This integral taught me Feynman's technique

Рет қаралды 10,102

Maths 505

Күн бұрын

Пікірлер: 58

@emanuellandeholm5657 7 ай бұрын

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

@primenumberbuster404 7 ай бұрын

Ngl, this should be a series.

@thiagorc25300 7 ай бұрын

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

@baptFulbion 15 күн бұрын

I knew this one with the king's property "∫f(x)dx = ∫f(a+b-x)dx" with the same result! It's nice seeing the same problem resolved by 2 different approaches!

@arunknown2177 7 ай бұрын

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

@SuperSilver316 7 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.

@intellix7133 7 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 7 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 7 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

@bandishrupnath3721 7 ай бұрын

Just wonderful integral and its explanation ❤sir.

@HenkVanLeeuwen-i2o 7 ай бұрын

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

@MrWael1970 7 ай бұрын

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

@bobbybannerjee5156 7 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?

@Czeckie 7 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!

@Ghaith7702 7 ай бұрын

very informative

@aliaujla110 7 ай бұрын

Maybe Third!

@adhamkassem3058 7 ай бұрын

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

@quite_unknown_1 7 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?

@CaioFalconieriLima 7 ай бұрын

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

@joshuaiosevich3727 5 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!

@giuseppemalaguti435 7 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...

@04-jayeshkumargupta8 7 ай бұрын

last step it should be pi/4 ln2

@maths_505 7 ай бұрын

Nope

@qetzing 7 ай бұрын

It’s 2I = …

@2kchallengewith4video 7 ай бұрын

Bro forgot division

@Ghaith7702 7 ай бұрын

(1/4)/2=1/8

@2kchallengewith4video 7 ай бұрын

Bro forgot division

@Mr_Mundee 7 ай бұрын

just do tan sub and some manipulation (no feynman)

@neg2sode 7 ай бұрын

awesommmeeee!

@merwan.houiralami 7 ай бұрын

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

@maths_505 7 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 7 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 7 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 7 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.