I do see all of your videos from India. I'm 63 year old. Whenever I see your video I always feel sorry for that I didn't get a teacher like you in my yearly life. You are such rare class of teacher who can make the learning fun and enjoyable.
@@aniruddhaghosh1303 thar is a nice tho Mmmm, well, a clue that I can give you is that, don't think on a function world accepted by school Just, plug the same varibles and use the quadratic formula for find out y [f(x)] That is something that works out
Hey bprp I am hosting a Integration Bee in my school and I included many integrals from your 100 integrals, so glad that you provide such good resources!
Feynman's technique works fine. I set up I(x) = the same integral where the integrand is multiplied by exp(-xt). Take the second derivative which is easy to find (just a bunch of exponentials to integrate), and then using that I''(x), I'(x) and I(x) approach 0 when x goes to infinity, you can find I(x) by integrating twice. If I didn't mess up, you get: I(x) = x•ln[x(x+2)/(x+1)²] + 2ln[(x+2)/(x+1)] which leads to the expected result I(0) = 2ln(2) = ln(4) (technically that final calculation is a limit calculation because of the first term, but it works fine).
So nice to recall my undergraduate calculus classes. I studied at the Technion, Israel and you are certainly on the level to teach there. Thank you so much for your wonderful explanations!
@@alphazero339you’d have to prove(or already know) that the function f(x,y) is integrable in the product measure space. Meaning that when you integrate |f(x,y)| over XxY with respect to the product measure on XxY(these things have to be properly defined using measure theory), then this gives a finite value. Then we can take the integral of f(x,y) (without absolute values) over the product measure space and evaluate it as a double integral and exchange the order in which we integrate. In practice, we can more easily use Tonelli’s theorem here: If f(x,y) is non-negative and measurable then we always have this equality, but the integrals may not be finite. In the video’s case, the exponential function is always non-negative and is measurable so this works and no need to even verify Fubini!
@@almightysapling if so then fair, but still something worth thinking about. And even Tonelli needs the basic justification that the functions are positive and measurable which they are
My calc 3 professor put this as a challenge problem, and you saved me! He might've watched your video lol to be reminded of this technique and give to us.
I love math, especially when such beautiful puzzles and solutions came out, that's just gorgeous, so beautiful and awesome, Please never stop seeking for such brilliants of math, that is indeed joyful thing)
Feynman's trick works really well if you parameterize the integrand as ((1-e^(-tx))/x)^2 Its derivative ends up being a constant function after an integral substitution, but you need to use feynman's trick a second time to find the constant
@@tommyliu7020 Yes, I did a first IBP letting dv = 1/t^2 and u = [1-e^(-t)]^2; then I've repeated this procedure with the resulting integral, obtaining -2 ∫_0^∞ [2e^(-2t)-e^(-t)] ln t dt
Integrate by parts (differentiate the top, integrate the bottom) to get I = ∫ (-2e^-2t + 2e^-t)/t dt. Let x = e^-t to get I = 2 ∫[0,1] (x-1)/(lnx) dx, which can be solved by Feynman's.
To use Feynman's technique you first need to Differentiate the numerator and Integrate the denominator using IBP, then use the technique on the resulting integral.
I used to think I was smart but don't think I would have ever thought of that solution method. It was originally thinking of utilizing hyperbolic cosine and sine
I solved the indefinite integral normally by expanding it out and got the 3 normal integrals by itself and they were actually quite easy to solve. You get a slightly complicated expression involving Ei function. I did it this way to simplify it, but when I took the limit from zero to infinity, the infinity part became 0 and the other zero part was quite tricky as I had to solve for lim as t->0 of 2Ei(-t)-2Ei(-2t), I had absolutely o idea how to do this as this was in the form infinity - infinity indeterminate. I used wolfram alpha and apparently it was -ln4, but I'm still pretty confused lol.
Integration by parts gives me 2\int_{0}^{\infty}\frac{(1-exp(-t))exp(-t)}{t}dt Integrand and interval of integration hints me to use Laplace transform so i calculate Laplace transform L((1-exp(-t))/t) plug in s = 1 and double the result To calculate L((1-exp(-t))/t) it is enough to calculate L(1-exp(-t)) and integrate the result
I just did it in ChatGPT, and it says that the answer is (pi²)/6 I told it that in bprp's video the answer was ln4, but GPT said that ln4 is the answer for the initial integral but without the square in the exponent
@@blackpenredpen It's pretty common occurance. Just in this video: 0:00 - 0:05 about 7 times 0:14 single one 0:33 single one 0:48 - 0:50 a couple And so on. It sometimes seems to correspond to movements of the mic, like the slight upward hand movement at 0:05. Could be a faulty connection or frayed wire.
Using Feynman's technique TWICE! (the integral of sin^3(x)/x^3 from 0 to inf)
ua-cam.com/video/weZLETAIDEk/v-deo.htmlsi=rns_1h9G4MbG5oDS
dude
Your photo...
=(
I like the one with 2 markets
=(
Awoms video
Also
BPRP please never stop posting you are the GOAT
Thanks!
Love you bro
I do see all of your videos from India. I'm 63 year old. Whenever I see your video I always feel sorry for that I didn't get a teacher like you in my yearly life.
You are such rare class of teacher who can make the learning fun and enjoyable.
Thank you so much for you nice comment! I am very happy to hear this!
Will you please make a video to find the coordinates of the points of intersection of two intersecting circles.
@@aniruddhaghosh1303 thar is a nice tho
Mmmm, well, a clue that I can give you is that, don't think on a function world accepted by school
Just, plug the same varibles and use the quadratic formula for find out y [f(x)]
That is something that works out
Distance formula at certain coordinates?
@@junkgum mmm
Sqrt[(y2-y1)^2+(x2-x1)^2]
If I am not bad
Hey bprp I am hosting a Integration Bee in my school and I included many integrals from your 100 integrals, so glad that you provide such good resources!
Glad to help!! 😃
I subbed to both bprp and Dr. PK Math, and watch both videos where they are using different methods, which are great
This channel has mostly everything one could need for calculus I to III
You are literally the only math youtuber that i will watch for fun
Feynman's technique works fine. I set up I(x) = the same integral where the integrand is multiplied by exp(-xt). Take the second derivative which is easy to find (just a bunch of exponentials to integrate), and then using that I''(x), I'(x) and I(x) approach 0 when x goes to infinity, you can find I(x) by integrating twice. If I didn't mess up, you get:
I(x) = x•ln[x(x+2)/(x+1)²] + 2ln[(x+2)/(x+1)]
which leads to the expected result I(0) = 2ln(2) = ln(4) (technically that final calculation is a limit calculation because of the first term, but it works fine).
I though of the same possible parameterization too. Nice work!
So nice to recall my undergraduate calculus classes.
I studied at the Technion, Israel and you are certainly on the level to teach there.
Thank you so much for your wonderful explanations!
Bprp: This integral is hard. To solve it, I'm going to make it WAY more intimidating first.
Wow I've suffered to solve this question but you really did it with the simplest way. Great 👍❤
I just watched Pk's response to this using complex analysis and Big O notation. Its pretty cool too.
In general we can’t always use Fubini so some justification in that step is required
What exactly would I have to prove here to be able to use it
@@alphazero339you’d have to prove(or already know) that the function f(x,y) is integrable in the product measure space. Meaning that when you integrate |f(x,y)| over XxY with respect to the product measure on XxY(these things have to be properly defined using measure theory), then this gives a finite value. Then we can take the integral of f(x,y) (without absolute values) over the product measure space and evaluate it as a double integral and exchange the order in which we integrate.
In practice, we can more easily use Tonelli’s theorem here: If f(x,y) is non-negative and measurable then we always have this equality, but the integrals may not be finite. In the video’s case, the exponential function is always non-negative and is measurable so this works and no need to even verify Fubini!
People often say "Fubini" to mean "Fubini-Tonelli" so no justification is needed IMO.
@@almightysapling if so then fair, but still something worth thinking about. And even Tonelli needs the basic justification that the functions are positive and measurable which they are
My calc 3 professor put this as a challenge problem, and you saved me! He might've watched your video lol to be reminded of this technique and give to us.
Finally my guy does hard stuff again. Would love to see some complex integration stuff via Residue Theorem.
Dr PK Math does it much
I love math, especially when such beautiful puzzles and solutions came out, that's just gorgeous, so beautiful and awesome, Please never stop seeking for such brilliants of math, that is indeed joyful thing)
You are the best mathematician 🎉
As ancient is my Calculus, I can still follow your work andvunderstand it. And watch woth excitement!
I think, first time I saw "reverse Feynman/Leibniz's rule" on the channel of Michael Penn.
That's a very nice approach 👍🏽
Incredibly incredible ❤❤
Actually this is a beautiful solve, I really enjoyed its simplicity.
this was really coool to watch. gj
Feynman's trick works really well if you parameterize the integrand as ((1-e^(-tx))/x)^2 Its derivative ends up being a constant function after an integral substitution, but you need to use feynman's trick a second time to find the constant
lovely solution
The kind of smile I had watching this video probably can't be achieved by any other entertainment thing in this world. What a nice way to solve it
great content
IBP twice and then Gamma function is your uncle (and your friend):
I = -2 ∫_0^∞ [2e^(-2t)-e^(-t)] ln t dt = 2γ - 2 ln 2 ∫_0^ e^(-v) = 2 ln 2 = ln 4
Are you differentiating the numerator and integrating the denominator?
@@tommyliu7020 Yes, I did a first IBP letting dv = 1/t^2 and u = [1-e^(-t)]^2; then I've repeated this procedure with the resulting integral, obtaining
-2 ∫_0^∞ [2e^(-2t)-e^(-t)] ln t dt
lovely tricks
Integrate by parts (differentiate the top, integrate the bottom) to get I = ∫ (-2e^-2t + 2e^-t)/t dt. Let x = e^-t to get I = 2 ∫[0,1] (x-1)/(lnx) dx, which can be solved by Feynman's.
To use Feynman's technique you first need to Differentiate the numerator and Integrate the denominator using IBP, then use the technique on the resulting integral.
VERY NICE
Mind Blown
OMG new bprp pfp and video? Lets go!
Triple integration is the bestest!
3 integral signs in a row😁👍
I used to think I was smart but don't think I would have ever thought of that solution method. It was originally thinking of utilizing hyperbolic cosine and sine
Amazing!
Love you
I think properies of laplace transform or residue theorem can be used
10nth grader from germany here.
First intuition is to do DI w/ (1-e^-t) , 1/t^2
Why are you telling your grade
Integration by parts , integrate 1 over t^2 differentiate numerator , after that feynman trick works
100 advanced trigonometric functions please
Very easy to solve!
All of our favorite math channels posted absolute bangers this week. Hiw did we math nerds get so lucky?
Funny part is that Chat-GBT says that the solution is Pi^2/6 , i'm never asking him about integrals again.
I solved the indefinite integral normally by expanding it out and got the 3 normal integrals by itself and they were actually quite easy to solve. You get a slightly complicated expression involving Ei function. I did it this way to simplify it, but when I took the limit from zero to infinity, the infinity part became 0 and the other zero part was quite tricky as I had to solve for lim as t->0 of 2Ei(-t)-2Ei(-2t), I had absolutely o idea how to do this as this was in the form infinity - infinity indeterminate. I used wolfram alpha and apparently it was -ln4, but I'm still pretty confused lol.
Can you use the Fubini Tonelli's Theorem at 6:04?
Yes, as the function is nonnegative and measurable, and the exponential function is nonnegative
He literally says that's what he's doing 4 seconds before that.
Are x and y greater than 0?
Yes bc those integrals go from 0 to 1
@@blackpenredpen ahaaaa tyyy
Im not even gonna try😂😂
Taking calc 1 rn
I would start with integration by parts
Then maybe Laplace transform
Integration by parts gives me
2\int_{0}^{\infty}\frac{(1-exp(-t))exp(-t)}{t}dt
Integrand and interval of integration hints me to use Laplace transform
so i calculate Laplace transform
L((1-exp(-t))/t) plug in s = 1 and double the result
To calculate L((1-exp(-t))/t) it is enough to calculate L(1-exp(-t)) and integrate the result
Cant we write this in the form of Exponential integral and the Gamma function?
Will you please make a video on how to get the coordinates of points of intersections of two intersecting circles.
Thank you.
Can't you just use Gamma Integral after opening the bracket and splitting the numerator??
Is there any additional condition to swap the integral inside out like that???
Please upload proof of Leibniz rule for differentiating under the integral sign
What about the power of n as it approaches infinity?
Which marker you use?! Please tell me us sir
The answer is nearly always ln(4) 😂
Only in the cases when it's not pi²/6 or the Euler Mascheroni constant. ;)
@@bjornfeuerbacher5514
Indeed.
New pfp!
Cool!
Exactly the same value as the integral of 1/t between 1 and 4! Can we get to that from ((1-e^-t)/t)^2 somehow?
Why do you say the integral without the square diverges at 0:24? The graph shows otherwise.
Hello teacher could you help me ? Limited X to infinity
2x/(1+x^2 )×tan[(πx+4)/( 2x+3)
I am sure that this easy question can be solved bey o1 within 30 seconds.
5:15 I didn’t understand, what theorem I’m missing?
Let t = x for mental simplicity sake and dt = dx.
Rip old pfp of bprp
I just did it in ChatGPT, and it says that the answer is (pi²)/6
I told it that in bprp's video the answer was ln4, but GPT said that ln4 is the answer for the initial integral but without the square in the exponent
There's been some crackling on the mic in all the latest videos.
Could you please provide me the time stamps? Thanks.
@@blackpenredpen It's pretty common occurance. Just in this video:
0:00 - 0:05 about 7 times
0:14 single one
0:33 single one
0:48 - 0:50 a couple
And so on.
It sometimes seems to correspond to movements of the mic, like the slight upward hand movement at 0:05. Could be a faulty connection or frayed wire.
Thank you so much for pointing those out! I will see what I can do to fix it!
I'm in class 10
eZ.
That's a brilliant approach 😳
Lol😊
Pls make a video on the Product integral and The Riemann Zeta function
This is too easy
Bro hellped me get through high school, next stop: college⚡💪
Please upload all your content to RUMBLE
First viewer
Bro is not pregnant but he never fails to delivery
lobter 🦞
Yee