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Beta, Gamma, and Zeta all in One!!

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This really was a lot of fun to solve, after all "there is no genius without a touch of madness" :)
Also check out a similar integral by ‪@maths_505‬ which inspired this one:
• This integral is actua...
And also check out my own induction proof referenced if you want:
• Conducting an Induction!

Опубликовано:

11 апр 2024

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Комментарии : 9

@ostireich 3 месяца назад

Is it an air raid alert in the background 😭💀

@GingerMath 3 месяца назад

If it is then we have much larger (and explosive) issues than the sound quality here...

@ostireich 2 месяца назад

@@GingerMathdamn man 😭😭

@asles 3 месяца назад

what level math is this?

@GingerMath 3 месяца назад

Pre-Algebra (in the state of Louisiana)? Nah just kidding - but I'm unsure about that one 😅

@giuseppemalaguti435 3 месяца назад

@GingerMath 3 месяца назад

I mean for odd n it is soooo I suppose that works... though it's a bit harder to make a video if that's just the answer lol - love the trivial solution but nottttt the best for views

@giuseppemalaguti435 3 месяца назад

@@GingerMathforse tu volevi scrivere..(ln....)^n...ma se è scritto ln(...)^n, allora diventa n*ln(.…) e quindi I=0

@GingerMath 3 месяца назад

La notazione potrebbe non essere chiara: it is ln^n((1-x)/(x)) rather than ln(((1-x)/(x))^n), that is to say the exponent is being applied to the log and not its argument. I'll try to make that clearer next time.