1 00:00:00,900 --> 00:00:03,820 So next section is improper integrals. 2 00:00:05,680 --> 00:00:08,320 An improper integral is just above the, let's 3 00:00:08,320 --> 00:00:11,150 see, I guess I should go to my slide. 4 00:00:11,150 --> 00:00:14,190 and an improper integral comes in two types. 5 00:00:14,190 --> 00:00:19,986 So, you might be interested in integrating a function over an unbounded interval 6 00:00:19,986 --> 00:00:25,480 so that would be an interval where at least one of the end points is infinite. 7 00:00:25,480 --> 00:00:26,494 Or the function 8 00:00:26,494 --> 00:00:30,706 could be unbounded, so I might be integrating something like 1 9 00:00:30,706 --> 00:00:33,870 over x and one of the end points would be zero. 10 00:00:33,870 --> 00:00:37,961 So, the limit as my function approaches zero is going to be infinite. 11 00:00:39,080 --> 00:00:41,093 And so these are going to be the conditions 12 00:00:41,093 --> 00:00:43,576 where we can make sense of integrals like that. 13 00:00:46,260 --> 00:00:52,720 So we'll start by considering the first type, so I want to integrate a function f 14 00:00:52,720 --> 00:00:59,600 of x over either the inter, interval minus infinity b or the interval a to infinity. 15 00:01:01,510 --> 00:01:04,408 And if I want to integrate a function over the entire 16 00:01:04,408 --> 00:01:07,375 real line so minus infinity to infinity I just think 17 00:01:07,375 --> 00:01:09,514 of it as I pick some point in the middle 18 00:01:09,514 --> 00:01:12,000 and go up to there and then finish the rest. 19 00:01:16,680 --> 00:01:20,100 The improper integral of f of x over a 20 00:01:20,100 --> 00:01:24,058 infinity Exists if and only if the limit, so 21 00:01:24,058 --> 00:01:26,846 basically all I'm going to do, I want to 22 00:01:26,846 --> 00:01:30,150 have an infinite, infinity up in the top here. 23 00:01:32,350 --> 00:01:34,345 But if I actually plugged in an infinity, 24 00:01:34,345 --> 00:01:36,592 most of the time something's going to go wrong. 25 00:01:37,850 --> 00:01:39,720 So I want to consider this as a limit. 26 00:01:41,610 --> 00:01:45,306 I'll evaluate this integral using a and t, and then take 27 00:01:45,306 --> 00:01:49,820 the limit of the resulting function as t goes to infinity. 28 00:01:49,820 --> 00:01:53,642 If that limit exists and it's finite, then we say 29 00:01:53,642 --> 00:01:58,010 that the improper integral of f of x over (a, infinity) 30 00:01:58,010 --> 00:01:59,550 exists. 31 00:01:59,550 --> 00:02:02,430 So now it's getting a bit trickier, so our 32 00:02:02,430 --> 00:02:06,900 definition of existing means the limit exists and it's finite. 33 00:02:10,120 --> 00:02:12,856 And so in particular, that means that if this 34 00:02:12,856 --> 00:02:15,448 integral, you know, as I take the limit as 35 00:02:15,448 --> 00:02:19,000 t goes to infinity, if I get infinity, Then 36 00:02:19,000 --> 00:02:25,020 f of x is not integrable over this this interval. 37 00:02:25,020 --> 00:02:29,202 And then we do exactly the same thing except now I have a t on the 38 00:02:29,202 --> 00:02:34,990 bottom so when I want the, the lower limit to be able to go to minus infinity. 39 00:02:34,990 --> 00:02:36,943 I'm just going to put t in the bottom and 40 00:02:36,943 --> 00:02:40,050 take the limit as t goes to minus infinity. 41 00:02:40,050 --> 00:02:41,770 And the condition has to be the same. 42 00:02:41,770 --> 00:02:45,130 For this to be well-defined, I need this limit to 43 00:02:45,130 --> 00:02:47,990 exist, and I need it to be a finite value. 44 00:02:52,460 --> 00:02:57,305 And so when those limits exist and they're finite, then I'm going 45 00:02:57,305 --> 00:03:01,980 to define the improper integral of f of x dx over a infinity. 46 00:03:01,980 --> 00:03:05,634 To be the value of this limit, so the, the limit s t 47 00:03:05,634 --> 00:03:09,620 goes to infinity integral from a to t of f of x d x. 48 00:03:09,620 --> 00:03:13,568 And then the, sort of the complementary piece for 49 00:03:13,568 --> 00:03:17,610 the lower limit, so I'm going to define the improper 50 00:03:17,610 --> 00:03:21,960 integral from minus infinity to b of f of x d x. 51 00:03:21,960 --> 00:03:25,509 To be the value of this limit, as t goes to minus 52 00:03:25,509 --> 00:03:28,950 infinite, the integral of t to b of f of x d x. 53 00:03:28,950 --> 00:03:35,682 So you can also add and subtract improper 54 00:03:35,682 --> 00:03:41,840 integrals, so all I'm doing here. 55 00:03:43,925 --> 00:03:47,705 Is I'm considering b greater than a so this is a normal 56 00:03:47,705 --> 00:03:51,905 way of setting up an integral if I think of if I think 57 00:03:51,905 --> 00:03:55,181 of an interval I have a lower end point of a and an 58 00:03:55,181 --> 00:03:58,710 upper end point of b so b has to be greater than a. 59 00:04:01,500 --> 00:04:05,800 And so if I intergrate. f of x, from a to infinity. 60 00:04:08,350 --> 00:04:13,550 The integral of f of x from b to infinity has to sort of fit inside of that. 61 00:04:13,550 --> 00:04:16,260 So what I've done, is I've said is a to infity. 62 00:04:16,260 --> 00:04:17,990 and then b to infinity. 63 00:04:17,990 --> 00:04:20,674 And if I do this subtraction, all I want to think about 64 00:04:20,674 --> 00:04:24,740 is, I'm taking a to infinity, infinnity is way over there. 65 00:04:24,740 --> 00:04:29,373 And b to infinity and I'm just sort of chopping of 66 00:04:29,373 --> 00:04:33,780 the tail at b, and so probably just as you would 67 00:04:33,780 --> 00:04:40,108 expect what's left is just going to be the integral of f of x from a to 68 00:04:40,108 --> 00:04:46,530 b, and then I can do exactly the same thing with the other tail. 69 00:04:46,530 --> 00:04:49,611 So if I have the integral from minus infinity up 70 00:04:49,611 --> 00:04:52,218 to b, so remember b is bigger than a, so 71 00:04:52,218 --> 00:04:55,141 this integral, minus infinity up to a of f of 72 00:04:55,141 --> 00:04:58,933 x, that's if you imagine what the area looks like, 73 00:04:58,933 --> 00:05:03,360 that's just going to fit in the first, the first area. 74 00:05:03,360 --> 00:05:04,110 And so what I did. 75 00:05:05,230 --> 00:05:08,002 When I do this subtraction, essentially all I'm 76 00:05:08,002 --> 00:05:10,048 doing is just chopping off the tail at 77 00:05:10,048 --> 00:05:13,570 a, and again I'm left with the integral from a to b of f of x dx. 78 00:05:18,870 --> 00:05:19,370 Okay. 79 00:05:23,910 --> 00:05:25,494 And now I want to be able to also 80 00:05:25,494 --> 00:05:28,730 make sensitive integral over the entire real line. 81 00:05:28,730 --> 00:05:34,426 So like I mentioned before, the integral from minus infinity to infinity of 82 00:05:34,426 --> 00:05:40,140 f of x, this is going to exist if and only if there's a point a such that both. 83 00:05:41,650 --> 00:05:45,572 The integral from minus infinity up to a, and the integral from 84 00:05:45,572 --> 00:05:48,976 a to infinity of f of x exists so in particular both those 85 00:05:48,976 --> 00:05:50,240 need to be finite. 86 00:05:57,510 --> 00:06:02,840 And then if that integral exists We're just going to define its value. 87 00:06:03,910 --> 00:06:06,925 So I, I'm just splitting up the real line out of point 88 00:06:06,925 --> 00:06:10,720 a so I have numbers bigger than a and numbers smaller than a. 89 00:06:10,720 --> 00:06:13,098 The value of this integral is just going to be the sum 90 00:06:13,098 --> 00:06:17,040 of these two integrals that we know how to compute already. 91 00:06:17,040 --> 00:06:22,713 And so I just wrote down the, the definitions down below, and the reason 92 00:06:22,713 --> 00:06:25,968 we need to be able to split this thing in the 93 00:06:25,968 --> 00:06:30,990 middle, you know, why, why not just define our limit like this 94 00:06:30,990 --> 00:06:34,152 so i take the limit as t goes to infinity of 95 00:06:34,152 --> 00:06:37,382 the integral from minus t to t of f of x dx. 96 00:06:37,382 --> 00:06:42,548 So that, that seems like it should work its going to get the whole real line but 97 00:06:42,548 --> 00:06:47,878 the problem we are running into, is I could do something like this, so what, 98 00:06:47,878 --> 00:06:51,076 what I'm doing here just imagining a line with 99 00:06:51,076 --> 00:06:54,274 a slop of 45 degrees so for positive values of 100 00:06:54,274 --> 00:06:57,636 x that's going to be above the x axis, for negative 101 00:06:57,636 --> 00:07:00,740 values of x it's going to be below the x axis. 102 00:07:04,690 --> 00:07:09,070 So if I integrate that from t to t, I'm always going to get zero. 103 00:07:11,550 --> 00:07:14,362 But if I split it up anywhere in the middle, I 104 00:07:14,362 --> 00:07:18,380 would end up getting a minus infinity and a plus infinity. 105 00:07:18,380 --> 00:07:21,916 And kind of like we saw in the first lecture when we're looking 106 00:07:21,916 --> 00:07:25,050 at this series of 1 minus 1 plus 1 minus 1 you know. 107 00:07:25,050 --> 00:07:29,027 If I have something that I can express as infinity 108 00:07:29,027 --> 00:07:34,000 minus infinity, I can't really make any sense out of that. 109 00:07:34,000 --> 00:07:36,964 Another good way to imagine what's happening 110 00:07:36,964 --> 00:07:41,068 with this is, I could also get an intergal from minus infinity to 111 00:07:41,068 --> 00:07:45,790 infinity by putting a 2t up here but then that would mean that. 112 00:07:45,790 --> 00:07:49,750 All of these integrals here, instead of integrating 113 00:07:49,750 --> 00:07:53,790 to zero, they would integrate to some, let's see. 114 00:07:53,790 --> 00:07:57,300 Well this one would depend on t, but 115 00:07:57,300 --> 00:08:02,097 if you're clever about the choice, you can make 116 00:08:02,097 --> 00:08:07,479 these all integrate to some non-zero value, so as you 117 00:08:07,479 --> 00:08:13,019 take the limit, then you can get any answer you want to. 118 00:08:13,019 --> 00:08:18,297 So if I, if I defined it this way, I would have the limit from minus t to 119 00:08:18,297 --> 00:08:23,666 t being 0, but then I'm not going to have the property, so I wanted to be 120 00:08:23,666 --> 00:08:28,936 able to do addition and subtraction with my indefinite integrals. 121 00:08:28,936 --> 00:08:31,776 And so now If I try to do that, if, if I want to add 122 00:08:31,776 --> 00:08:36,035 two together, I need to be able to break it up somewhere in the middle. 123 00:08:36,035 --> 00:08:38,939 And that's gona give me, so if I take a being zero for 124 00:08:38,939 --> 00:08:43,910 instance, I'm going to get minus infinity plus infinity and that's not defined. 125 00:08:46,960 --> 00:08:51,200 On the other hand, if you know for some reason, so for instance 126 00:08:51,200 --> 00:08:54,840 if you know that f of x belongs to a family of function. 127 00:08:54,840 --> 00:09:00,720 So I'm thinking of probability density functions, where this integral does exist. 128 00:09:00,720 --> 00:09:03,560 Then you're free to evaluate it however you like. 129 00:09:03,560 --> 00:09:06,940 So if you know that this integral exist basically what that's 130 00:09:06,940 --> 00:09:10,930 telling you is if you evaluated correctly and get an answer. 131 00:09:10,930 --> 00:09:12,730 That's going to be the correct answer. 132 00:09:12,730 --> 00:09:16,240 So you can evaluate it however you like, you'll get the same answer. 133 00:09:16,240 --> 00:09:18,380 So you can evaluate it however you like. 134 00:09:23,260 --> 00:09:24,996 So this formula if it was useful, you 135 00:09:24,996 --> 00:09:27,290 could use it to evaluate an improper integral. 136 00:09:29,860 --> 00:09:38,440 Okay, so now This is probably going to be a little bit strange for some of you. 137 00:09:38,440 --> 00:09:41,720 But, I want to, for my example, show that for an 138 00:09:41,720 --> 00:09:46,148 integer value of alpha, that's greater than or equal to zero, 139 00:09:46,148 --> 00:09:50,248 this integral, so the integral from zero to infinity, of x 140 00:09:50,248 --> 00:09:53,940 to the alpha, times e to the minus x squared exists. 141 00:09:56,050 --> 00:09:59,622 And so what's going to happen with, with this particular 142 00:09:59,622 --> 00:10:02,814 argument if I'm pretty sure the way you would 143 00:10:02,814 --> 00:10:06,158 actually evaluate this, if you wanted to get if 144 00:10:06,158 --> 00:10:08,890 you wanted to get a number for an answer. 145 00:10:10,340 --> 00:10:12,709 You need to use integration by parts. 146 00:10:13,820 --> 00:10:16,687 And when you use integration by parts, you're going to 147 00:10:16,687 --> 00:10:19,290 end up with an integral where alpha is 1 smaller. 148 00:10:20,500 --> 00:10:21,172 And so if alpha 149 00:10:21,172 --> 00:10:24,630 was, say, 10, you would need to use integration by parts ten times. 150 00:10:26,274 --> 00:10:31,400 but I want to get a result that's valid for any alpha that's greater than zero. 151 00:10:31,400 --> 00:10:33,820 And so this is kind of a strange situation. 152 00:10:33,820 --> 00:10:38,870 Well remember, I said this integral exists if the limit is finite. 153 00:10:38,870 --> 00:10:41,014 So I can show that the limit is finite, 154 00:10:41,014 --> 00:10:44,670 without actually finding out what that limit is specifically. 155 00:10:44,670 --> 00:10:46,290 But if I can say whatever limit it 156 00:10:46,290 --> 00:10:51,090 is, it's less than a finite number, then I can say that this integral exists. 157 00:10:51,090 --> 00:10:52,810 This indefinite integral exists. 158 00:10:56,080 --> 00:11:03,720 So, what I want to show, is that if I integrate this function from zero to t. 159 00:11:03,720 --> 00:11:07,670 And then I take the limit as t goes to infinity. 160 00:11:07,670 --> 00:11:09,830 I have to get a finite number. 161 00:11:09,830 --> 00:11:12,314 And so another sort of shorthand for finite 162 00:11:12,314 --> 00:11:15,170 is you just write strictly less than infinity. 163 00:11:15,170 --> 00:11:17,502 So that means some real number. 164 00:11:17,502 --> 00:11:18,890 some finite real number. 165 00:11:23,020 --> 00:11:28,940 So, we can recall form the, from the first lecture when I was talking about limits. 166 00:11:28,940 --> 00:11:36,510 If p of x is a poloynomial and e to the x is the exponential function. 167 00:11:36,510 --> 00:11:42,110 Then if I take the limit of this as x goes to infinity I get 0. 168 00:11:42,110 --> 00:11:47,450 So in particular, I have this minus x squared here. 169 00:11:47,450 --> 00:11:48,062 So if I have 170 00:11:48,062 --> 00:11:51,920 e to a negative power I can also write that as 1 over e to that power. 171 00:11:53,160 --> 00:11:56,264 So I can rewrite the integrand here as x 172 00:11:56,264 --> 00:12:01,660 to the alpha Divided by e to the x squared. 173 00:12:01,660 --> 00:12:06,764 And so, by using those rules of limits, you can also show that this is 174 00:12:06,764 --> 00:12:12,140 going to the limit of this as x goes to infinity is also going to be 0. 175 00:12:12,140 --> 00:12:13,340 And so, you might 176 00:12:13,340 --> 00:12:18,430 wonder why this plus 2 is up there, because there's no plus 2 in the intregran 177 00:12:19,490 --> 00:12:24,566 And the answer is you know I just put it there because I want it 178 00:12:24,566 --> 00:12:30,330 there. So it, it'll pop out later on. 179 00:12:30,330 --> 00:12:33,186 And so often when you're trying to work these things out, so I know 180 00:12:33,186 --> 00:12:37,030 how to solve this problem already, so I can put the plus 2 there. 181 00:12:37,030 --> 00:12:38,515 Otherwise I wouldn't have put the plus 182 00:12:38,515 --> 00:12:40,540 2 there, I would have done more work. 183 00:12:40,540 --> 00:12:42,700 I would have realized, hey, it would be really 184 00:12:42,700 --> 00:12:45,980 handy if there's another x squared in this equation. 185 00:12:45,980 --> 00:12:48,260 So then I would have gone backwards to this line. 186 00:12:48,260 --> 00:12:50,210 I would have added 2 to this exponent. 187 00:12:52,110 --> 00:12:56,110 But the point is, it's not it's not meant to be equal to this. 188 00:12:56,110 --> 00:12:59,960 It's basically, what's going to end up happening is that 189 00:12:59,960 --> 00:13:04,162 I can now Write this as something times x squared and 190 00:13:04,162 --> 00:13:09,310 have my limit here, or, my integrand here. And so 191 00:13:09,310 --> 00:13:14,386 this, the, if, if I consider a sequence as the x goes to 192 00:13:14,386 --> 00:13:19,611 infinity, this is going to be equal to 0. And so because 193 00:13:19,611 --> 00:13:24,525 I know that limits equal to 0, I can now use the 194 00:13:24,525 --> 00:13:29,817 definition to say that there is a value M so capital 195 00:13:29,817 --> 00:13:36,117 M such that, so this is my, my integrand again this value is 196 00:13:36,117 --> 00:13:42,669 less than 1 whenever x is greater than M, so this is basically 197 00:13:42,669 --> 00:13:48,465 this epsilon delta argument except you know, for 198 00:13:48,465 --> 00:13:54,811 the, for the, for the sequence that is going to infinity. 199 00:13:54,811 --> 00:13:57,241 So the epsilon is 1 and then that 200 00:13:57,241 --> 00:14:01,021 means because I know the limit exists, there must 201 00:14:01,021 --> 00:14:03,991 be some value such that when I'm beyond 202 00:14:03,991 --> 00:14:07,321 that value I'm within epsilon of what I want 203 00:14:07,321 --> 00:14:12,811 it to be and so I'm going to choose 1, I can choose any number I want to by the 204 00:14:12,811 --> 00:14:18,945 definition of the limit but I'm going to choose 1 because its easy to work with. 205 00:14:18,945 --> 00:14:20,010 So generally when 206 00:14:20,010 --> 00:14:23,631 you are free to choose a number, if you choose anything other 207 00:14:23,631 --> 00:14:27,397 than zero or one, you better have a really good reason for it. 208 00:14:27,397 --> 00:14:30,477 So now it should be clear why I wanted this x 209 00:14:30,477 --> 00:14:34,173 squared here, so if this quantity has to be less 1 210 00:14:34,173 --> 00:14:37,693 one then I can just divide both sides by x squared 211 00:14:37,693 --> 00:14:42,233 and so I'm considering the limit as x goes to infinity. 212 00:14:42,233 --> 00:14:45,149 you know, at some point this M is going to be 213 00:14:45,149 --> 00:14:47,071 a number bigger than zero. 214 00:14:47,071 --> 00:14:51,230 And x, you know, I'm looking at values of x greater than that M. 215 00:14:51,230 --> 00:14:53,600 So x is not going to be 0. 216 00:14:53,600 --> 00:14:56,920 So I can divide by x as many times as I want to. 217 00:14:56,920 --> 00:15:01,680 So I'm going to divide by x twice. So my alpha over 2 becomes an alpha. 218 00:15:02,720 --> 00:15:05,780 And my 1 becomes a 1 over x squared. 219 00:15:05,780 --> 00:15:10,324 And so the important thing that I've done here is I've taken the integrand and I've 220 00:15:10,324 --> 00:15:15,792 bounded it by one over x squared. So that means for every value of x greater 221 00:15:15,792 --> 00:15:21,900 than M, this thing is going to be less than one over x squared. 222 00:15:21,900 --> 00:15:23,250 And what that's going to let me do. 223 00:15:24,870 --> 00:15:30,070 Is now if I consider values of t that are greater than M, I'm 224 00:15:30,070 --> 00:15:35,374 going to split my integral so my integral zero to t into a piece 225 00:15:35,374 --> 00:15:36,518 from zero to M 226 00:15:36,518 --> 00:15:37,038 [COUGH] 227 00:15:37,038 --> 00:15:39,920 and another piece from M up to t. 228 00:15:44,830 --> 00:15:48,104 And now what can I say about the, the first term here? 229 00:15:53,040 --> 00:15:56,280 Though it's not pretty, but 0 is a constant and M is a 230 00:15:56,280 --> 00:16:01,589 constant, so whatever, you know whatever number this is it's going to be constant. 231 00:16:02,870 --> 00:16:05,642 And so, t is the only thing that's changing now, 232 00:16:05,642 --> 00:16:08,280 this is the, what I'm taking to be the limit. 233 00:16:09,330 --> 00:16:12,060 So I can just rewrite this as a constant. 234 00:16:12,060 --> 00:16:15,590 It's no longer really important. If it's constant, it's finite. 235 00:16:15,590 --> 00:16:18,174 And I'll I'm trying to do, is show that this 236 00:16:18,174 --> 00:16:20,150 thing is not going to be infinite. 237 00:16:24,190 --> 00:16:32,430 And also I have said that this integrand can be bounded by 1 over x squared. 238 00:16:33,650 --> 00:16:35,694 So if I were to replace this with 1 239 00:16:35,694 --> 00:16:39,400 over x squared, like I've done in the second line. 240 00:16:39,400 --> 00:16:44,670 It means it's everywhere for t bigger then M, x squared is bigger than that. 241 00:16:44,670 --> 00:16:48,780 So c, this thing, plus the integral from m to t. 242 00:16:49,790 --> 00:16:52,286 You remember one of the properties of the integral was if I 243 00:16:52,286 --> 00:16:55,146 had to functions and one of those functions was always bigger than 244 00:16:55,146 --> 00:16:58,992 the other function, If I integrate both of those functions over the 245 00:16:58,992 --> 00:17:01,138 same integral, then whichever function is 246 00:17:01,138 --> 00:17:03,620 always bigger, it's integral is also bigger. 247 00:17:05,150 --> 00:17:09,560 So that allows me to bound this integral by some constant, 248 00:17:09,560 --> 00:17:13,430 and a func, and the integral of one over x squared. 249 00:17:13,430 --> 00:17:14,833 And now one over x squares, 250 00:17:14,833 --> 00:17:17,730 that's something that is actually easy to integrate. 251 00:17:19,840 --> 00:17:22,956 So that just becomes minus 1 over x so 1 over x 252 00:17:22,956 --> 00:17:26,564 squared, I can think of that as x to the minus 2 power 253 00:17:26,564 --> 00:17:30,746 and I can use my anti-power rule and that just means its going 254 00:17:30,746 --> 00:17:34,995 to come from, its the derivative of minus x to the minus 1. 255 00:17:36,580 --> 00:17:43,070 So I get minus x to the minus 1 here evaluated at from M to t. 256 00:17:44,280 --> 00:17:44,980 So again 257 00:17:44,980 --> 00:17:50,280 I can rewrite that now as a constant minus this thing here. 258 00:17:55,170 --> 00:17:58,010 So this is just to summarize what I've done on the other slide. 259 00:17:58,010 --> 00:18:03,652 I now have bounded this integral from zero to t of this very unhappy looking 260 00:18:03,652 --> 00:18:10,050 integrand, by this expression over here that's a much simpler expression of t. 261 00:18:12,990 --> 00:18:18,400 And so I have c plus 1 over M so these are just constant. 262 00:18:18,400 --> 00:18:22,050 And now remember what I was trying to do is show that the limit 263 00:18:22,050 --> 00:18:26,550 of this guy over here as t goes to infinity is approaching a finite value. 264 00:18:30,630 --> 00:18:37,838 And so 1 over t, for t is positive. So, this can never be, this can 265 00:18:37,838 --> 00:18:45,510 never be bigger than c plus 1 over M. So, I can make an upper bound of c. 266 00:18:45,510 --> 00:18:48,800 So, that's the integral of my integrand from 0 to M. 267 00:18:50,090 --> 00:18:52,930 That's a constant value, so, it's going to be finite. 268 00:18:52,930 --> 00:18:55,912 And then M is a really big number, and I have 1 over 269 00:18:55,912 --> 00:18:59,520 M, so I have to add just a little bit more to my constant. 270 00:18:59,520 --> 00:19:03,570 And that gets me an upper bound, for no matter what value of t 271 00:19:03,570 --> 00:19:08,010 I choose, this is it's going to be smaller c plus 1 over M. 272 00:19:10,690 --> 00:19:15,080 So that means I can say that the limit as t goes to infinity Of the 273 00:19:15,080 --> 00:19:20,040 integral from 0 to t of x to the alpha, e to the minus x squared. 274 00:19:20,040 --> 00:19:23,390 That's going to be less than c plus 1 over M. 275 00:19:23,390 --> 00:19:24,840 And that's going to be a finite number. 276 00:19:24,840 --> 00:19:27,270 So I write less than infinity. 277 00:19:27,270 --> 00:19:31,146 And so without actually evaluating this integral, and finding out 278 00:19:31,146 --> 00:19:34,370 the, the number, I was able to show that, let's see. 279 00:19:36,800 --> 00:19:40,230 That this indefinite integral, from 0 to infinity, of x 280 00:19:40,230 --> 00:19:43,430 to the alpha, e to the minus x squared, dx. 281 00:19:43,430 --> 00:19:44,300 This exists.