This integral taught me Feynman's technique

Рет қаралды 8,103

4 ай бұрын

This is an integral from the 2005 Putnam exam. It's the first integral I had ever solved using Feynman's trick of differentiating under the integral sign and I think it's one of the best examples on how to apply the technique.
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Пікірлер: 56

@daddy_myers 4 ай бұрын

Ah! I remember this one from the good ol' days of mobile 505 🥺

@maths_505 4 ай бұрын

Can't believe I got a friend and a brother like you from literally uploading an integral video 🥺 yo speaking of brothers where tf is man stuck in a box?

@aravindakannank.s. 4 ай бұрын

@@maths_505 i still remember that day I was tired from a long hard day at school scrolling yt for fun which i actually found that day luckily i know the substitution of tanu function will because it was there in my 12 th syllabus last year but u introduced me to the legendary weierstrass substitution and u didn't stop there shown me a magic which u called Feynman integration technique which opens a lot of doors in a lot of problems for me and I enjoy every time u use it to develop solutions for crazy looking integrals which I like very much idk what would I have done if I didn't click that video on that period of time thanks mate now my interest in maths is much higher than the past years😊❤

@maths_505 4 ай бұрын

@aravindakannank.s. bro reading comments like this one always feels awesome cuz I feel like the channel is actually delivering on some front🔥

@emanuellandeholm5657 4 ай бұрын

I love when the derivative jumps over the integral sign and becomes a partial. Never gets old! :D

@primenumberbuster404 4 ай бұрын

Ngl, this should be a series.

@SuperSilver316 4 ай бұрын

I recently got introduced to a similar integral except the integrand is Int(ln(1+x^2)/(1+x)) from 0 to 1. You should try this one with Feynman’s trick, the choice and placement of parameter might surprise you.

@thiagorc25300 4 ай бұрын

Great video. Finally i have understood the famous Feynman's technique.

@arunknown2177 4 ай бұрын

had my math exam today, and your channel really helped!

@bandishrupnath3721 4 ай бұрын

Just wonderful integral and its explanation ❤sir.

@MrWael1970 4 ай бұрын

Thank you for your featured effort. Instead of log/2, it should be log2/2.

@intellix7133 4 ай бұрын

The integral / derivative switch up doesn't only require the integral to converge for certain values of alpha, normally you also want the partial derivative with respect to x to be continous by parts and the partial derivative with respect to alpha to be continous and smaller than an integrable function depending only of x between 0 and 1 (in that case x / (1+x²) wouldve worked)

@anaymulay7285 4 ай бұрын

This problem was in our college exam but we were told to do it by trigo Since by seeing your videos i tried it by Feynman and got answer really quick Thanks bro

@mystik4957 4 ай бұрын

i think a simple tanx = u, then applying integral (f(x) = integral f(a+b-x) will work where the limits are a to b

@Ghaith7702 4 ай бұрын

very informative

@user-yz3he2jm4o 4 ай бұрын

Usually log is the base 10 logarithm and ln is the natural logarithm.

@neg2sode 4 ай бұрын

awesommmeeee!

@joshuaiosevich3727 2 ай бұрын

I tried defining I(a)=integral from 0 to 1 of (x+1)^a/(1+x^2) dx, since I’(a=0) gives the integral in this video, alas the resulting integral is no better than what we started with, or at least I got stuck on it!

@adhamkassem3058 4 ай бұрын

Great solution development, Please solve Bee integrals in the next videos

@bobbybannerjee5156 4 ай бұрын

Brilliant. Can you tell us what's the name of the software/app into which you write ✍️ so that it shows up on your computer screen?

@CaioFalconieriLima 4 ай бұрын

Is possible to integrate 1/(1+x^5) with thia method?

@quite_unknown_1 4 ай бұрын

Could you solve this with 1/(1+x^2) = sum_(k=0) (-1)^k x^{2k} and then switch sum & integral, and then you get x^(2k) * log(1+x), solvable with IBP?

@pandavroomvroom 4 ай бұрын

nice

@Czeckie 4 ай бұрын

can I use this method for computing the integral with variable upper bound and parameter? I mean I(t, a) = integral of log(1+ax)/(1+x^2) from zero to t. I'm getting a formula, but it's numerically definitely wrong. EDIT: I see where's the problem. The observation from 9:33 doesn't work in general, so I only computed I(a,a)=arctan(a)*log(1+a^2)/2 which is no good!

@aliaujla110 4 ай бұрын

Maybe Third!

@giuseppemalaguti435 4 ай бұрын

L'ho fatto con le serie I=(πln2)/2-(1-1/3(1/3-1/2+1)+1/5(1/5-1/4+1/3-1/2+1)-1/7(1/7-1/6+1/5-1/4+1/3-1/2+1)....)..il risultato è corretto,ma non ho voglia di raggrupparli...

@Mr_Mundee 4 ай бұрын

just do tan sub and some manipulation (no feynman)

@merwan.houiralami 4 ай бұрын

you should really try the integral from 0 to infinity of sqrtx ln(1+x) / (1+x)^2

@maths_505 4 ай бұрын

That's a bit too easy since one trig sub and some trig manipulations yeilds a couple integrals that I've evaluated quite a large number of times.

@merwan.houiralami 4 ай бұрын

@@maths_505 that’s like the lamest answer ive ever seen in my life. first of all id love to see what magic trig sub would be useful here because there isn’t any. second of all actually try the integral instead of saying it’s too easy and calling it a day.

@maths_505 4 ай бұрын

@@merwan.houiralami ohhhh f**k 🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️ sorry bruh I thought the square was on the x 🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️and I didn't see the sqrt(x)🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️ my bad....the integral does look cooler so I'll give it a shot.

@maths_505 4 ай бұрын

Aight here it goes: Sub sqrt(x)=u You get 2 * int(0, infty) ( u²log(1+u²) )/(1+u²)² du Now expand the u² term in the numerator as: (u²+1)-1 and split the integrand into 2 terms You get: 2 * times int(0, infty) ( log(u²+1)/(u²+1) - log(u²+1)/(u²+1)² )du Now perform the sub u=tan(z) and the first integral will be easily reduced to Euler's famous log trig integral and the other integral i.e. Int (0,π/2) ( log(sec²(z))/sec²(z) dz is easy to solve. The sec²(z) in the denominator is a cos²(z) in the numerator and then expand cos²(z) using the double angle formula for the cosine function. The resulting integrals are again Euler's log trig integral plus an integral that can be solved trivially using integration by parts. Final result: π/2 + πlog(2) Integral's not half bad but honestly a bit too easy for a video since all I needed was algebra, Euler's log trig integrals and IBP. Thanks though.