Can you solve this integral?
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- Опубліковано 5 лип 2022
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π² also cancels with the 9
Ah yes, engineering
Noooo it's pi^2 = g = 10
@@boringextrovert6719 but 10/9 is one(why am i commenting this)
heheh
@@aweebthatlovesmath4220 10/9 is 1 in java B)
Thumbnail should be x^3 instead of x, probably because of font colour(makes it hard to see)
Bruuuuh I'm looking for the error in my calculations for like an hour now. Because I thought it's ln(x)/(x+1) and according to my calculations, that would not converge. Bruuuuuuuh. I can't believe this. Anyhow, now at least I know where the problem was xd
Starting to think he does it for the algorithm, haha. People will comment on the video just to point out typos = higher engagement.
Has anyone ever seen Michael and the "Integral Suggestor" in the same room at the SAME TIME??? Hmmmmm........
I've never heard of the dominated convergence theorem. FUCK this was a great problem
thanks as always. I'm gonna buy some merch if you have any now
Ah no way a question I can actually answer!
No, I cannot solve this integral
Amazing juggling with infinite series in the end!
I like this idea of integral inversion through substituting u = 1/x. I would enjoy a video that gave some more examples of that strategy, and/or other types of inversion. It reminds me of the pole/polar inversion of a circle by (iirc) Apollonius.
Why do we need a substitution? Just use integration by parts directly: integrate x^m, then differentiate ln x and both are just simple power integration. Kind of an overkill at the start.
We then have to find lim x-0 x^(m+1).lnx
@@advaykumar9726 You can just use l'hôpital on that with lnx/x^-(m+1)
@@tijmenvanderree487 * ln(x)/x^[-(m + 1)]
You don't need a substitution, either method works. I don't see it as overkill, either pathway is a bit troublesome.
What does he mean by invert..didnt anyone else find that unclear and indont see AT ALL WHY anyone would think of thst..and I thought of integration by parts..hiw with x3 plus q is not tbe same as x^m ..ingues because you can transform x^3 + 1 into (×^3 +1)^-1..is that what ppl are thinking I surmise?
Thank you, professor.
Another way to do the second integral of the tool is to recognized that is the mean of the exponential distribution.
It's a little tricky since it's not normalized but surely it is simpler than "integration by part" methods
There is a mistake in the thumbnail. In the thumbnail, the denominator of the integrand is x+1. It should be x^3+1.
Oh, he changed it. I was so sure that just yesterday it said x + 1 and not x^3 + 1 !
Let F(m) = int(x^m) from 0 to 1. Differentiate with respect to m and get the int(x^m ln(x)) from 0 to 1. Since F(m) is easy to integrate you can then take its derivative and get the integral identity desired = to -1/(m+1)^2.
Nice! Classic application of Feynman's integral trick.
Great intégration 😛👌 and so tricky 🤓🦊
You are amazing 🔥
I naturally went to separate the integral in 2 parts so x would be between 0 and 1 so I could use the sum of a convergent geometric series. I even remembered the dominated convergence theorem from 2 decades ago to justify why you could swap the sum and integral. But then I could not sum the series. I went to write the n=0,1,2,3 terms to get a feel of the sum, but could not get a good enough picture. I thought about grouping, adding and subtracting terms in the hope to get to the inverse squares sum. Too vague an idea. So here I am, watching the video to see the trick. Figuring the partition of odd and even numbers was the key to completing the sum! From there I got the result (even though in a less elegant way as I set to calculate the even and odd parts instead of doing some last grouping. Cumbersome, but I got there). That was very technical, thanks!
تحياتي لكم.
صراحة مايكل أستاذ صبور واصل.
Greate video! I got answer - 4π²/27 using residuals. Could you try same example using complex variables and residuals, please!
I miss the empty squares beside the lemmas which you fill in 🙁
You can prove the identity by a simple integration of parts. \int x^m*log(x)dx=x^{m+1}log(x)/(m+1)-\int x^m/(m+1)dx
I really love your videos with integrals.
I would like to see your approach to solving the integral (e^(2x)-1)/sqrt(e^(2x)-(x+1)^2)
There's a link where you can send your suggestions under the video. If it's not late.
@@thesecondderivative8967 well, thanks, but I have figured it out
I decided to turn this into a beast of a problem and evaluate the generalized version of this integral, being the integral of log(x)/(x^n+1) from 0 to infinity for integers n>1. I did it using contour integration from complex analysis. The result is -pi^2/n^2 * csc(pi/n) * cot(pi/n). I had to evaluate the integral of 1/(x^n+1) from 0 to infinity as part of it, which is pi/n * csc(pi/n). It was interesting to see how he evaluated this using entirely real methods in the video.
I wasn't thinking about it at the time, but this is actually valid for any real n>=2, not just integers.
I actually stumbled upon this random integral by my own messing around. Apparently you can also prove it with the digamma and gamma functions
When you had the 1-x factor it made me want to factor the denominator, cancel out 1-x and then use partial fractions
yes, but the denominator factors as (x+1)(x^2 - x + 1) so no cancellation. Close though!
The forbidden tabular method: maths teachers are malding everywhere
4:46 how did you erase the board? Please teach me how to do it
For the substitution of x^3 into the summation, x must be less than 1 but the integration goes to 1. How do you resolve this please?
The thumbnail doesn't fit the integral in this video
integral clickbait.
Someone who tried to use complex analysis and the residue theorem? When I tried the integral cancelled itself.
Why is it important to add and subtract those summand???is it a must?
19:54
Is this a special different integration technique with conditions? To convert integrand into a sum ( or an antiderivative) and switch order of integration and summation (or order of integration)?
It follows from dominated convergence theorem
Can also be solved by taking doing a contour integral of ln^2(z)/(z^3+1)
Use the Keyhole contour with a branch cut along the positive real axis, and compute the residues at the three poles. The integrals along the small and large circles go off to zero after the proper estimations are made.
You can collect the ln(x)/(x^3+1) integral and you also get the integral of 1/(x^3+1) for free after playing around with the integrals along the real axis. After that take real and imaginary parts of both sides and you should be good to go!
It can be done with a simpler one third circle contour. This is particularly desirable for higher powers of z in the denominator, and doesn't require a higher logarithmic power.
Oh yeah I forgot about that one, yeah that one is a lot easier to work with and two less residues to include!
you can do countour integration on f(z) = log(z)/(z^3+1) with a indented circular sector of opening 2Pi/3, again this will give as a byproduct the integral of 1/(x^3+1) (the only residue you need to compute is the one in e^(i Pi/3) and it is pretty simple)
I’m not super skilled with complex analysis, so how did you know this could be done? I can work through the math for myself, but it’s difficult for me to see which contours to use or when it’s better to use complex tricks.
@@JM-us3fr most indefinite integrals on unbounded domains can be evaluated with complex contours with the help of a result called Jordan's lemma (which holds for semicircular domains). For other kind of results (like for instance the one considered here) one needs to do it by hand, but the number of "useful" contours it is not so wide. You just need practice in recognising possible symmetries of the functions. A good reference is Serge Lang's book "Complex Analysis", where in Chapter VII there are several examples of this kind (the integral in this video is Exercise 8 in paragraph VII.4).
Very cool! If you replace the 3 with an n, then you get that the integral of ln(x)/(x^n+1) from 0 to infinity = C_n*pi^2, where C_n is a constant depending on n. For example, this video shows C_3 = -2/27. And following the video with different n shows C_2 = 0, C_4 = -1/(8sqrt(2)), C_5 = (-1/125)(5+3sqrt(5)).
And in general, it seems that C_n = (-1/n^2)cot(pi/n)csc(pi/n), though I can't work out the sum yet for general n like you did in the video. I suspect some Taylor/Laurent series trick?
Where did this integral come up for you? I'd love to see if there is any interpretation to C_n.
Another fun thing you can prove is that as n goes to infinity, C_n goes to -1/pi^2.
I think this works: Take the Laurent series about x = 0 of cot(x)csc(x), which has a single term with negative index: 1/x^2.
Plug in x = pi/n and we get (smaller order terms) + n^2/pi^2. So the coefficient C_n = (-1/n^2)cot(pi/n)csc(pi/n) has Laurent series (negative index terms) - 1/pi^2, so as n goes to infinity, C_n goes to -1/pi^2.
Which is cool because this means that as n goes to infinity, the integral from 0 to infinity of ln(x)/(x^n+1) approaches -1.
But it's not like a point function, it's a basically flat for x > 1 and equal to ln(x) for x < 1. And the integral from 0 to 1 of ln(x) is exactly -1.
Found a video of someone working this out via complex analysis! ua-cam.com/video/Sj8IJOBK33w/v-deo.html
I still wonder whether you could somehow convert the sums you get into C_n...
@@sk8erJG95 I found a series representation of the integral with the n using Feynman's trick and some regular calculus techniques. And from there, with the help of Wolfram Alpha, using Herwitz Zeta function and an identity linking the Trigamma function and Trigonometric functions, I actually showed that
C_n = (sec^2 (pi/2n) - csc^2 (pi/2n))/(4n^2)
Which should be equal to your formula using trig identities.
8:30 that's a good place to...start 😳
Why not integration by parts directly
If you want substitution anyway why not substitution which leads us to the Gamma function
The general solution for m>1 (m is integer) & logx being the natural log
Integral( logx / (1+x^m) ), x=0 to infinity =
- (pi/m)^2 cot(pi/m) csc(pi/m) + constant
a calc problem with an ‘emu’ !
I thought the thumbnail had x² in the denominator and not x cubed... so I worked it out in my head cuz it was only two integration by parts and felt like a genius :p
Fabulos!
hey micheal, viewer from india ,really love the work you are doing, just one suggestion pls try to take up some problems on counting aka permutations and combinations and elementary probability(conditional probability, bayes theorem etc) , thanks
this is also ∫_{-∞}^∞ exp(x)*x/(exp(3x)+1)dx. try the residue theorem to evaluate this one.
Hi,
8:30 : good place to start a next board.
great
Is the Integral Suggestor the only person whose problem suggestions get turned into videos? I can’t recall hearing that a problem was suggested by someone else…
if you have good ear you can ear a child playing twinkle twinkle little star in the background ...
The title of the video does not match the actual integral? In the video title, the cube is missing.
1:33 I didn't make a comment, but I did go through the whole process. I instead used u=x^k, k=(m+1)/2 as my substitution, taking advantage of the fact that I can raise the inside of the natural log to that power if I divide the outside by it, and x^k•x^(k-1)=x^m
Then, the integral becomes (1/k²) times the integral from 0 to 1 of u•ln(u) which is pretty simple to solve. I did guess and correct
thumbnail?
Do your kids play the violin?
thumbnail says ln(x)/(x+1)
Condition : m not equal to -1
emu
That's a yikes from me. Just generalize to ln(yx)/(x^3+1), take the derivative and exchange the integration and derivation. Bam, the logarithm is gone, now just shut up and calculate.
You don't "solve" integrals. Definite integrals are *evaluated.*