this innocent integral hides Euler’s constant

Added: 2026-04-29

[00:00:00]Today we're going to look at a fairly quick but interesting integral. We're going to evaluate the integral from 0 to 1 of the fractional part of 1 /x. So what's the fractional part of a number?

[00:00:13]Well, it's exactly what it sounds like.

[00:00:15]So it's going to be not the floor of the number, which is the largest integer smaller than the number, or the ceiling of the number, which can be defined similarly. It's going to be that distance between the floor and the actual number. So in other words, if we have something like this 1.79 and we take the fractional part of that, that's just going to be equal to 0.79.

[00:00:42]And I think it's pretty clear that we can take a number x and or maybe I shouldn't use x here because we've got an x up here in our integral. If we take a number t, we can decompose it into its floor and its fractional part. That's because if we think about what's going on on a number line, if we've got our number t right here, well, the floor of t is simply going to be the biggest integer that's smaller than t. And then this distance between the floor of t and t will be this fractional part of t. So perhaps that's the way to think about this that's the most intuitive.

[00:01:25]But anyway, let's get back to this fractional part of 1 /x thing. But now let's observe the following. If we've got x between 1 / n + 1 and 1 / n. And so we can break up this interval from 0 to 1 into pieces like that for sure.

[00:01:43]Then that's equivalent to saying that 1 /x is between n and n + 1. But then we can say that the floor of 1 /x is n in this scenario.

[00:02:00]Which means we have 1 /x decomposing as the floor of 1 /x plus the fractional part of 1 /x. In other words, putting together this fact that we have this 1 /x= n, we have the fractional part of 1 /x is simply equal to 1 /x - n. And so that's a decomposition which I think we'll find pretty helpful. Okay, so now let's go back up to our integral. And what I want to do is decompose this interval that we are integrating over into infinitely many pieces. And those pieces are going to be between reciprocals of natural numbers. So in other words, we can rewrite this as the sum as n goes from 1 to infinity of the integral from 1 / n + 1 up to 1 / n of the floor of 1 /x dx. So we've got something like that. I think that's pretty clear. That's like the integral from a half to 1 and then the integral from a/3 to a half and then a/4 to a/3 so on and so forth. But now we can use this observation that we made down here to replace that fractional part of 1 /x with 1 /x minus n. So let's do that. So we've got this is the sum as n goes from 1 up to infinity of this integral from 1 / n + 1 up to 1 /n of 1 /x - n dx. Okay, good. But now I can evaluate this integral that's on the inside pretty easily using the fundamental theorem of calculus. So that's going to give me this sum as n goes from 1 to infinity.

[00:03:53]Then I'll have the natural log of x evaluated from 1 / n +1 up to 1 / n. And then minus n * x evaluated from 1 n +1 all the way up to 1 / n. Okay. But then that's going to give me the sum as n goes from 1 up to infinity of the natural log of 1 /nus the natural log of 1 / n + 1. And then let's see that's going to be minus. So n * 1 / n is obviously equal to 1. And then we'll have - n / n + 1. Okay. So that's all within my infinite sum. And now the trick here is I want to use the fact that an infinite series is defined as the sequence of partial sums. And I'm going to bring this out into the limit as capital n goes to infinity of my sequence of partial sums. But I've got a huge advantage once I change this to a sequence of partial sums. And that is I can split this sum into pieces. So let's do that. So my first piece will be the sum as n goes from 1 to infinity of the natural log of n + 1 / n. And just FYI, I got that from putting these two terms together using logarithm rules. I think there's not really a ton to that. That shouldn't be that big of a deal. And then we can do a bit of a calculation right here as well. Notice that this is n +1 / n +1 minus n / n +1 just by writing that one as obviously n +1 / n +1. But if we do that difference, we'll get 1 / n +1. But that's wrapped up in a subtraction here. So I'm going to rewrite this as the sum as let's see that's going to be little n goes from 1 to capital n of 1 over little n + 1. So something like that. But now what I'm going to do is take this whole thing and I'm going to add one to it. And I'm going to add one to it and then change this sum so that it starts at zero because notice the zeroth term of that is one but that's attached to a minus sign. So that means to undo that I need to add a one to that. So check it out.

[00:06:17]I've got a sum another sum which is finite and then I've got just my number one. And those are all within this limit. So now what I'll do is I'll bring that plus one outside of the limit because that doesn't have anything to do with my limiting variable. And then I'll use logarithm rules on this first part.

[00:06:37]That'll turn a sum into a product. So that's going to give me the natural log of the product as little n goes from 1 to capital n of n + 1 / n. Good. And then from that we are subtracting this thing right here which is the sum of all of the reciprocals of natural numbers up to capital n plus one. So this is going to be 1 + 1/2 + 1/3 ending at 1 / n + 1 like I just said. Good. But we can do a bit of a simplification procedure on this product. It's actually pretty nice how this simplifies. So let's write out the first couple of terms and maybe the last couple of terms. Notice that the n=1 term gives us 2 over 1. The n=2 term gives us 3 over 2. Then 4 over3. So on and so forth. And notice the last term, the n minus one term, or maybe the next to last term is n / n minus one. And then the last term is capital n +1 / n. But we've got this telescoping action occurring. Notice that this two cancels with this two.

[00:07:52]This three right here cancels with this three right here. This four cancels with something that comes afterwards. This n minus one cancels with something that comes before. And then this n cancels with this n right here. And so that means that all we're left with in here is simply this capital n + one. Okay. So let's see what we've got. So we've got 1 plus the limit as capital n goes to infinity of the natural log of capital n + 1 minus this sum 1 + a half all the way up to capital n + 1 in the denominator. But next up, what I want to do is reindex this sum a little bit. I'm going to replace n with n minus one, which is totally okay because I've got this limit as n goes to infinity. And I'm going to factor a minus sign out as well. So that's going to give me 1 minus the limit as capital n goes to infinity of 1 + 1/2 + 1/3. Now it's ending at 1 / n.

[00:09:00]And then from that I'm subtracting the natural log of n. But that limit has a famous value that we've seen on the channel before. And that is the oiler masheroni constant gamma. So in the end we have a final value for this integral of 1 minus gamma.