A Rewarding Putnam Integral

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  • เผยแพร่เมื่อ 10 ม.ค. 2025

ความคิดเห็น • 6

  • @profenevarez
    @profenevarez 3 วันที่ผ่านมา

    Really nice insights! I like the way you motivated the steps and how you brought back everything to connect seamlessly at the end. All done in 8 minutes and some change. Well done! Thank you.

    • @ProfOmarMath
      @ProfOmarMath  3 วันที่ผ่านมา

      Thanks so much, I’m glad you enjoyed! It really is a rewarding one

  • @drynshock1
    @drynshock1 3 วันที่ผ่านมา

    J=I by King's rule if I'm not mistaken.

  • @jyfortin
    @jyfortin 3 วันที่ผ่านมา

    Nice use of symmetries to compute this integral. I would have naively parametrized the problem by replacing log(1+x) with log(1+ux), then derived with respect to u to obtain a rational form which can be integrated over x, and then integrate again over u between 0 and 1 (the integral is zero when u=0). Of course this would have taken more time and is less elegant.

    • @ProfOmarMath
      @ProfOmarMath  3 วันที่ผ่านมา +1

      I actually like this approach as well. If you take it further it can become elegant when recognizing that expression as an integral u.