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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. 〈 Berger | Dillon 〉‏ @InertialObservr 10 May 2019
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      Go ahead, square the integrand. π doesn't mind.pic.twitter.com/Cx5HCElH4q

      11 replies 51 retweets 257 likes
      Show this thread
    2. 〈 Berger | Dillon 〉‏ @InertialObservr 10 May 2019
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      Here's my proof of the equalitypic.twitter.com/HZTyokc1Su

      6 replies 12 retweets 81 likes
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    3. Wood Croft‏ @wood_croft 10 May 2019
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      Replying to @InertialObservr

      You forgot to define "I". I(0) = 0 => I(alpha) = alpha*pi => I(1) = pi

      1 reply 0 retweets 2 likes
    4. 〈 Berger | Dillon 〉‏ @InertialObservr 10 May 2019
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      Replying to @wood_croft

      I define I on the first line

      1 reply 0 retweets 0 likes
    5. Wood Croft‏ @wood_croft 10 May 2019
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      Replying to @InertialObservr

      You define it as a function of alpha, then use it as a constant. On the last line you should redefine it to I(1), or just write I(1).

      2 replies 0 retweets 2 likes
    6. 〈 Berger | Dillon 〉‏ @InertialObservr 10 May 2019
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      Replying to @wood_croft

      Oh I see what you’re saying.. Twitter publishing luckily isn’t going into a journal

      1 reply 0 retweets 2 likes
    7. Colin V. Parker‏ @ColinVParker 10 May 2019
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      Replying to @InertialObservr @wood_croft

      I’m wondering if you can dispense with the parameter altogether and make an equivalent change with integration by parts?

      1 reply 0 retweets 1 like
    8. 〈 Berger | Dillon 〉‏ @InertialObservr 10 May 2019
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      Replying to @ColinVParker @wood_croft

      there's always more than one way to prove something :D

      1 reply 0 retweets 1 like
    9. Colin V. Parker‏ @ColinVParker 10 May 2019
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      Replying to @InertialObservr @wood_croft

      Yes, but I was actually flailing and forced into contour integration until I saw this.

      1 reply 0 retweets 1 like
      〈 Berger | Dillon 〉‏ @InertialObservr 11 May 2019
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      Replying to @ColinVParker @wood_croft

      I don’t see why ibp wouldn’t work tbh

      1:35 AM - 11 May 2019
      1 reply 0 retweets 0 likes
        1. New conversation
        1. Colin V. Parker‏ @ColinVParker 11 May 2019
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          Replying to @InertialObservr @wood_croft

          It does, it’s just that showing the two are equal is a trick I didn’t think of. Showing either is equal to pi using IBP is where I got stuck.

          1 reply 0 retweets 0 likes
        2. 〈 Berger | Dillon 〉‏ @InertialObservr 11 May 2019
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          Replying to @ColinVParker @wood_croft

          Oh I see.. here’s a hint rewrite 1/x as an integral from 0 to infinity of e^{-xy} dy

          1 reply 0 retweets 1 like
        3. 1 more reply

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