Solving Integral: ∫ sin x / x dx

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  • čas přidán 28. 12. 2023
  • ❓Solving Integral: ∫ sin x / x dx
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Komentáře • 48

  • @lavonoff
    @lavonoff Před 2 měsíci +27

    You forgot to add +C :)

    • @Sedi3D
      @Sedi3D Před 21 dnem +1

      Never forget to add the constant C

  • @darcash1738
    @darcash1738 Před 6 měsíci +21

    Nice. Also, Feynman works for the definite bounded one 👍

  • @jamilshirinov2931
    @jamilshirinov2931 Před 6 měsíci +5

    What program did you for the creation of this video? The animations are just next level!

  • @SRINIX4356
    @SRINIX4356 Před 2 měsíci +5

    next time, dont forget a constant +C

  • @user-he3bx5ds7u
    @user-he3bx5ds7u Před 3 měsíci +5

    +C

  • @aadityakiran07
    @aadityakiran07 Před 2 měsíci +1

    cool. looks just like 3b1b

  •  Před 6 měsíci +3

    Muy buen video Genio!

  • @moutaa4
    @moutaa4 Před měsícem +1

    How do you evaluate that sum tho

    • @mohammadsenanali
      @mohammadsenanali  Před měsícem +1

      You can't :)

    • @moutaa4
      @moutaa4 Před měsícem

      @@mohammadsenanali Oh yeah
      I didn't pay enough attention
      I thought you were doing Dirichlet integral so I thought that method is another way to go that isn't Feynman technique

    • @moutaa4
      @moutaa4 Před měsícem

      @@mohammadsenanali That sum is basically Si(x)

  • @itachu.
    @itachu. Před 3 měsíci +3

    by parts?

    • @antoine35210
      @antoine35210 Před 2 měsíci

      does it work ?

    • @jacobstarr9010
      @jacobstarr9010 Před 2 měsíci +2

      No, if you try it the values of 1/x will just infinitely decrease and the values of sinx will just cycle between sinx and cosx indefinitely. Since you can’t get back to the original function, integration by parts won’t work. Also, the integral is non-elementary which is why the creator used a maclaurin series to “evaluate” it.

    • @antoine35210
      @antoine35210 Před 2 měsíci

      @@jacobstarr9010 yeah i was confused bcs my man up there said he did it by parts

  • @Agentsecret
    @Agentsecret Před 6 měsíci +5

    0:25 why ? justifies

    • @mohammadsenanali
      @mohammadsenanali  Před 5 měsíci +4

      The integration of a sum is equal to the sum of the integrations. Cheers!

    • @xninja2369
      @xninja2369 Před 3 měsíci +1

      @@mohammadsenanali that took me a while to understand 😂

    • @castagnos509
      @castagnos509 Před 3 měsíci +4

      that's not allways the case for infinite sums

    • @antoine35210
      @antoine35210 Před 2 měsíci

      just bcs integral is linear and so is sigma

    • @realcirno1750
      @realcirno1750 Před 24 dny

      @@castagnos509 its true because the series converges absolutely

  • @bjornfeuerbacher5514
    @bjornfeuerbacher5514 Před 6 měsíci +7

    Don't you have to prove that this power series actually converges?

    • @user-ze7jv1vw4b
      @user-ze7jv1vw4b Před 4 měsíci

      I also want to know

    • @phylI
      @phylI Před 4 měsíci

      It converges with Leibniz

    • @jacobstarr9010
      @jacobstarr9010 Před 2 měsíci

      No, it is an indefinite integral so we don’t need to prove it. Also, the Maclaurin series of sinx converges for all real numbers, so it will converge regardless of the values.

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 Před 2 měsíci

      @@jacobstarr9010 What does my question have to do with this being an indefinite integral?
      And how does "the integral of the series converges" follow from "the series itself converges"?

    • @jacobstarr9010
      @jacobstarr9010 Před 2 měsíci

      @@bjornfeuerbacher5514 well, sinx converges for all x. And all we change is the removal of 1 power of x. This doesn’t change much, since x^-1 is a polynomial and not a series. So we can assume the same. In fact, if we want to prove it, we can use the ratio test. By using the ratio test, we find the limit to be equal to zero, and zero times x^2 will be zero for all x. That means no matter what value of x we use, the integral will converge. And the reason I say it doesn’t matter for an indefinite integral is because we treat x as a variable. If we were to say we want to evaluate the integral from 0 to 1, then convergence would matter because we would want to verify that these x values are within the interval of convergence. But since the radius of convergence is infinity, the x values will converge for all x. So in summary, it doesn’t matter unless we have values of x we want to plug in.

  • @danielc.martin1574
    @danielc.martin1574 Před 5 měsíci +22

    This is not elementary 😮😂

    • @mohammadsenanali
      @mohammadsenanali  Před 5 měsíci +4

      Exactly

    • @filipeoliveira7001
      @filipeoliveira7001 Před 2 měsíci

      You learn this in AP Calculus BC though

    • @danielc.martin1574
      @danielc.martin1574 Před 2 měsíci

      @@filipeoliveira7001 do you prove it?

    • @filipeoliveira7001
      @filipeoliveira7001 Před 2 měsíci

      @@danielc.martin1574 You learn the tools to prove it: Taylor Expanion (Mclaurin representation of sin(x)), Summation Integration, and how to determine convergence of an integral and of a power series.