WOOOOOOOOOOO 1K!!!!

@OscgrMaths

10 күн бұрын

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.

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

@OscgrMaths

14 күн бұрын

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

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

@OscgrMaths

11 күн бұрын

Thank you!!

Discovered yet another KZread treasure!

@OscgrMaths

Күн бұрын

@@MohdAbuolwan Thank you!

@MohdAbuolwan

Күн бұрын

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

@OscgrMaths

Күн бұрын

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

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

14 күн бұрын

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

@suzum0978

13 күн бұрын

prepa?

wow this is so fancy and elegant

@OscgrMaths

12 күн бұрын

Thank you!

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

@OscgrMaths

14 күн бұрын

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

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, $I_n = \int \frac{Px + Q}{(ax^2 + bx + c)^n} dx$) 😃.

Awesome!

@OscgrMaths

5 күн бұрын

@@mertaliyigit3288 Thanks so much!

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

@OscgrMaths

14 күн бұрын

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

14 күн бұрын

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

14 күн бұрын

@@franolich3 Nicely done!!

@RamblingMaths

14 күн бұрын

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

14 күн бұрын

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

Damn son, where'd you find this?

Superb!

@OscgrMaths

2 күн бұрын

Thanks for the comment!

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

@OscgrMaths

10 күн бұрын

Thanks so much!

Beta function enters the room

very nice

@OscgrMaths

2 күн бұрын

Thanks so much!!

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.

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

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

14 күн бұрын

Nice! Love that method, great approach.

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

Using residue theorem is also nice for this one

@OscgrMaths

10 күн бұрын

Definitely agree!

can you make more logarithmic integrals

@OscgrMaths

14 күн бұрын

Yeah definitely, thanks for the suggestion!

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

@OscgrMaths

14 күн бұрын

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.

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

13 күн бұрын

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

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

@OscgrMaths

14 күн бұрын

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

@dogackoca8426

14 күн бұрын

​@@OscgrMaths love to see complex integral

asnwer=1 oo sin cos tan

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

@OscgrMaths

13 күн бұрын

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

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

@OscgrMaths

2 күн бұрын

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!

You should teach the chain rule differently lol

@OscgrMaths

12 күн бұрын

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

@coreymonsta7505

12 күн бұрын

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

@coreymonsta7505

12 күн бұрын

@@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

12 күн бұрын

@@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

12 күн бұрын

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

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

12 күн бұрын

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.

@user-eb6mn3dw1v

12 күн бұрын

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

@t1hunna429

10 күн бұрын

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

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

@OscgrMaths

14 күн бұрын

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

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

2 күн бұрын

@@paulpinecone2464 😂😂😂 Great response!

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

11 күн бұрын

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!

very nice

@OscgrMaths

12 күн бұрын

Thank you!!