It took me 3 hours to do this integral!

Integral of ln(x^2+1)/(x+1) from 0 to 1. Feynman's technique for differentiation under the integral sign, partial fraction decomposition, and so many parts. Oh yeah, this is a fun one.
Proof that the sum equals π²/12: • Sum of Alternating Inv...
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[@blackpenredpen]

Whoaaaa!!!! I enjoy this very much. Btw, I never thought of doing partial fraction for ln(t+1)/(t*(t+1)) since there’s that ln(t+1) on the top. Very well done!!

[@MuPrimeMath]

I made a video on this problem a month or two ago, but the audio was corrupted, so I've remade it with better quality!

[@jitendratiwari6886]

wow

[@BluePi3142]

Even with you erasing parts of your work and simplifying them within the same step, it’s still very easy to follow your process. It only gets a little bit tricky to read the work at the very bottom, and even then you do a great job at explaining. Keep up the good videos :)

[@MuPrimeMath]

Yes, a few people have commented about my handwriting at the bottom of the board, so I will try to work on that!

[@BluePi3142]

Mu Prime Math I love how now I started watching more of your videos, the old ones are popping up in my recommended lol. I really do like your style, and I do the same thing with erasing/updating parts when I solve problems :P

[@BluePi3142]

What a fine integral, man... :)

[@MurshidIslam]

I see what you did there. :D

[@txikitofandango]

That was wild! Great presentation, and Feynman's technique is very clear now. Thanks!

[@slavinojunepri7648]

This is a wonderful result for a fascinating integral. Thank you for sharing.

[@spintwohalves]

Good job! Thanks for the fun video.

[@desmondhutchinson6095]

I like the effort you put in your thumbnails !

[@juanfco14]

That was a really nice video , thank you so much.

[@chideraachinike7619]

Duuuuuuuude! You're so cooooool! ❤️ Good math! ✌️

[@mamadetaslimtorabally7363]

Wow! Tremendously fascinating!

[@azmath2059]

Absolutely sensational mathematics!

[@user-wu8yq1rb9t]

Just beautiful .... Great I enjoyed a lot (actually I wanna watch it again, definitely!) Thank you so much

[@carlosgiovanardi8197]

Great!! Very interesting evaluation!

[@Homayoun197250]

Great video. Thank you.

[@andrzejpietraszkiewicz7538]

Beautiful

[@3manthing]

"Oh it definetly works", that made me laugh😆 great video☺

[@victormanuelpatosilva1637]

Brilliant.

[@valentinochiola1827]

Nice video, u are a genious, dude.

[@diegopimentelfonseca5379]

Wooaaah, new suscriber man! (;

[@eliyasne9695]

Awesome! When i tried to solve it i didn't use Feynman integration, i just turned ln(x^2+1) into a power series. It didn't work, i couldn't solve it, but at least i got stuck in a nice result. I showed that the integral will be equal to ln(2)^2 - (sum from 1 to infinity of (-1)^(n+1)*H(2n)/n ) Were H is the harmonic sum. If i had no mistakes (i am not sure about that) then it means i can now use your result to evaluate this bizarre sum. :)

[@TheWisator]

have you tried approximating that bizarre sum with a computer?

[@pbj4184]

Harmonic number sums can usually be evaluated using dilogarithms. Try doing that, it might lead to the result

[@jesusandrade1378]

@@pbj4184 The Maple Calculator App shows indefinite integral answer with dilogarithms

[@jesusandrade1378]

@@pbj4184 ... and the definite integral result in decimal form agrees with (3/4)*(ln(2)^2)-π^2/(48)

[@F1U7R2Y9]

Cool man ! 👍

[@violintegral]

You can also use integration by parts on the integral at the beginning, and then use Feynman's trick for the resulting integral. But I'm not sure if this is actually easier or not. I had to use partial fraction decomposition twice and then evaluate the integral of ln(t+1)/t from 0 to 1, just as you did. I was actually able to solve for the original integral I(1) after I integrated again with respect to t because I(1) popped back up again.

[@andreamonteroso8586]

very well done...

[@muradbashirov6435]

That was great

[@user-lr8od4uz1n]

Love from Korea

[@blackloop1861]

Nice video

[@ortho5387]

New subscriber :)

[@marinmaths3826]

Interesting!!!

[@asp2194]

Really interesting

[@HamzaMessaoudi-lu3km]

a nice idea is to do a U substituon, by letting x=t-1/t+1

[@khushalpatil711]

oh my goodness!!!!!

[@holyshit922]

Leibniz rule comes to dilog which can be reduced to Basel problem

[@hoodedR]

Great proof for ln2 > π/6

[@grantplunkett8410]

*gets integral with T’s* ok that wasn’t too bad I guess *video is halfway done* monkaS

[@robertflynn6686]

In a minor change this problem was on the Putnam Math Competition. Lee jung woo solved it in 10 min on board. Note. He was 6yo. Check out his x subst. Method. On the Putnam 3 hours is 1/2 of the morning session... 6 hours.

[@jesusandrade1378]

Lee Jung Woo solve a similar, but different (indefinite) integral: The integrand was ln(1+x)/(1+x^2)

[@jesusandrade1378]

I made a mistake, it was a definite integral from 0 to 1, but for the integrand I said above.

[@nafishsarwar2077]

Using the Maclaurin series I solved it within 1.5 hours, and my answer is [{(3log2)/5} - (6/19)] approximately equal to 0.1001

[@kaustavchakraborty811]

integration 1/sqrt(1-x^4) Between limits 0 to 1.. Please do this.

[@fourier07able]

Is it licit exchanges the summation by the integral? Yes!

[@dinosaric4862]

Awesome video! I would love if you could write a little less messy..

[@azzteke]

Would it be possible to use the residue theorem here?

[@saraqostahterra4548]

How long does it take to get to this level?

[@Manuel-pd9kf]

Damn

[@farkasmaganyos]

Nice job! I'd like to know why did you use 0 and 1, when you did the integrating in the variable t. Could you explain it?

[@MuPrimeMath]

We choose the upper bound of 1 because I(1) is the desired original integral, and the lower bound of 0 because we found earlier that I(0) = 0 which makes evaluation easy.

[@farkasmaganyos]

@@MuPrimeMath Ok! Really many thanks!

[@bernardofabrica]

3/4[ln(2)]^2 - pi^2/48 = 1/48*(6ln(2)+pi)*(6ln(2)-pi)

[@MuPrimeMath]

Interesting!

[@gateaspirant3912]

Let me tell u one more thing if u interchange the x and x^2 u will get answer as {πln2/8}

[@kishoresaldanha815]

I would just use Trapezoid rule

[@dcwadu6126]

Actually, breaking into power series i actually got quite close approximate to your answer, but yeah it isn't accurate.

[@spiritualisautembellator8399]

(3/4)ln(2)² simplifies to (3/2)ln2.

[@MuPrimeMath]

Remember it's not ln(2²) = 2 ln(2), it's [ln(2)]²

[@spiritualisautembellator8399]

@@MuPrimeMath Oh, yes!, Excuse me!. Good exercise!

[@zahari20]

I have one important remark: This technique, differentiating with respect to a parameter, does not originate from Feynman. It is much older. The guy who did this first was called Gottfried Wilhelm Leibniz. He lived three centuries before Feynman and did some other nice things too. Look up "Leibniz integral rule" ans see the examples on Wiki. No need to bring Feynman here.

[@denislalaj9345]

This is indeed Leibniz rule for integration. The reason that people refer to it as Feynman's trick is because of how much Feynman used it throughout his life. I'm linking below a great article about how this came to be called Feynman's trick: medium.com/cantors-paradise/richard-feynmans-integral-trick-e7afae85e25c

[@tiziocaio101]

Can I ask you an integral question?

[@MuPrimeMath]

Sure

[@tiziocaio101]

Mu Prime Math thank you and good video

[@tiziocaio101]

Mu Prime Math I’m 16 y/o and I’m learning calculus. I found this integral but I can’t solve it

[@tiziocaio101]

Mu Prime Math integral from 0 to -1 of (sqrt(x+1)-2(sixth root(x+1)))/(cbrt(x+1)+1)

[@tiziocaio101]

Mu Prime Math I tried by substituting u=sixth root(x+1) and then long division but it didn’t work.

[@NinoCuteCute]

It took me less than 3 hours after this video

[@SatyanarayanaMudunuri]

where is dt in the second step? a vey cluttered presentation

[@jserink1]

Its Leibnitz's rule, not Feynman's rule. Cheers, jim

[@Observer_detector]

F(s)=Laplace{ln(x^2+1)/(x+1)}{s} lim (s→0) F(s) Easy? lol

[@MoonLight-sw6pc]

This bro is really cool! Carry on making good stuff..... "Checking the channal $_$"...,.