Feynman's overpowered technique on iterated integrals

[@Bobi006]

This is absolutely insane stuff for me but I’m understanding more with each video - Keep up the amazing work!

[@cinemaipswich4636]

The greatest thing about Feynman was that he invented a visual language. It was not words or mathematical script, it was a new invention of "diagrams" to explain atomic and particle physics. At first his method was described a "doodles". Now they form part of every student's text books.

[@MrWael1970]

Thank you very much for your powerful effort. Note that when you made a shift from N to N+1, it is appropriate to make N tends to N-1

[@manstuckinabox3679]

After almost a year of consuming Maths 505 content, I finally can see how easy these kinds of integrals have become.

[@insouciantFox]

Point of rigor, when you split up the sum, the two series must absolutely converge. It works here since you know it will resolve, but in general this is a dangerous thing to do.

[@mathalysisworld]

absolutely fantastic man!!! keep up the good work. You are helping me improve a lot!! Grateful !!

[@joniiithan]

Tbh I have never seen this application of feynman’s trick. Well played bro!

[@hyperpsych6483]

7:37 shouldn't we be getting -1/(t+2) here for the second term? edit: I see it is fixed a moment later

[@mtl.wraith9981]

what integration technique would you use for the integral from 0 to 1 of sqrt(-lnx)? i couldn’t find any results on wolfram alpha or youtube, so i figured i would ask the best math channel on this platform

[@maths_505]

Made a short based on it: kzbin.info_WdsyxcW9qY?si=BeCdMfSqd-qTEnr7

[@anthonyjohnfinch6870]

How do I find the description below? I can’t see anything to press. Incidentally can you integrate x^( n-1)/ (x-1) without using contour integration? This would help to explain Euler’s reflection formula which you often use.

[@giuseppemalaguti435]

A me risulta ,dopo aver impostato I(a)=…..I=I(0)=2Σln(k+1/k)-Σ(1/k+1/k+1)...e li mi sono bloccato,non saprei come muovermi con il ln e anche il +'della 2'serie è un problema...forse I=1-2γ...ma non sono sicuro

[@jessedutton3112]

7:53 The difference of ln^2 is still a difference of squares. You can combine them, but not that way.

[@Bobi006]

You can since the term inside the log is squared and not the log itself

[@satyam-isical]

From where do you get these integrals? Btw your videos on feynman technique is very cool and explanation is very easy to understand👍👍👍👍 Love from India..

[@Anonymous-Indian..2003]

I'm also From India, BTech CSE 3rd year, You ?

[@satyam-isical]

@@Anonymous-Indian..2003 I'm in class 12th preparing for jee

[@Anonymous-Indian..2003]

@@satyam-isical Lag hi Raha tha..... High level integrals agar koi india se kar raha tha jarur IIT se sambhandh rakhta hoga....😂 Lekin JEE ke syllabus mein double integrals aur digamma function jaisi cheezein nahin hai.....

[@satyam-isical]

@@Anonymous-Indian..2003 oh good to see you I'll try to come iit nxt year Maths is love😊😋

[@GeoffryGifari]

wait... i noticed that the integral is *almost* symmetric with respect of x and y, aside from the additional x factor. Because of how the log terms are written and the limits of integration, if: I = ∫∫ *x* ln(x)ln(y)/(1-xy)ln(xy) dxdy, x and y from 0 to 1 we'll get the same result if x and y are interchanged! I = ∫∫ *y* ln(x)ln(y)/(1-xy)ln(xy) dxdy, x and y from 0 to 1 we can add those together 2I = ∫∫ (x+y) ln(x)ln(y)/(1-xy)ln(xy) dxdy, x and y from 0 to 1 so that I = (1/2)∫∫ (x+y) ln(x)ln(y)/(1-xy)ln(xy) dxdy, x and y from 0 to 1 a prettier form. Maybe this helps?

[@comdo777]

asnwer=1 dx 🤣

[@BorisNVM]

coooool

[@gavasiarobinssson5108]

And he looks cool

[@yoav613]

Cool!