Raabe's Integral
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- Опубліковано 10 лют 2025
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The “oookkk cool” has a veneer of suppressed anxiety to it, and makes me feel like some nasty unsolicited integral is about to jump out from the shadows at any moment.
Never stop sharing your brilliant solutions Kamaal
the last term of the results can be written as ln(H(N-1)), where H(n) is the hyperfactorial function.
Rewriting the solution as: I(α) = log [√(2π) (α/e)ᵅ], it reminds Stirling's formula, log n! ~ log[ √(2πn) (n/e)ⁿ ]. Could've been a great opportunity to mention it.
That’s exactly what I was thinking.
Always nice to hear from you Slava!
0:17 Oh man. I didn't even know he was sick.
Yeah really unfortunate, apparently he caught Raabes.
@@naturallyinterested7569 💀💀💀
Thank you for your fruitful results.
Ooook cooolll😅
12:31 no differential dz at end of integral, 0 marks
😅😅😅
May i suggest a question for a future video?
Int(0,1) (x³/sqrt(1-x³))dx
It took me like 1 week to solve using your tricks but I couldn't find an antiderivative for it.
Also you the GOAT
You forgot about the natural log on the thumbnail, my dear. More general is indefinite integral via Barnes G function 😁
That was what I thought when I saw the thumbnail lol. I thought it was a different integral.
I don't know what you're talking about...there's definitely a log there😂
@@maths_505 he meant you wrote log instead of ln, in some placed we're taught that log is decimal while ln is neperiene
@@MaxCE No, there wasn’t a log in the thumbnail. He edited the thumbnail later.
You could have written the integral as the riemann sum, and then using the properties of logarithm you’d have I = lim (n->inf) (1/n)log(F(a)F(a+1/n)F(a+2/n)…F(a+1-1/n)) and then using the multiplication theorem easily derive that I = lim( n-> inf) log(2pi)/2-alog(n)+log(F(na))/n = log(2pi)/2 + alog(a)-a.
Very enticing
Hi,
12:33 : missing dz .
"ok, cool" : 0:33 , 2:16 , 3:09 , 5:58 , 7:32 , 11:22 ,
"terribly sorry about that" : 0:59 , 1:33 , 3:49 , 5:50 , 7:49 , 9:35 .
@@CM63_France Just wanted to let you know that you are loved and appreciated, man
"What matters is that the guy this integral is named after is dead."
Fucking savage OMG 😂
Here I have a very interesting integral that involves the trigamma function.
\int_0^1 \frac{x\sqrt{x}\ln x}{x^2-x+1}dx
answer=\frac{2\sqrt{3}\pi^2}{9}-4 you can try
What app do You use to write the math in the video?
J. L. Raabe was German, so if you're uncertain on how to say his name, use the German pronunciation.
hi, you have any videos on reducing Σeⁱⁿˣ to 1/(1-eⁱˣ) in the context of a larger encompassing summation? is it considered «legal»?
வணக்கம் broo ❤🎉
Wont it be \alpha + 1 after its Gamma(\alpha + 1)/Gamma(\alpha + 1)
Yes, you did butcher his name.
The correct German pronunciation would be ['ʀa:bə], so a voiced uvular trill, a long "ah" sound, a pretty standerd "b" sound and a "u" sound, as in "under", where the first syllable is stressed. If you can't do the uvular trill, which is acceptable, I guess, you can do the normal English "r" sound, but the rest should be doable.
I am getting too diZ....😵💫
clickbait 😭