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
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!
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.
I have visited this integral on your channel quite a number of times over the last few years and each time the evaluation, the experience of so many Cal 2 topics in one integration problem and the beauty of your explanation never ever fail to impress me. This is how calculus should be taught. Absolutely fantastic evaluation and explanation.
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.
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.
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 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.
Please note. Fubini is exchanging the order of 2 nested integrals. Exchanging infinite sum with integral requires Lebesgue dominated convergence theorem.
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.
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!
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).
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
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 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.
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
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.
20:30 when you've been doing calculus and such for so long you forget how to quickly resolve fractional problems just immediately multiply all sides by 4 to get 1=4+2b-1, isolate B: -2=2b, b=-1
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
You lost me at 18:01 with the "cover-up" method... By imposing n=-1 you are making the fraction tend to infinity, how would that help us find A? Plus, when I try to determine A,B and C with the "usual" method (writing the common denominator of the three partial fractions and comparing to the original fraction to determine A,B,C) I find a system of 4 equations in 3 unknowns, which turns out to have no solutions. What am I missing??
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.
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
If you do the integration by parts the other way around then you can avoid separating the partial fractions (but you do have to integrate ln(x)). Also, I think that proving ζ(2)=pi^2/6 is more difficult than all of the work that precedes it.
At the final board where bprp finds the solution, he makes a mistake after he has calculated ∑_{n=0^ꝏ} [1/(n+1) - 1/(n+2)] = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + 1/4 - ... = 1, since all terms cancel by pairs with the exception of the first one, i.e. 1. So far is all okay, but now, below, he puts again the term [-1/(n+2)], which already was considered in the above sum. Fortunately this term does not lead to error as it tends to zero as n tends to infinite. BTW, the sum 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... = ∑_{n=1^ꝏ} 1/n^2 = ζ(2) is the famous Riemann Zeta function ζ(s) for s = 2. The great mathematician Leonhard Euler was the first out in calculate ζ(2) = π^2/6.
I don't think this problem falls within the Calculus 2 scope though. But it has a lot of Calculus 2 elements in it, that's for sure. I remember I did problems like "replacing integrals by series vice-versa" in a introductory real analysis class....
Ok so let me just say this, havent watched the video yet. Idk if im correct here but lets see. We can see that when we plug in 0 and 1 into lnxln(1-x) we get For 0 -> ln0 times ln1 For 1 ->ln1 times ln0 so in our case in this function 0=1 so boundary can be edited and integral will be _ Integral from 1 to 1 of f(x) which is 0 unless y'know weird subs
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.
In germany calc 2 is Topology, multivariable calculus and vector analysis. We have to do this kind of stuff at the end of calc 1. Had no idea that the curriculum is that different in the US. maybe i'm gonna come over :D
This might be a dumb question but Idc I'm in eighth grade so I'm allowed to ask dumb questions😂😂 Anyway: when u integrated the power series, couldn't u just have taken away the first term(make n start at 1) and divide by n? Just asking
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 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?
@turbo potato: yes it converges. The factor ln(1-x) has a simple root at x=0, which cancels the pole of 1/x, leaving the integrable singularity of ln(x).
We can exploit the labor of our slave: bprp :-) Compared to the video you have an extra 1/x. Everything up to 8:04 goes through the same. Instead of x^(n+1) we have x^n. Then bprp dives into the evaluation of the x-integral of x^(n+1)ln(x). This goes through with n+1 replaced by n, but we have to pay attention at the limit for x=0 discussed at 11:40. It's fine because he has an extra factor of x to spare before the limit becomes problematic. The result bprp shows at 16:38 is in our case modified to +sum_{n=0}^infinity 1/(n+1)^3. Now that's simpler than the video! We re-index the sum to sum_{n=1}^infinity 1/n^3, and directly recognize zeta(3), where zeta is the Riemann zeta function. en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function#Odd_positive_integers This is the result, so we shout "Done!".
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
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
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?
This was the best thing I've ever seen
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!
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.
Wow awesome result, this is why maths makes me happy, integral of logs multiplied together - final answer has pi in it
I have visited this integral on your channel quite a number of times over the last few years and each time the evaluation, the experience of so many Cal 2 topics in one integration problem and the beauty of your explanation never ever fail to impress me. This is how calculus should be taught. Absolutely fantastic evaluation and explanation.
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
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 video has to be my favorite one yet. Thanks for making math enjoyable for all 😁
"DONE!!"
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!
That was epic, excellent job!
Harlequin314159 thank you!!
Watched this at 7 am in the morning. Made my day😍😍🔥🔥🔥🔥🔥
Mohammed Motorwala yay!
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.
Wow. Amazing integral. Beautiful answer to see that identity at the end. Truly awesome for anyone who enjoys calculus
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.
I can't believe you've done it, I'm actually interested in calc for the first time in my life.
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...
Bro the most unexpected result wtf!! Math is really dark magic... And logic ofc🔥❤️🔥
Now that was sensational! Who would have thought that the answer would simplify down to 2 - pi^2/6 , and using Euler.
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"
24:40 the best part
MrQwefty definitely!
And the hardest
Most random thing ever
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!
Till now that is the most beautiful integration I have ever seen.
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.
I see "calc 2 review" in my recommended, and instantly start having flashbacks.
Ohh man this is so hardcore :/
Yea.....
And I think people can see why this is a "calc2 review"
This might just be one of my favorite integrals that I've seen solved
Absolutely amazing! Loved the video, thank you for showing such great math alongside such great work.
Please note. Fubini is exchanging the order of 2 nested integrals. Exchanging infinite sum with integral requires Lebesgue dominated convergence theorem.
Great content as usual. I wasnt on best terms with Frobenius back in college, but this was great.
I have also done by power series i.e. taylor series of ln (1-x) to get the same answer!!!!!
Yay!!!!!
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.
seems useful in physics, but not here
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!
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!!!!!
曹老師你的鬍子好有型啊
Btw this one is pretty insane
Aegis iStatic lol!! Thanks!!
You’re just so cool , I really appreciate your work ... Keep on going ❤️
I'm addicted to this channel
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).
Didn't get formal education for this level of math! But I love this! Had to watch two times to understand properly😍
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
This is so impressive and clever, an awesome video!
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
Damn, that was a great series of things you did on the board 😎
👏calc 2👏review👏
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
Holy oreo! Such twisted integral.. I'm stunned.
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.
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
Good job, you deserve a beer ;)
Seriously, you looked like you needed to unwind :)
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.
20:30 when you've been doing calculus and such for so long you forget how to quickly resolve fractional problems
just immediately multiply all sides by 4 to get 1=4+2b-1, isolate B: -2=2b, b=-1
Ryan Roberson I remember. But I was afraid to run out of space
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.
You lost me at 18:01 with the "cover-up" method... By imposing n=-1 you are making the fraction tend to infinity, how would that help us find A? Plus, when I try to determine A,B and C with the "usual" method (writing the common denominator of the three partial fractions and comparing to the original fraction to determine A,B,C) I find a system of 4 equations in 3 unknowns, which turns out to have no solutions. What am I missing??
Filippo Malgarini before imposing n=-1 he multiples everything by n+1 thats a known method to decompose franctions
Best math video I saw so far!
Hi . Look at this : SUM 1/ [n*(n+1)^2] from n=1 to inf is [ 2 - pi^2/6 ] just the same result you have found.
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
When your professor decides that the final will only have one question.
Beautiful video!!
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
If you do the integration by parts the other way around then you can avoid separating the partial fractions (but you do have to integrate ln(x)). Also, I think that proving ζ(2)=pi^2/6 is more difficult than all of the work that precedes it.
Holy crap didn’t expect this to be the solution. And how did you find out the value for 1+1????
Pfft, I just did it in my head. Super easy!
I enjoyed the journey this took me through
At the final board where bprp finds the solution, he makes a mistake after he has calculated ∑_{n=0^ꝏ} [1/(n+1) - 1/(n+2)] = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + 1/4 - ... = 1, since all terms cancel by pairs with the exception of the first one, i.e. 1. So far is all okay, but now, below, he puts again the term [-1/(n+2)], which already was considered in the above sum. Fortunately this term does not lead to error as it tends to zero as n tends to infinite. BTW, the sum 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... = ∑_{n=1^ꝏ} 1/n^2 = ζ(2) is the famous Riemann Zeta function ζ(s) for s = 2. The great mathematician Leonhard Euler was the first out in calculate ζ(2) = π^2/6.
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
Hey can you go through Calc 1 - beyond, your vids are amazing
I don't think this problem falls within the Calculus 2 scope though. But it has a lot of Calculus 2 elements in it, that's for sure. I remember I did problems like "replacing integrals by series vice-versa" in a introductory real analysis class....
Can we get a Calc 3 review question like this in the future?
A true review indeed.
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.
I like your presentation and content
If we integrate by parts first we can use geometric series expansion
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?
Ok so let me just say this, havent watched the video yet.
Idk if im correct here but lets see.
We can see that when we plug in 0 and 1 into lnxln(1-x) we get
For 0 -> ln0 times ln1
For 1 ->ln1 times ln0 so in our case in this function 0=1 so boundary can be edited and integral will be _
Integral from 1 to 1 of f(x) which is 0 unless y'know weird subs
It seems i was incorrect lol im stupjd
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.
Where do you find such integrals? This was fun!
In germany calc 2 is Topology, multivariable calculus and vector analysis. We have to do this kind of stuff at the end of calc 1. Had no idea that the curriculum is that different in the US. maybe i'm gonna come over :D
This video feel like you fight hardess boss in RPG game. Excellent!
I'm gonna take calculus 2 this semester, do you think they will give problems like this for tests? This thing is so hardcore..
Could you please show some double substitution integrals done by the wolfram app. Can wolfram app do double and triple integrals?
This might be a dumb question but Idc I'm in eighth grade so I'm allowed to ask dumb questions😂😂
Anyway: when u integrated the power series, couldn't u just have taken away the first term(make n start at 1) and divide by n? Just asking
You truly are impressive
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 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?
I have also a question integration of ln(x)*ln(1-x)/x from 0 to 1
let u=lnx
NestorV S i will not work either
Are you sure that it converges?
@turbo potato: yes it converges. The factor ln(1-x) has a simple root at x=0, which cancels the pole of 1/x, leaving the integrable singularity of ln(x).
We can exploit the labor of our slave: bprp :-)
Compared to the video you have an extra 1/x. Everything up to 8:04 goes through the same. Instead of x^(n+1) we have x^n. Then bprp dives into the evaluation of the x-integral of x^(n+1)ln(x). This goes through with n+1 replaced by n, but we have to pay attention at the limit for x=0 discussed at 11:40. It's fine because he has an extra factor of x to spare before the limit becomes problematic. The result bprp shows at 16:38 is in our case modified to +sum_{n=0}^infinity 1/(n+1)^3. Now that's simpler than the video! We re-index the sum to sum_{n=1}^infinity 1/n^3, and directly recognize zeta(3), where zeta is the Riemann zeta function. en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function#Odd_positive_integers
This is the result, so we shout "Done!".
Brutal
I would have been so tempted to get rid of all the n+1 by making n start at 1 instead
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?
BpRp,I want to know about connection of limit to integral .
Please,make videos and explain about this section.
Thanks. .
Very impressive my friend
Very interesting!
Superb !!!!!!!
I now desire u as my calc 2 teacher this semester
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
18:00
Wait, no, the 3rd term only needs to be n+2, not (n+2)^2
Good explanation sir