This integral taught me Feynman's technique
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- Опубликовано: 7 мар 2024
- 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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Ah! I remember this one from the good ol' days of mobile 505 🥺
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?
@@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😊❤
@aravindakannank.s. bro reading comments like this one always feels awesome cuz I feel like the channel is actually delivering on some front🔥
I love when the derivative jumps over the integral sign and becomes a partial. Never gets old! :D
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.
Great video. Finally i have understood the famous Feynman's technique.
Ngl, this should be a series.
had my math exam today, and your channel really helped!
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)
Just wonderful integral and its explanation ❤sir.
Thank you for your featured effort. Instead of log/2, it should be log2/2.
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
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
very informative
Usually log is the base 10 logarithm and ln is the natural logarithm.
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?
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!
awesommmeeee!
Great solution development, Please solve Bee integrals in the next videos
nice
Is possible to integrate 1/(1+x^5) with thia method?
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!
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?
Maybe Third!
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...
just do tan sub and some manipulation (no feynman)
you should really try the integral from 0 to infinity of sqrtx ln(1+x) / (1+x)^2
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.
@@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.
@@merwana.2278 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.
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.
Probably fourth
last step it should be pi/4 ln2
Nope
It’s 2I = …
Bro forgot division
(1/4)/2=1/8
Bro forgot division
you're a genius!!!
Nah bro I just have great subscribers♥️
3:54 since when did partial fractions feel hospitable lol
I hate partial fractions
Me too which is why we have Wolfram alpha 😂
Do you understand all things you studided in math?
This was a really cool video but instead of using feynmans trick substituting x=tan(t) is also quite a simple solution!
First
Second
@@maths_505 🙏
@@maths_505 sir where do you live.
As I want to meet you oneday.
@@fahadibrar379 I live in Pakistan but plan to move abroad after my masters.
@@maths_505 I live very close to you, India😅
Great but what have we have learned. Zero