How Richard Feynman would evaluate this monster log integral

Comments · 336

• MA

  @maths_505 · 3 years ago (edited) · pinned

  At the <a href="https://www.youtube.com/watch?v=XnvFr2w2gUI&amp;t=1225">20:25</a> mark I forgot the modulus operator on the the argument of the natural logarithm. However, it didn&apos;t affect the solution as we end up multiplying complex conjugates anyway. However, I should not have omitted it as it leaves a hole in the solution development. <br>The modulus operator will remain and on adding I(i) and I(-i) the moduli of two complex conjugate numbers will be multiplied (due to the logarithms) giving us exactly the same result.

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• MA

  @manstuckinabox3679 · 3 years ago

  Deciding between contour and Feynman&apos;s is liek deciding between nuking and nuking harder...

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• UB

  @ublade82 · 3 years ago

  Feynman&apos;s Technique: Knowing the answer to everything

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• TI

  @TimothyOBrien6 · 3 years ago

  This technique was developed by Leibniz, one of the inventors of calculus (whose notation we still use today). It&apos;s silly to call it the Feynman technique when the inventor of calculus used it.

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• SH

  @shashidharnrao · 13 days ago

  This is a brilliant solution using yet another place where a parameter can be introduced while using Feynman&apos;s technique. &nbsp;I wish to express my sincere thanks to the author of this channel for explaining the problem in such great detail and in a very understandable fashion.

• RI

  @riadsouissi · 3 years ago

  I used log(x^4+t^4) instead to avoid dealing with complex values of t. Got the same value.

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• MA

  @matthew.y · 3 years ago

  I was eating dinner when I found this video. Now my dinner is cold, but I just found a new magical technique!

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• PA

  @pablosarrosanchez460 · 3 years ago

  The antiderivatve in <a href="https://www.youtube.com/watch?v=XnvFr2w2gUI&amp;t=898">14:58</a> can be done easier by noticing that (1-t) can be written as (1+sqrt(t))(1-sqrt(t)), and this last one cancels with the numerator, leaving us with the integral of dt/[sqrt(t)·(1+sqrt(t))]<br>Now perform a substitution making u = sqrt(t), du=dt/2sqrt(t) =&gt; int of dt/[sqrt(t)·(1+sqrt(t))] = int of 2·du/(1+u) = 2·ln(1+u) + C = 2·ln(1+sqrt(t)) + C

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• JO

  @josefhamilton9907 · 1 year ago

  <a href="https://www.youtube.com/watch?v=XnvFr2w2gUI&amp;t=1330">22:10</a> e to the i-pi-OVER-four

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• TW

  @twistedcubic · 3 years ago

  The Residue Theorem is clearly more overpowered, since you brought up complex numbers.

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• AM

  @amrendrasingh7140 · 3 years ago

  The flow of the solution was awesome and stimulating. Good work kamaal 👌

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• WI

  @wilurbean · 3 years ago (edited)

  Prof Fred Adams, &quot;If you use it once its a trick, if you use it twice its a technique&quot;

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