I used a double integral to solve a single improper integral

I know we haven't done a hard integral for a while now, so let's integrate ((1-e^(-t))/t)^2 from 0 to infinity. I didn't use Feynman's technique of integration for this since the denominator is t^2, but I used a double integral to solve this one! Surprised? Watch the video to find out my solution!
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[@blackpenredpen]

Using Feynman's technique TWICE! (the integral of sin^3(x)/x^3 from 0 to inf) kzbin.info/www/bejne/rZa9fXiKdq53e80si=rns_1h9G4MbG5oDS

[@leonardobarrera2816]

dude Your photo... =(

[@leonardobarrera2816]

I like the one with 2 markets =(

[@leonardobarrera2816]

Awoms video Also

[@yuri117_br]

BPRP please never stop posting you are the GOAT

[@blackpenredpen]

Thanks!

[@AlokPatil-sz7er]

Love you bro

[@aniruddhaghosh1303]

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.

[@blackpenredpen]

Thank you so much for you nice comment! I am very happy to hear this!

[@aniruddhaghosh1303]

Will you please make a video to find the coordinates of the points of intersection of two intersecting circles.

[@leonardobarrera2816]

@@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

[@junkgum]

Distance formula at certain coordinates?

[@leonardobarrera2816]

@@junkgum mmm Sqrt[(y2-y1)^2+(x2-x1)^2] If I am not bad

[@cdkw2]

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!

[@blackpenredpen]

Glad to help!! 😃

[@umylten4142]

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).

[@MichaelMaths_]

I though of the same possible parameterization too. Nice work!

[@klerulo]

Bprp: This integral is hard. To solve it, I'm going to make it WAY more intimidating first.

[@paya4030]

This channel has mostly everything one could need for calculus I to III

[@Happy_Abe]

In general we can’t always use Fubini so some justification in that step is required

[@alphazero339]

What exactly would I have to prove here to be able to use it

[@Happy_Abe]

@@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]

People often say "Fubini" to mean "Fubini-Tonelli" so no justification is needed IMO.

[@Happy_Abe]

@@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

[@mathnerd5647]

I subbed to both bprp and Dr. PK Math, and watch both videos where they are using different methods, which are great

[@allmight801]

Finally my guy does hard stuff again. Would love to see some complex integration stuff via Residue Theorem.

[@domedebali632]

Dr PK Math does it much

[@alex_ramjiawan]

I just watched Pk's response to this using complex analysis and Big O notation. Its pretty cool too.

[@Intu_369]

Wow I've suffered to solve this question but you really did it with the simplest way. Great 👍❤

[@newname5205]

You are literally the only math youtuber that i will watch for fun

[@joedapotatowater]

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.

[@rockstarayan1959]

You are the best mathematician 🎉

[@dudl2945]

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

[@ДанилоФилонов]

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)

[@redrosin99]

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!

[@kaidenpink1771]

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

[@IoT_]

I think, first time I saw "reverse Feynman/Leibniz's rule" on the channel of Michael Penn. That's a very nice approach 👍🏽

[@tambuwalmathsclass]

Incredibly incredible ❤❤

[@عَمرُبنوليدالمسلم]

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.

[@richardhole8429]

As ancient is my Calculus, I can still follow your work andvunderstand it. And watch woth excitement!

[@benjoshuayip2520]

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.

[@nuclearrambo3167]

I think properies of laplace transform or residue theorem can be used

[@Mario_Altare]

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

[@tommyliu7020]

Are you differentiating the numerator and integrating the denominator?

[@Mario_Altare]

@@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

[@fdileo]

Can you use the Fubini Tonelli's Theorem at 6:04?

[@kristopherwilson506]

Yes, as the function is nonnegative and measurable, and the exponential function is nonnegative

[@almightysapling]

He literally says that's what he's doing 4 seconds before that.

[@Bhattarai_hari]

Why do you say the integral without the square diverges at 0:24? The graph shows otherwise.

[@bain8renn]

the graph tending to 0 as t->infinity does NOT mean that the area under that curve converges, a good example of that is the integral from 1 to inf (1/t)dt, which simplifies to (ln(t)) from 1 to infinity, which is ln(infinity) - ln(1), or just infinity.

[@walidability]

Actually this is a beautiful solve, I really enjoyed its simplicity.

[@pianoprobability6490]

this was really coool to watch. gj

[@Szynkaa]

lovely tricks

[@neriya-bd]

lovely solution

[@UsmaniaOfficialStudio]

Please upload proof of Leibniz rule for differentiating under the integral sign

[@6612770]

Mind Blown

[@aniruddhaghosh1303]

Will you please make a video on how to get the coordinates of points of intersections of two intersecting circles. Thank you.

[@De_Firma]

What about the power of n as it approaches infinity?

[@sovietwizard1620]

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.

[@RoyalYoutube_PRO]

Can't you just use Gamma Integral after opening the bracket and splitting the numerator??

[@rauladrianbringasjimenez8656]

5:15 I didn’t understand, what theorem I’m missing?

[@nmaedu.100]

great content

[@namangoyal1297]

Cant we write this in the form of Exponential integral and the Gamma function?

[@chayapholtopar5992]

Is there any additional condition to swap the integral inside out like that???

[@stevemonkey6666]

3 integral signs in a row😁👍

[@actualRocketScientist]

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

[@cdkw2]

OMG new bprp pfp and video? Lets go!

[@sadi_supercell2132]

Integration by parts , integrate 1 over t^2 differentiate numerator , after that feynman trick works

[@Nain115]

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

[@perost1227]

Are x and y greater than 0?

[@blackpenredpen]

Yes bc those integrals go from 0 to 1

[@perost1227]

@@blackpenredpen ahaaaa tyyy

[@yurfwendforju]

10nth grader from germany here. First intuition is to do DI w/ (1-e^-t) , 1/t^2

[@alphazero339]

Why are you telling your grade

[@HiddenKey_210]

Triple integration is the bestest!

[@gabest4]

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?

[@Patapom3]

Amazing!

[@OpPhilo03]

Which marker you use?! Please tell me us sir

[@UnTipoSinNombre]

VERY NICE

[@wise_man-goted]

100 advanced trigonometric functions please

[@TechnoBeats1251]

Funny part is that Chat-GBT says that the solution is Pi^2/6 , i'm never asking him about integrals again.

[@user-mf7li2eb1o]

Im not even gonna try😂😂 Taking calc 1 rn

[@danielntoko2117]

Very easy to solve!

[@AlokPatil-sz7er]

Love you

[@holyshit922]

I would start with integration by parts Then maybe Laplace transform

[@holyshit922]

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

[@nornsoriya1257]

Hello teacher could you help me ? Limited X to infinity 2x/(1+x^2 )×tan[(πx+4)/( 2x+3)

[@guitaristxcore]

All of our favorite math channels posted absolute bangers this week. Hiw did we math nerds get so lucky?

[@weishanlei8682]

I am sure that this easy question can be solved bey o1 within 30 seconds.

[@davidbrisbane7206]

The answer is nearly always ln(4) 😂

[@bjornfeuerbacher5514]

Only in the cases when it's not pi²/6 or the Euler Mascheroni constant. ;)

[@davidbrisbane7206]

@@bjornfeuerbacher5514 Indeed.

[@toeknee3900]

New pfp!

[@scottleung9587]

Cool!

[@Silvar55x]

There's been some crackling on the mic in all the latest videos.

[@blackpenredpen]

Could you please provide me the time stamps? Thanks.

[@Silvar55x]

@@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.

[@blackpenredpen]

Thank you so much for pointing those out! I will see what I can do to fix it!

[@namangoyal1297]

Pls make a video on the Product integral and The Riemann Zeta function

[@matei_woold_wewu]

Rip old pfp of bprp

[@ValidatingUsername]

Let t = x for mental simplicity sake and dt = dx.

[@kift.]

eZ.

[@Someone19471]

I'm in class 10

[@aashishkumar9658]

That's a brilliant approach 😳

[@ikerluqup5661]

Lol😊

[@rehanalam3900]

First viewer

[@stonecrane167]

This is too easy

[@LoneWolfAndKub]

Please upload all your content to RUMBLE

[@kyriakoskourkoulis1159]

Bro hellped me get through high school, next stop: college⚡💪

[@AlokPatil-sz7er]

Bro is not pregnant but he never fails to delivery

[@keymind117]

lobter 🦞

[@fsjyxnutfgh1493]

Yee