the bottom instead of one half. But the important thing to notice is that when we used integration by parts on i two, it reduced the problem down to i naught. Likewise, if we run across one that looks like i sub three or i sub four or i sub five, if we have i sub n, if we integrate by parts, it will relate it to the one that's too smaller. So now that we have an answer for i naught and i one, i two is related to i naught, i three is related to i one, and so on. So if someone, for some reason, gives you i sub seventeen, an integral of x to the seventeenth times a Gaussian, you could repeatedly use integration by parts to work your way back down to one that we already know. Or if you go to an integral table, of course, you can find a closed form expression for these integrals that are obtained by a process like this. And because this integration by parts drops by two and we have different behavior for the odd integrals than we do for the even integrals, then it turns out we have two different answers for this integral. If n is an odd number, like i sub one, then the answer we get is this collection of terms of one-half, a factorial that hasn't reared its head yet in the i one term. But as we bootstrap our way down from i seventeen to fifteen to thirteen to eleven and so on, that's going to involve some factorials showing up. And then a number of powers of alpha in the denominator that is increasing as n gets larger and larger. So that's for the odd n terms. And if n is even, on the other hand, then what we have is a different number of powers of two, a different style of factorial. And here's the one where we have the square root of pi showing up over, again, some number of powers of alpha. That's when n is even. And it's worth pointing out here that that's not a typo. I have intentionally written two exclamation points there. That is called a double factorial. If you've not run across one of those before, it's different than a normal factorial in that, for example, if I take five double factorial, that's not five times four times three times two times one, I have to skip terms. That's five times three times one, where I've skipped dropping by two instead of just dropping by one. So the double factorial is different than a regular factorial, so keep that in mind. So point here is to illustrate the technique of how to evaluate these Gaussian integrals in case you need to run across one that you need to evaluate. But the answer that you can get by plugging into these closed form expressions that you can look up in an integral table is just as good if you don't want to do the work yourself. So there is what we have for Gaussian integrals.