In this video, I calculate the integral of the fractional part of 1/x from 0 to 1. The end result is really cool, as it leads us to define the Euler-Mascheroni constant
you say you don't know the proof that "this thing" is a finite negative constant, but this is the proof. The fractional part of 1/x is positive and less then or equal to 1 between 0 and 1, so the integral has to be less than or equal to 1. And by the way, that fudge that you did with the M+1 is pretty easily explained, when you swap M for M-1 throughout the limit, you can add and subtract a 1/M term, one of these gets tagged on to the sum and the other can be pulled outside of the limit as it goes to zero.
In other words, the fractional part must vary exclusively between 0 and 1 (not inclusive) on *any* interval of nonnegative numbers, so the result must be less than the constant function y=1 evaluated on that same interval. The integral of this is of course a box with area 1, so we can use 1 as an upper bound for the integral. In a less rigorous way, a lower bound for the integral is 0 because the fractional part of a positive number is positive, and 1/x is positive on (0, 1], so the infinite sum must be greater than 0. Alternatively, because the measure of the domain is nonzero and the range is nonzero at an infinite amount of points, the infinite sum must be strictly greater than 0. Therefore (call the result of the integral R) 0 < R < 1 and we know (from the video) that: limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) + 1 = R so 0 < limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) + 1 < 1 which implies that -1 < limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) < 0 Substitute M = M-1 to get: -1 < limit as M-1 -> inf of (ln(M)-sum from n=1 to n=M of 1/n)) < 0 but M-1 approaches inf at the same rate as M approaches inf, so we can safely replace the "M-1 -> inf" with "M -> inf" without changing the result, leaving us with: -1 < limit as M -> inf of (ln(M)-sum from n=1 to n=M of 1/n)) < 0
Integrating a fractional part of a function is always interesting and non-trivial.It remainds me your previous video on integrating the fractional part of tan x over 0 to pi/2.
if we consider the area under 1/x between n and n + 1, minus the corresponding area of the rectangle with base 1 and height 1 / (n + 1), with n natural going from 1 to infinity, we obtain the same final result: Σ n=1 to inf [ ln (1+1/n) - 1/(n+1) ) ] = 1-ϒ
I tried to generalize this for the fractional part of 1/x^s for s not 1 (which converges as the fractional part is strictly smaller than 1 and the interval is compact). There arises an expression that is the difference of something resembling the Riemann-Zeta-function evaluated at 1/s and the limit as N goes to infinity of N/(1+N)^(1/s). So I'm wondering: are these constants c(s) known constants that I can look up somewhere or do I have to calculate them myself?
This is so cool! I would think adding rational numbers together would give another rational number but weird things happen when you do things infinite times :D
So I have the problem. In physics, the magnetic pull on an electron at a distance d (assumed to be 1 since calculus allows any distance to map to it) while a current flows (again 1) is twice the integral from 0 to infinity of 1/(x^2+1). This integral, you may recognize, is just arctan. But in four dimensions, the formula is 1/(x^3 +1), and I've been trying for a few hours to integrate it, but nothing seems to work. Can you do this? Is it possible to generalize to any dimension? (For 2d, the answer is infinity though, so maybe not)
I know it's really late haha, but isn't that negative sum going from 1/2 up to 1/(M+1)? Instead of 1/M? Then you need an extra correcting factor of +1/(M+1) (i know it goes away in the limit but still)
Here's a simple, related proof for the finiteness of the macaroni constant. Take the left and right Riemann sums of 1/x from 1 to m with the delta being 1. The true area is ln(m)-ln(1)=ln(m) and is bounded by our Riemann sums. The Riemann sums are actually the Harmonic series and the Harmonic series minus 1. I'm going to start writing H_m to indicate the m-th Harmonic number. So, we know that H_m-1
Great video! I wondered what would happen if you use the Taylor series of ln at 1, and I arrived at the expression Gamma = 1- Sum{k=2 to infinity} (sum {i=2 to infinity} (1/(i•k^i))). I then checked this using Excel for k up to 100, and i upto 14, and arrived at a value of 0.582 for gamma. So It seems correct, but it converges quite slowly. I never saw this expression (without ln) for gamma, so could anyone tell me if this is right?
What if we use the original function of 1/x which is ln (|x|) then we have f(x)=1/x then F(x)=ln(|x|) then the integral from 0 to 1 of 1/x turns into ln(1)-ln(0) then ln(1) is 0 which means that the integral is -ln(0) we know that cos(π/2)=0 then we use the Euler is Formula we get -ln(0) equals to ln(i)=(π/2)i
Hi Dr. Peyam, I’m a sophomore in Stuyvesant High School who doesn’t know much of...well, anything about calculus. But I have a question: why could you not make the integral from 1 to infinity and replace the 1/x with just x? Why wouldn’t that give you the same eventual result? There is a one-to-one correspondence in the real numbers in (0,1] and [1,infinity), coming from the reciprocal function, so is there a reason this wouldn’t work? Thanks a ton.
That’s because then you’d have to replace dx with d(1/x), and dx is not equal to d(1/x), you need a technique called u-substitution for that! Also OMG, I know Stuyvesant High School! I went to the Lycée Français de New York, which is close!
Dr. Peyam's Show That’s so cool! I just proctored a math team from the Lycée at the Downtown Mathematics Invitational around a month ago- they were really nice kids. Also, thanks for the explanation!
Nice video, By the way, the decompression in the fractional part of tan(x) video(ua-cam.com/video/0Ris20r3svE/v-deo.htmlm20s ) is still one of my all times favorites things in math
I think there is a mistake. Although you got the correct result, you forgot to add the 1/(M+1) term at 8:38. You are lucky because this term tend to zero as M tends to infinity and that is why you got the correct result.
I came in expecting something a little more mellow after the previous integral of a fractional function video. Then I spent 10 minutes fretting over the limit of the natural logarithm and the summation and the Euler-Mascheroni constant came in.
8:38 and 8:50 small inconsequential mistake. The sum is 1/2 + 1/3 + ... + 1/(M+1) In the limiting process, it makes no difference but just wanted to point it out.
you say you don't know the proof that "this thing" is a finite negative constant, but this is the proof. The fractional part of 1/x is positive and less then or equal to 1 between 0 and 1, so the integral has to be less than or equal to 1.
And by the way, that fudge that you did with the M+1 is pretty easily explained, when you swap M for M-1 throughout the limit, you can add and subtract a 1/M term, one of these gets tagged on to the sum and the other can be pulled outside of the limit as it goes to zero.
beautiful
Goddess Haruhi has spoken.
as expected from the SOS brigade leader
In other words, the fractional part must vary exclusively between 0 and 1 (not inclusive) on *any* interval of nonnegative numbers, so the result must be less than the constant function y=1 evaluated on that same interval. The integral of this is of course a box with area 1, so we can use 1 as an upper bound for the integral.
In a less rigorous way, a lower bound for the integral is 0 because the fractional part of a positive number is positive, and 1/x is positive on (0, 1], so the infinite sum must be greater than 0. Alternatively, because the measure of the domain is nonzero and the range is nonzero at an infinite amount of points, the infinite sum must be strictly greater than 0.
Therefore (call the result of the integral R) 0 < R < 1 and we know (from the video) that:
limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) + 1 = R
so
0 < limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) + 1 < 1
which implies that
-1 < limit as M->inf of (ln(M+1)-(sum from n=1 to n=M+1 of 1/n)) < 0
Substitute M = M-1 to get:
-1 < limit as M-1 -> inf of (ln(M)-sum from n=1 to n=M of 1/n)) < 0
but M-1 approaches inf at the same rate as M approaches inf, so we can safely replace the "M-1 -> inf" with "M -> inf" without changing the result, leaving us with:
-1 < limit as M -> inf of (ln(M)-sum from n=1 to n=M of 1/n)) < 0
Well, one sidedly between 0 and 1 (zero included), and that bit isn't very important, so yes. :)
By the way, this video is actually a proof of the finiteness of this constant because 0< {1/x}
I love math so I try to learn and improve my English language with your math videos
Is there anything impossible for Dr. Peyam?
Integrating a fractional part of a function is always interesting and non-trivial.It remainds me your previous video on integrating the fractional part of tan x over 0 to pi/2.
Shanmuga Sundaram
NO! Do not bring that up
I really enjoy in your videos how you manage to cancel an infinite number of terms or combine them into a simple product/sum!
You made a spoiler in the description! You should have warned me about that!
Omg, you realy did this video. Thank you!
Others only mimic but this channel discover its own.that's why I like this channel most
Congrats on 10K subss
Einfach grandios, wunderprächtig erklärt. Vielen Dank, Dr. Peyam. :-DDD
Awesome!!! Thank u, a little detail, the first big parenthesis have to be put in front the limit In your final result
It'd cool a dedicated video for the Euler-Mascheroni constant
if we consider the area under 1/x between n and n + 1, minus the corresponding area of the rectangle with base 1 and height 1 / (n + 1), with n natural going from 1 to infinity, we obtain the same final result: Σ n=1 to inf [ ln (1+1/n) - 1/(n+1) ) ] = 1-ϒ
I tried to generalize this for the fractional part of 1/x^s for s not 1 (which converges as the fractional part is strictly smaller than 1 and the interval is compact).
There arises an expression that is the difference of something resembling the Riemann-Zeta-function evaluated at 1/s and the limit as N goes to infinity of N/(1+N)^(1/s).
So I'm wondering: are these constants c(s) known constants that I can look up somewhere or do I have to calculate them myself?
This is so cool! I would think adding rational numbers together would give another rational number but weird things happen when you do things infinite times :D
But none of the ln(M)'s will be rational.
@@TheRealSamSpedding technically you're wrong. Exactly one of them will be rational, namely ln(1).
No wait... ln(M) is in the limit term as M goes to infinity and not in the sum. How the hell does this thing converge...
It converges because both the sum and the ln go to infinity as M goes to infinity, and so their difference can converge
So I have the problem. In physics, the magnetic pull on an electron at a distance d (assumed to be 1 since calculus allows any distance to map to it) while a current flows (again 1) is twice the integral from 0 to infinity of 1/(x^2+1). This integral, you may recognize, is just arctan.
But in four dimensions, the formula is 1/(x^3 +1), and I've been trying for a few hours to integrate it, but nothing seems to work. Can you do this? Is it possible to generalize to any dimension? (For 2d, the answer is infinity though, so maybe not)
Dr Peyam has made a video about the integral of 1/ (x^n+1), basically, the general case, just look for it
Little error in the sum, you forget 1/(M+1) and no more problem with ln(M+1) and ln(M).
Nice vid!
beautiful
I know it's really late haha, but isn't that negative sum going from 1/2 up to 1/(M+1)? Instead of 1/M? Then you need an extra correcting factor of +1/(M+1)
(i know it goes away in the limit but still)
8:36
Shouldn't it be the sum to 1/(m+1) instead of just 1/m?
Is this how the numerical value of that constant is determined? By numerically doing this integral rather than just finding the harmonic numbers?
Harmonic is easier, I think! The integrand is pretty badly behaved!
Here's a simple, related proof for the finiteness of the macaroni constant. Take the left and right Riemann sums of 1/x from 1 to m with the delta being 1. The true area is ln(m)-ln(1)=ln(m) and is bounded by our Riemann sums. The Riemann sums are actually the Harmonic series and the Harmonic series minus 1. I'm going to start writing H_m to indicate the m-th Harmonic number. So, we know that H_m-1
Nice to be the first to see this :-)
👍👏👏👏
Great video! I wondered what would happen if you use the Taylor series of ln at 1, and I arrived at the expression
Gamma = 1- Sum{k=2 to infinity} (sum {i=2 to infinity} (1/(i•k^i))).
I then checked this using Excel for k up to 100, and i upto 14, and arrived at a value of 0.582 for gamma.
So It seems correct, but it converges quite slowly.
I never saw this expression (without ln) for gamma, so could anyone tell me if this is right?
You are correct, that expression converges exactly to gamma, according to Wolfram Alpha.
@@kikones34 Thank you!
Very nice video Dr 3.14159 .......m.Thanks with sincere regards.
What if we use the original function of 1/x which is ln (|x|) then we have f(x)=1/x then F(x)=ln(|x|) then the integral from 0 to 1 of 1/x turns into ln(1)-ln(0) then ln(1) is 0 which means that the integral is -ln(0) we know that cos(π/2)=0 then we use the Euler is Formula we get -ln(0) equals to ln(i)=(π/2)i
Superb
Is the gamma constant inspired by this integral?
Nope
I Love You, Brazil
Spoiler in description!
Amazing video though!
🙏🌺🙏
Preety nice
Dr. Tigre Peyam !!!
Didn't get much 😅 as I'm a beginner btw really enjoyed this video ...😌
Thanks @blackpenredpen 🔥🎆
Hi Dr. Peyam, I’m a sophomore in Stuyvesant High School who doesn’t know much of...well, anything about calculus. But I have a question: why could you not make the integral from 1 to infinity and replace the 1/x with just x? Why wouldn’t that give you the same eventual result? There is a one-to-one correspondence in the real numbers in (0,1] and [1,infinity), coming from the reciprocal function, so is there a reason this wouldn’t work? Thanks a ton.
That’s because then you’d have to replace dx with d(1/x), and dx is not equal to d(1/x), you need a technique called u-substitution for that!
Also OMG, I know Stuyvesant High School! I went to the Lycée Français de New York, which is close!
Dr. Peyam's Show That’s so cool! I just proctored a math team from the Lycée at the Downtown Mathematics Invitational around a month ago- they were really nice kids. Also, thanks for the explanation!
Great to hear they have a math team now, that’s so cool!!!
Or you can just break the integral into 0 and 1/n and 1/n 1 and do the limit as n->infinity
O this shouldn’t be that bad compared to fractional tangent
*shudders just thinking of that flashback*
I got it right.
Oh 7:18. When you get annoyed and have to delete it multiple times but imagine it being on the paper... then its just a mess that you... dispose
Nice video,
By the way, the decompression in the fractional part of tan(x) video(ua-cam.com/video/0Ris20r3svE/v-deo.htmlm20s ) is still one of my all times favorites things in math
I think there is a mistake. Although you got the correct result, you forgot to add the 1/(M+1) term at 8:38. You are lucky because this term tend to zero as M tends to infinity and that is why you got the correct result.
Hello Doctor,
Thank you for everything,
Can you solve the
Integral from 0 to 1
{1/x}. {1/x-1}
At least give an aproximate value for gamma :( I feel blueballed now
0.577215665...(oeis.org/A001620 )
Fractional Fart of 1/x
I came in expecting something a little more mellow after the previous integral of a fractional function video.
Then I spent 10 minutes fretting over the limit of the natural logarithm and the summation and the Euler-Mascheroni constant came in.
8:38 and 8:50 small inconsequential mistake. The sum is
1/2 + 1/3 + ... + 1/(M+1)
In the limiting process, it makes no difference but just wanted to point it out.
Prove that the constant is irrational pls.
Open problem
@@drpeyam heh.. I hoped that if you didn't know about that you would try to prove it)