Integral of so many things! (great for calculus 2 review)
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- Опубликовано: 12 янв 2018
- I will just call this "THE CALC2 REVIEW" since this involves sooooo many skills from calc 2. I integrate a power series, did integration by parts (with the DI method, of course), used the L'Hoptial's Rule, Partial fractions, saw a telescoping series, and ended up with some pi^2/6.
this explains "build up the power" part: • This is how we partial...
integral of ln(x)/(x-1), also uses series, • integral of ln(x)/(x-1... ,
sum of the reciprocals of squares,
by Max. Z: • Proof by intuition don... ,
by Dr. Peyam: • Video ,
And here's the Fubini's Theorem, en.wikipedia.org/wiki/Fubini%... . In fact I made a mistaking quoting it to switch the integration and the summation. We can switch the integration and the summation because of the power series is uniformly convergent. Thus, the integral of the sum is the sum of the integrals, even with infinitely many terms.
ENJOY THIS RIDE!
Integral of ln(x)*ln(1-x) from 0 to 1,
blackpenredpen,
Math for fun,
trying to solve it by symmetry with a u-sub of u=1-x lets you conclude that the integral is exactly equal to itself. Amazing.
@@TejasKd221B You forgot to distribute a negative sign, I did the same.
Tejas Acharya wtf, it's easy to see that's wrong. Both of the factors are negative in the interval so the product is positive hence the integral also is
#horseshoemaths
@@rayquaza1vs1deoxys I remember that meme. Lmao
huh i got the integral of ln(u)*ln(-u) when i let x --> x-1/2
I would start by looking at it and doing a U sub. Then cry and go to sleep
The derivative of sadness with respect to frustration is failure.
I changed one of the xes to a y, and did a double integral dxdy like the way you'd integrate the normal function. I got the same answer. Took like 2 minutes.
wait, thats illegal!
No u.
This was the best thing I've ever seen
An integration marathon! I love examples like this that demonstrate several different methods of problem-solving. Well done, sir!!
alkankondo89 my pleasure!
Agree totally! Problems like this are what helps to keep your mental agility on the ball!
Thanks! By the way, does your username have any kind of mathematical significance?
"DONE!!"
The two most complex things in the universe:
A: The meaning of life the universe and everything = 42
B: The value of the above integral = 2 - π²/6
the most complex thing in the universe is none other than i.
Hey man, may I ask you a stupid question? Where did you get pi and ^2 on the keyboard?
I got them from google and cut and paste :)
Thanks, man
mrBorkD that sounds like a girl would say: bcs girls are just so confusing O.o
Wow awesome result, this is why maths makes me happy, integral of logs multiplied together - final answer has pi in it
Excellent job!
Much appreciated you allowed us to witness the ongoing exhaustion and lack of concentration without covering it up by jump cuts etc. This shows that maths is sometimes real work. But you can still succeed in the end, if you don't give up.
Thank you for your comment!!
It's kinda tricky. I have edited many of my videos previously mainly to shorten them with a hope that more people would like to click on it and watch it. Right now I am trying just the raw files to see how my viewers react to them. I thank you for your input!
After hearing the idea of series expansion, I was able to solve till the end and achieved the same answer! Which is nice. I can't remember solving such definite integrals. The only such thing I clearly remember from the university is how to calculate the Euler integral.
this is seriously one of your best vids! i love it, keep up the good work, you seriously are one of my favourites youtubers man!
24:40 the best part
MrQwefty definitely!
And the hardest
Most random thing ever
This video has to be my favorite one yet. Thanks for making math enjoyable for all 😁
Absolutely amazing! Loved the video, thank you for showing such great math alongside such great work.
Ohh man this is so hardcore :/
Yea.....
And I think people can see why this is a "calc2 review"
Wow. Amazing integral. Beautiful answer to see that identity at the end. Truly awesome for anyone who enjoys calculus
That was epic, excellent job!
Harlequin314159 thank you!!
Definitely do more vids like this where you include a bunch of techniques and explain them all briefly!
brilliantly done....hats off....what an idea especially using power series...I love the way you said "DONE"
👏calc 2👏review👏
You’re just so cool , I really appreciate your work ... Keep on going ❤️
just finished calc 3... Imma still watch it for funsies
Yay!
Calc 3 was so much fun. I thought it was even more fun than Calc 1 and Calc 2!
I see "calc 2 review" in my recommended, and instantly start having flashbacks.
I used this method to solve your previous video's integral
Ln(1+x)/(1+x^2).
Compared to your trig substitution and FlamMath parametric method, plugging a series felt like brut force, but interesting things happened...
I believe I like this about the most of all your integration videos. There are a lot of principles used here - which I last considered over three decades ago.
i love it when you do challenging problems. lately you have done a lot of easier problems. but great video!
This is so impressive and clever, an awesome video!
I guess this is why teachers give 5 hours to do 4 problems. LMAO!
Don't understand any of this, but I enjoyed watching it.
EDIT: Now that I have partially finished AP Calculus AB, I sort of understand what you are doing now.
Hey bprp, I love your videos, really big fan! Hope that you can always post amazing videos like this. I recently learnt about the gamma function and I found that i factOREO is really hard to solve. Could you sometimes do a video on it? Love from Hong Kong
Wow, this was a crazy combo of concepts. First, reverse engineering that series expansion from the derivative of ln(1-x), flipping the order to make em both under the integrand to do IBP, L'Hop shenanigans, and partial fractions to telescope and famous result the way to an answer. I just saw that one video where you proved the pi^2/6 series, so that was pretty cool.
Great content as usual. I wasnt on best terms with Frobenius back in college, but this was great.
Please note. Fubini is exchanging the order of 2 nested integrals. Exchanging infinite sum with integral requires Lebesgue dominated convergence theorem.
Beautiful video!!
This might just be one of my favorite integrals that I've seen solved
Watched this at 7 am in the morning. Made my day😍😍🔥🔥🔥🔥🔥
Mohammed Motorwala yay!
Damn, that was a great series of things you did on the board 😎
I can't believe you've done it, I'm actually interested in calc for the first time in my life.
I have also done by power series i.e. taylor series of ln (1-x) to get the same answer!!!!!
Yay!!!!!
Now that was sensational! Who would have thought that the answer would simplify down to 2 - pi^2/6 , and using Euler.
I'm addicted to this channel
Till now that is the most beautiful integration I have ever seen.
Pfft, I just did it in my head. Super easy!
I dropped a like just as a token of appreciation for the few microseconds you were begging for death through that 25 minutes. I salute you.
Holy oreo! Such twisted integral.. I'm stunned.
Best math video I saw so far!
Didn't get formal education for this level of math! But I love this! Had to watch two times to understand properly😍
Bro the most unexpected result wtf!! Math is really dark magic... And logic ofc🔥❤️🔥
I like your presentation and content
I enjoyed the journey this took me through
Good job, you deserve a beer ;)
Seriously, you looked like you needed to unwind :)
Very interesting!
Superb !!!!!!!
A true review indeed.
Hey can you go through Calc 1 - beyond, your vids are amazing
Mind blowing! Very nice, B-Pen....Euler is smiling right now
BTW, i solved it in a different way as usual :)
I'm not really going to write each and every step tho!
Anyway, i used integration by parts on the main integral. After 3 (fun) limits and another integration by parts and one u-sub i came up with:
INT [ln(x)*ln(1-x)] = 2- INT [ln(1-x)/x] (all INT from 0 to 1 of course)
Finally it was quite straightforward to solve the last INT with the power serie.
A very challenging problem indeed!
Yay!!! Thank you for your nice comment!! I am smiling now too!!!!!
We can roughly approximate the function y = ln(x)*ln(1-x) by a semicircle.
The graph of the function looks like the top half of a circle with center 1/2 and radius 1/2 with an area of 1/2 * pi * (1/2)^2 = pi/8
In fact, if we multiply the semicircle by the constant C = (48-4*pi^2)/(3*pi) which is approximately 0.9, then this integral would have the same area as the semicircle.
Can we get a Calc 3 review question like this in the future?
Thank you.
曹老師你的鬍子好有型啊
Btw this one is pretty insane
Aegis iStatic lol!! Thanks!!
Good explanation sir
Thanks!
Very impressive my friend
Brutal
Great 👍👍👍💯
I'm just an fresh highschool student in Korean, and I have a quite interest in math..
but thanks to your video, I quite likely understanded this video
When your professor decides that the final will only have one question.
Now, could we find another way to calculate this integral, and then deduce the famous result sum(1/n²) = π²/6? Some trig substitution maybe?
I think I have found the easier way. First, notice that Ln(x) Ln(1 - x) = -0.5*((Ln(x) - Ln(1-x))^2 - Ln(x)^2 - Ln(1-x)^2). Integrals from the logs squared are trivial and one can take it by integration by parts e.g. \int_0^1 Ln(x)^2 dx = - \int_0^1 (x 2 Ln(x) /x ) dx = - 2 \int_0^1 Ln(x) dx = 2. Same answer is for the integral from Ln(1-x)^2. Somewhat tricky part is the integral from -0.5*(Ln(x) - Ln(1-x))^2 = -0.5*(Ln(x/(1-x)))^2 = -0.5 d^2(x^t (1 - x)^(-t))/dt^2 at t = 0. In the last line I used that Ln(x) = d(x^t)/dt taken at t = 0. As we have the second power of the log, we need the second derivative. But the integral \int_0^1 x^t (1 - x)^(-t) dx = pi*t/sin(pi*t) where I used the definition of the beta-function. In order to take the second derivative, notice that sin(z)/z = 1 - z^2/6 + ... i.e. z/sin(z) = 1 + z^2/6 +... As we only need the second derivative at zero argument, it is enough. So, we get \int_0^1 -0.5*(Ln(x) - Ln(1-x))^2 dx = -0.5* d^2(\int_0^1 x^t (1-x)^(-t) dx)/dt^2|(t=0) = -0.5 d^2(pi*t/sin(pi*t))/dt^2|(t = 0) = -0.5*pi^2/3 = -pi^2/6.
No you only have to show why 1/1^2 + 1/2^2 + 1/3^2 + ... equals pi^2/6 :D
That would have been another 20 minutes.....
I was trying so hard to fit everything on just one board and I was happy because I did it!
*erasing the c=0 part doesn't count.
I was joking anyway :D
Ok, here is the deal. Whenever I write something serious in your comment section I explicitly say so :D
AndDiracisHisProphet : what about this comment itself?
Good question
Another approach
Substitute t = -ln(x)
Calculate Laplace transform of ln(1-e^{-t})
Use derivative of original formula L(tf(t)) = -d/ds F(s) where F(s) is Laplace transform of f(t)
Plug in s = 1
You truly are impressive
blackpenredpenbluepen
This integral is the convolution of ln(x) and ln(x) at 1. Thus, we can use Laplace-transform to calculate the integral, as the inverse Laplace-transform of the square of ln(x)'s Laplace-transform. I think this method is much easier.
This video feel like you fight hardess boss in RPG game. Excellent!
Beautiful.
I think instead of using integration by part, it is possible to transform the integral to a gamma function with a change of variable y=-ln(x).
Where do you find such integrals? This was fun!
a great source for integral solutions, including unique solutions and definite integral solutions is a book called the Table of Integrals, Series and Products by Gradshteyn and Rhyzhik
This video made me realize just how much I forgot of calc 2. Hopefully I don't need it too much in calc 3 and diff eq
How is the interchange of limiting operations at 7:06 justified? I’m pretty sure term-by-term integration is only valid for power series, and the integrand in the integral preceding 7:06 does not appear to be a power series. Am I overlooking something?
similar method but faster in my opinion: turn ln(1-x) into a series to get sum of 1/n the integral of x^nln(x)
now name that integral I_n or something you can find a recurrence relation between I_n+1 and I_n
(obviously since I_n keeps a constant sign it is not identically zero) by that you can deduce that I_n = -1/(n+1)^2
going back to the initial result we get sum of 1/n*(n+1)^2
then finally by decomposing we end up getting our result 2- pi^2 /6
I want to ask.. why is 1/(n+1) constant such that you can take it out of the integral? Since there’s a range of values of n for the summation. Thank you
Nice solution
Good job! Maybe I miss something, is it possible to make a video on that part of the partial fractions? I didn't well understand why you divided the fraction adding the terms (n+1) and (n+2)^2
Me to.
I made a video today on that. It will be up by Wed or so.
Hey love your videos, I have a problem which i think would be a cool video. It goes like this "find the real and imaginary parts of (sinx)^i, I have don't know how to do it! Keep up the great work
How to show that sinh=(e^x-e^-x)/2? can't find it anywhere
Thanks Sir
Ved Prakash my pleasure!
Wow, I got a bit lost when you calculated the numerators of the series. I guess in engineering we don't learn how to solve series properly. May you recommend any book or source that I could use to gain insight into series' calculus.
So great video anyway.
Drop the marker! Well done!
beautiful
Thank you!!!
BpRp,I want to know about connection of limit to integral .
Please,make videos and explain about this section.
Thanks. .
I now desire u as my calc 2 teacher this semester
Well, I have a question.
I know that lim [x-> 0] ( x^x ) =1
but, can I say that 0^0 =1 or is it indeterminate?
Could u solve it using Laplace transforms
Holy crap didn’t expect this to be the solution. And how did you find out the value for 1+1????
Like music to my eyes and math to my ears.
The 1st thing, if it strikes in your mind then question is simple and if it doesn't then it is really hard to solve it.
Can this integral be done by transforming it into a double integral?
Goodness me, that was satisfying
what is the theorem at 6.46??? I couldnt hear claerly
14:39 LOL
I love you 😻