 So, one integral that shows up over and over in physical chemistry is an integral of the form e to the minus x squared with perhaps some constant up in the exponent. And this is an integral that doesn't yield to the normal techniques, normal integrals that you've memorized the answer to or used substitution and so on don't work here. In fact, the integral itself is only solvable directly if it's a definite integral with certain limits. For example, definite integral from zero to infinity can be solved. And again, because it shows up all the time in PCAM, it's worth discussing how to do this integral. So, just so that we have a visual picture of what this integral is doing, a Gaussian function e to the minus alpha x squared has this general shape, so e to the minus alpha x squared. So, if we've asked for the integral from zero to infinity of that function, that's the area under this integral on the right side of the y-axis. So, we want to know the value of this integral. And again, the reason this integral shows up so often is because we're very commonly taking the exponentials of things perhaps because we're doing Boltzmann probability distribution, e to the minus some energy over kT causes an integral to look like this or there's also certain quantum mechanical problems that have Gaussians integrals in them as well. So, how do we do this integral? Let's go ahead and define, give this integral a name, let's call it I naught. The reason I'll call it I sub zero or I naught is because it's related to some other forms. If I had some powers of x inside here, those would be I sub one or I sub two and so on. But for now, we'll stick with this integral I zero. And because this Gaussian function is symmetric, if I don't want to know just the area on the right hand side of the curve, if I want to know the area under the entire curve, that's the integral of the Gaussian from negative infinity to infinity. And that's exactly twice the area under the right side of the curve. So, this integral itself is not terribly tractable. It's easy to get the answer to. The trick, however, is let's write this integral down twice. So, what I'm going to do is I'm going to write down the integral that we've already written down, e to the minus alpha x squared, integrated everywhere, and I'm going to multiply that by the same integral. And now instead of using x as my integration variable, I'll use y, so this should be a y, this is alpha y squared, integrated over all values of y. So, I've just written down twice I naught and then I've written down the same thing, multiplied them together. So, that's two I naught quantity squared. So, if I can do this integral, I'm finding the value of four I naught squared. And I can think of that integral, of course, as a double integral. e to the minus alpha x squared times e to the minus alpha y squared. The product of those two exponentials is e to the sum of their exponents. So, e to the minus alpha x squared with a negative sign, e to the minus alpha y squared, also with a negative sign. I have to integrate that over both dx and dy. So, both of those variables are running from negative infinity to infinity. But now that I've written it out as a two-dimensional integral, I realize, especially after seeing x squared plus y squared, I can think of that two-dimensional integral in Cartesian coordinates, like I have it written here. Or, I can think of doing that same integral in polar coordinates, where the integral would be with respect to already already theta. So, if I rewrite the exponent e to the minus alpha and then rewriting x squared plus y squared as r squared in polar coordinates, then I have this integral to do. That looks an awful lot like the integral I started with, which I said was not an easy integral to do. The difference is that now we have this extra r that comes in from the integration variable in polar coordinates. So, let me point out that the r integral radius goes from zero to infinity, and the angular variable goes from zero to two pi. Let's go ahead and do the angular integration. So, there's no theta's anywhere except the d theta. Integrating d theta from zero to two pi gives me two pi. So, I've got r times e to the minus alpha r squared times dr integrated from zero to infinity. So, that's actually a much easier integral to do, especially once we notice that r looks like the derivative of an r squared. So, if we use u substitution, so that's continuing from this result, if I say let's let u be maybe alpha r squared so that du is equal to twice alpha r dr, then let me go ahead and say the r dr that I have, this r dr that I have in the integral. If I bring the two alpha over the other side, that'll be one over two alpha times du. So, if I use those definitions to rewrite this integral, my integral now looks like two pi integral from zero to infinity. r dr looks like one over two alpha du. e to the minus alpha r squared looks like e to the minus alpha, I'm sorry, e to the minus u. And that's the integral of a normal exponential function. Integral of e to the u is e to the u. I've brought the two alpha out. Integral of e to the minus u is minus e to the minus u. I'm now evaluating that from zero to infinity. My two's cancel. Quantity and parentheses, I have a e to the minus infinity. And that's zero. So, I've got a negative zero for the first time. And when I stick the zero in for you, when u is equal to zero, e to the zero is one. So, I have, I'm subtracting a negative one. And so, I get pi over alpha multiplied by just one. And the thing that this is equal to that what I'm solving for is this four i naught squared. So, we were interested in the value of i naught. So, let's go ahead and take the square root. So, the two i naught is going to be square root of pi over alpha. And if I move the two to the other side, i naught is going to turn out to be one-half of the square root of pi over alpha. So, that's the answer we were looking to, looking for. The area under the right-hand side of the integral is i naught, one-half squared of pi over alpha. Just as commonly, if we're interested in the area under the full, let's say in perhaps a different color, if I want the entire area of the integral, that full area, in other words, integral of e to the minus alpha x squared dx from negative infinity all the way to infinity, that's actually the problem we run into most often. That full area under this Gaussian curve is twice the area under half the curve, so that's just square root of pi over alpha. Equation is actually probably worth memorizing because we see these Gaussian integrals very often. So, next time you see an integral of a Gaussian, you can say that the integral under the entire Gaussian is the square root of pi divided by the exponent in the Gaussian, or if you only need half of that integral, just throw a one-half in. So, this is useful both for times that this integral shows up in problems, and it's also useful because it helps us evaluate a related set of integrals with additional polynomials in front of the Gaussian, which we'll consider in a separate video.