Contour integration with a STUNNING result!

Comments · 76

• TO

  @Tosi31415 · 2 years ago

  the phi jumpscares are getting out of hand

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• SH

  @sherinodak7 · 2 years ago

  „2 is an even number the last time I checked“ XD

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• AR

  @arctan_x-o3e · 2 years ago

  Wake up babe, new Maths 505 dropped.

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• CM

  @CM63_France · 2 years ago

  Hi,<br><br><a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=735">12:15</a> : note that tan^-1 (-2) = 2 tan^-1 phi where phi is the golden ratio.<br><br>&quot;ok, cool&quot; : <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=17">0:17</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=162">2:42</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=187">3:07</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=275">4:35</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=340">5:40</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=396">6:36</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=421">7:01</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=523">8:43</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=589">9:49</a> , <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=735">12:15</a> ,<br><br>&quot;terribly sorry about that&quot; : <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=887">14:47</a> .

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• NO

  @NorthMavericks-ow7jk · 2 years ago

  This truly is complex.

  5

• KI

  @kingzenoiii · 2 years ago

  ah yes, the complex realm

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• AN

  @ananyapamde4514 · 2 years ago (edited)

  <a href="https://www.youtube.com/watch?v=W25v7wcO2SM&amp;t=231">3:51</a> yes bro, we like it thicc

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• LI

  @LinaWainwright · 2 years ago

  Another, more elementary, method would be to go off of the addition formula of arctangents [i.e. arctan(a) + arctan(b)=arctan((a + b)/(1 - a b)), for ab&lt;1)]. After some work, you can show that with a = (√φ x + 1)/φ and b = (-√φ x + 1)/φ, then (a + b)/(1 - a b) = 2/(1 + x^2), so we have that arctan((√φ x + 1)/φ)+arctan((-√φ x + 1)/φ) = arctan(2/(1+x^2)), since [(√φ x + 1)/φ][(-√φ x + 1)/φ] = (1 - φ x^2)/φ^2 ≤ 1/φ^2 &lt; 1.<br>Now we see that we have split the original arctangent of a rational function into a sum of arctangents of linear functions, which is much easier to work with. Using integration by parts, we see that ∫arctan(x)dx = x arctan(x) -1/2 ln(x^2 + 1) + C, so after substituting x = a u + b, we get the following result:<br> ∫arctan(a u + b)du = (u + b/a) arctan(a u + b) -1/(2a) ln((a u + b)^2 + 1) + C.<br>After plugging in everything and simplifying a bit, we get:<br>∫arctan(2/(1 + x^2))dx = x arctan(2/(1 + x^2)) + (1/√φ) [arctan((√φ x + 1)/φ) + arctan((√φ x - 1)/φ)] + (√φ/2) ln(((√φ x - 1)^2 + φ^2)/((√φ x + 1)^2 + φ^2)) + C (feel free to check with WolframAlpha).<br>At x = 0, everything is equal to 0, and for the limit as x→∞, the first and last terms go to zero, leaving is with the middle terms, which give us (1/√φ)[π/2 + π/2] = π/√φ. ■

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• CA

  @cadmio9413 · 2 years ago

  What an amusing video, nevertougth contour integration had such elegance when solving integrals, love it

• OR

  @orionspur · 2 years ago

  § f-¹(f&apos;) vibes

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• BA

  @bahiihab-y2r · 2 years ago

  thank you for this wonderful contour integration

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• PE

  @petterituovinem8412 · 2 years ago

  the elusive contour integration video!