The Power Of Recursion: An UNSOLVABLE Integral

Hope everyone enjoyed! I thought this integral was particularly unique and love how theres so many approaches - please comment all of yours down below!! Please comment with any questions or suggestions for new topics, and as always, subscribe to stay updated. ~ Thanks for watching!Still looking for your answer to my last two video challenges ‪@blackpenredpen‬ ... Thanks again for all the recent support and please keep sharing and interacting!!#maths #mathematics #integrals #MIT #Cambridge #JEE #recursion #problemsolving #whoknew #fascinating #functions #euler #funproblems #proofs #functions #physics #sums #series #limits #whiteboard #math505 #blackpenredpen #integral #trig #trigonometry

[@maths_505]

WOOOOOOOOOOO 1K!!!!

[@OscgrMaths]

I KNOW!!! Thanks so much for everything - couldn't have got anywhere near without the help or the inspiration to start a channel in the first place.

[@redpepper74]

Wow, it’s neat how many different kinds of mathematical tools we used to solve this one

[@OscgrMaths]

That's why I thought it was such a good one to share!

[@futuure]

I’d just like to answer to the title of the video: in France, we are being taught the general method for integrals of 1/(ax^2+bx+c)^n where a, b, c are the coefficients of an irreducible polynomial considered in the real space. If the polynomial is of an order above 2, or it’s reducible, you can use partial fraction decomposition. Actually that last method is how you would start solving any rational fraction integration problem!

[@OscgrMaths]

Thanks! That's really interesting to hear and a great problem to solve.

[@suzum0978]

prepa?

[@niom9446]

wow this is so fancy and elegant

[@OscgrMaths]

Thank you!

[@maths_505]

Nice one, I remember solving the trig case as a question on my cal2 finals. A wonderful exercise indeed!

[@OscgrMaths]

Thank you!!

[@spiderwings1421]

inf*0 is not necessarily zero. its indeterminate without further work

[@OscgrMaths]

Yeah I was thinking of leaving it as a challenge at the end... you can gauge roughly it ends up at 0 because the power of 2n for cosine scales more rapidly than tan but of course there is a longer rigorous proof. I didn't want to include it in the video because it was already a bit long but feel free to share it here!

[@franolich3]

One can avoid the issue of the indeterminate form by using a hyperbolic substitution instead: x = sinh(u) dx = cosh(u).du x=0 => u=0 x-->inf => u-->inf Letting sinh(u,n) = (sinh(u))^n etc: J[n] = Integral[0 to inf: 1 / (1+x^2)^n] = Integral[0 to inf: cosh(u) / (1+sinh(u,2))^n] = Integral[0 to inf: cosh(u) / cosh(u,2n)] = Integral[0 to inf: 1 / cosh(u,2n-1)] J[n] = Integral[0 to inf: cosh(u,2) / cosh(u,2n+1)] = Integral[0 to inf: (1+sinh(u,2)) / cosh(u,2n+1)] = Integral[0 to inf: 1 / cosh(u,2n+1)] + Integral[0 to inf: sinh(u,2) / cosh(u,2n+1)] = J[n+1] + Integral[0 to inf: sinh(u) . sinh(u) / cosh(u,2n+1)] = J[n+1] + (-1/2n).[0 to inf: sinh(u) / cosh(u,2n)] - (-1/2n).Integral[0 to inf: cosh(u) / cosh(u,2n)] = J[n+1] - (1/2n).[0 to inf: tanh(u) / cosh(u,2n-1)] + (1/2n).Integral[0 to inf: 1 / cosh(u,2n-1)] = J[n+1] - (1/2n).[(1/inf) - (0/1)] + (1/2n).J[n] = J[n+1] + (1/2n).J[n] => J[n+1] = J[n].(2n-1)/2n

[@OscgrMaths]

@@franolich3 Nicely done!!

[@RamblingMaths]

You can rewrite tan x = sinx/cos x. If n > 0 then the cos x in the denominator is cancelled by a cos x in the numerator and you end up with a positive power of cos x multiplied by sin x. Then cos x->0 as x->pi/2, and sin x->0 as x->0, and there are no infinities.

[@OscgrMaths]

@@RamblingMaths Yes this is what I was thinking too... Thanks for sharing, great solution!

[@oraz.]

I had never heard of recursive integrals. That's amazing.

[@OscgrMaths]

Thanks so much!

[@mohamedanirelkarta7962]

Great video ! Would love to see more series in the channel

[@OscgrMaths]

Absolutely - series are some of my favourites too, thanks so much the comment!

[@sapwho]

You are so easy to follow!! Thank you so much for your content ❤️

[@OscgrMaths]

@@sapwho Thanks so much! Really appreciate that.

[@bachvaroff]

This one is well known in BG as well, it's the notorious "14000th integral" ('coz ~14k freshmen got F on their Calculus 1 exam in a single day, one of the problems being the general form of this one, $I_n = \int \frac{Px + Q}{(ax^2 + bx + c)^n} dx$) 😃.

[@MariusMeunier]

if you consider just the "indefined integral" of f_n : x -> 1/(1+x²)^n, you can let u(x) = f_n(x) ; u'(x) = -2nx/(1+x²)^(n+1) ; v(x) = x ; v'(x) = 1 and then you can reconsider the integral as the integral of u(x)v'(x)dx ; now you can make an integration by part and the obtained formula is In+1 = 1/2n * (xf_n(x) + (2n-1)In) ; now you want to integrate between 0 and +\infty, just reconsider the "xf_n(x)" term as [xf_n(x)] evaluate in zero and +\infty : in zero, you get 0, in +\infty, same. you can conclude between 0 and +\infty : In+1 = (2n-1)/2n * In i guess this method is more simple than just doing substitution tricks thanks for the video ! please continue ! Marius

[@jamiewalker329]

The integral is just pi*i*residue at z=i. (complete a contour in the UHP), and using the evenness of the function. The residue is pretty straight forward as 1/(1+z^2)^n = 1/(z+i)^n(z-i)^n. So we have a pole of order n. So we just need to do 1/(n-1)! * the n-1'th derivative of (z+i)^-n evaluated at z = i. This is 1/(n-1)! * (-1)^(n-1) n(n+1)(n+2)...(2n-2) (2i)^(2n-1). Multiplying by pi*i and a bit of cleaning up gives the answer.

[@OscgrMaths]

Yes!! I was hoping somebody would take a contour approach... This is a beautiful solution, very nicely done.

[@MohdAbuolwan]

Discovered yet another KZbin treasure!

[@OscgrMaths]

@@MohdAbuolwan Thank you!

[@MohdAbuolwan]

@@OscgrMaths I only recommend that you take it slowly on the viewer because not all people math geniuses like yourself😅

[@OscgrMaths]

@@MohdAbuolwan Sorry!! I'll keep that in mind next time. Thanks again for the comment.

[@booshkoosh7994]

Superb!

[@OscgrMaths]

Thanks for the comment!

[@nickmcstuffins1836]

fantastic video

[@OscgrMaths]

@@nickmcstuffins1836 Thank you! Really glad you enjoyed.

[@NonTwinBrothers]

Damn son, where'd you find this?

[@aleksandervadla9881]

Using residue theorem is also nice for this one

[@OscgrMaths]

Definitely agree!

[@hyperkahler]

Perhaps another way to solve it is to consider F(t) = sum_{n=1}^{inf} I_n t^n, and then by exchanging the order of integration and summation, we find F(t) = t/sqrt(1-t) pi/2 . Taylor expanding this around 0 we recover the result.

[@holyshit922]

And this result can be useful when orthogonalizing base of polynomials {1,x,..,x^n} to get base of Chebyshev polynomials {T_{0}(x),T_{1}(x),...,T_{n}(x)} And it seems that result is just nth term of expansion of 1/sqrt(1-x)

[@OscgrMaths]

@@holyshit922 Great! Thanks for the comment.

[@mertaliyigit3288]

Awesome!

[@OscgrMaths]

@@mertaliyigit3288 Thanks so much!

[@clementp7648]

Let n be a positive integer. The integral converges by majoration with the case when n = 1. For all x in the set of real numbers: 0 ≤ 1 / (1 + x²)^(n + 1) ≤ 1 / (1 + x²)^n We can integrate this relation because we have shown that these integrals converge. Thus, the sequence (Iₙ) is positive and decreasing, which means it is convergent. Next, we can integrate by parts on the interval [0, A] where A > 0, to establish the recurrence relation between Iₙ and Iₙ₊₁, leading to the same result! Using trigonometry, as demonstrated in the video, is also a nice approach Great video on what would be a classic prep class exercise in France! :)

[@OscgrMaths]

Nice! Love that method, great approach.

[@harrygaming9090]

Hi, how do you get better at answering these type of questions. Do you have to be naturally gifted. I have just finished A Level maths.

[@IbraheemMatanmi]

very nice

[@OscgrMaths]

Thanks so much!!

[@theupson]

this very specific class of problem was standard in all textbooks at least through the 90s- i definitely did this problem in high school. i feel like it may have been dropped from some curricula as not being worth the time but i wouldn't call this an arcane topic.

[@yassinenajar4369]

Haha I love that the unsolvable integral you have has literally been given to us as an exam question to get into uni 😂

[@OscgrMaths]

@@yassinenajar4369 Nice!

[@Akhulud]

using beta function, the integral is trivial after subing x=tan(theta)

[@OscgrMaths]

Yeah I mentioned at the start - beta function is definitely a shortcut after one substitution, but I wanted to offer a more accessible second method. Great spot though! That's part of what makes the beta function so useful.

[@noicemaster5173]

Beta(n+1/2,1/2) sneaking up

[@holyshit922]

14:10 Another content ? Maybe series expansion of f(t) = 1/sqrt(1-2xt+t^2) Here binomial expansion can be useful but after using binomial expansion twice I get n+k as powers of t but this is generating function for sequence of some polynomials in x so I need n as powers of t

[@lbonts]

and I thought I knew how to integrate…

[@dogackoca8426]

Can you work on complex numbers next time, I really like to see a problem with that.

[@OscgrMaths]

Definitely - do you mean a complex integral (like contour integration) or just anything complex in general?

[@dogackoca8426]

​@@OscgrMaths love to see complex integral

[@renesperb]

Could you check your calculation ? Mathematica gives the result I(n) =√π *Gamma[n -1/2)/(2*Gamma(n)) .The two results differ slightly numerically.

[@renesperb]

Also,the value for n=1 should be π/2 .If I copied your formula correctly it gives π/4.

[@renesperb]

I found the mistake I made when copying your result.Your result is the same as the one given by MATEMATICA.

[@Ben-wv7ht]

Beta function enters the room

[@attica7980]

Involving trigonometric functions is unnecessary. One can start with the identity 1/(1+x^2)^n=x x/(1+x^2)^(n+1)+1/(1+x^2)^(n+1). Integrating both sides, the first term on the right can be integrated by parts. Rearranging the result. one obtains a recursive formula expressing the integral of 1/(1+x^2)^(n+1) in terms of the integral of 1/(1+x^2)^n. This works for indefinite integrals, or for definite integrals with any limits.

[@aiGeis]

Me after (hopefully) barely passing my summer 6 week calc 2 class....Jesus Christ

[@rafaelcalderon5272]

4:48 isn’t 0 times infinity indeterminate?

[@OscgrMaths]

@@rafaelcalderon5272 Hey thanks for the comment! There is more work you do here - there's a more rigorous proof for the limit if you want to try it! Didn't have time to fit in my video so just went off the fact that I knew it worked

[@rafaelcalderon5272]

@@OscgrMathsokay I really appreciate the response this made things make a lot more sense thank u ❤

[@OscgrMaths]

@@rafaelcalderon5272 No problem, glad it helped!

[@bahiihab-y2r]

can you make more logarithmic integrals

[@OscgrMaths]

Yeah definitely, thanks for the suggestion!

[@adityavikramsinha408]

Another “simplification” is making the (2n)!/(n!)(n!) => 2nCn, neat

[@OscgrMaths]

Wow that's a great point, makes the answer even more satisfying!

[@paulpinecone2464]

They don't teach you this technique for good reason. This derivation is highly exothermic and releases hard beta particles when the terms are separated. Advanced integrations should only be performed in controlled environments with appropriate radical dampers. This sort of irresponsible how-to is a reason why the internet has acquired such a reputation for postings of questionable accuracy.

[@OscgrMaths]

@@paulpinecone2464 😂😂😂 Great response!

[@eevee8856]

Bold of you to assume anyone learns integrals from school (unless they want to mess up their thinking ability and hamper their originality)

[@leif1075]

I thinkna lot of ppl might nkt thinknofntrig dub so why not do a regular sub like u equals 1 plus x^2 ..wouldnt that work? Thanks for sharing

[@OscgrMaths]

@@leif1075 For this one I wouldn't chose that substitution because it'll be hard to put dx in terms of du - you'll end up with du=2x dx and there's no x in the integrand to sort that out. Hope this makes sense!

[@Charky32]

coudlnt u do Integration by parts, with u = 1/(x^2 + 1)^n and v' = 1?

[@OscgrMaths]

You could try it! I'm not sure how it would work because the derivative of 1/(1+x^2)^n is messy but let me know if you get somewhere!

[@МартинТемелакиев]

I didn't know that this simple technique, that I have found by experimenting(things they don't teach at school, but also subtract from students) with unfamiliar integral by myself for 3 mins btw, is SO HARD that it deserves a whole video of explanation :)))). Lol, we learn something everyday don't we?

[@OscgrMaths]

Glad you enjoy this kind of maths! Maybe check out some of the other videos on my channel like the one from the BMO if you want a different kind of question.

[@МартинТемелакиев]

I have a textbook, you know? Absolutely no need for videos, bruh

[@t1hunna429]

bro stfu like how can you be this much of a virgin. lol ig we learn something new every day

[@niom-nx7kb]

does this also work if n is a negative integer?

[@OscgrMaths]

I don't believe it will. Thanks for the comment!

[@niom-nx7kb]

@@OscgrMaths 👍thanks

[@coreymonsta7505]

You should teach the chain rule differently lol

[@OscgrMaths]

Sorry! I was hoping everyone watching would know it already. Would you like me to do a separate video teaching it?

[@coreymonsta7505]

The derivative of the outside function evaluated at the inside function times the derivative of the inside function

[@coreymonsta7505]

@@OscgrMaths That’s one of the better things to explain, if you do try to explain it, I wouldn’t do it the way you did

[@OscgrMaths]

@@coreymonsta7505 Yeah I think given the full time I'd use the fact that substitution u for inner function allows you to find dy/du and du/dx, then show the cancellation comes after that giving dy/dx. But sometimes with these longer videos I try and rush over the basic stuff - I'll keep it in mind next time!

[@coreymonsta7505]

@@OscgrMaths I mean just explain how the rule works during the video in a different way

[@NonameBozo88]

Don't use hand to wipe the board, oil on the skin ruins the pens and the board

[@OscgrMaths]

Thanks for the tip!! I'll make sure to always use my rubber.

[@comdo777]

asnwer=1 oo sin cos tan

[@worldnotworld]

Fantastic. [[ Technical comment: is it possible to direct the microphone more directly towrad your voice? You're picking up an awful lot of marker-squeak, which, though not as bad as fingernails on a chalkboard, is somewhat annoying...

[@OscgrMaths]

Yes, I'm thinking of getting a lapel mic which I'm hoping will pick up more voice and less board. Thanks for the comment!

[@television-channel]

very nice

[@OscgrMaths]

Thank you!!