A wonderful generalised integration result
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- Опубликовано: 3 окт 2024
- Been a while since I derived cool results for generalised integrals so here's one to make up for lost time. A nice surprise here is just how clean the results are for integer values of the parameter.
I also love that this result leads to the integral of sech^s(x)!!
I was looking for somebody to mention hyperbolic trig functions :)
Great! You had the opportunity in another video to use it and you didn't. That kind of disappointed me 🤣 because you did really have the great opportunity to name the constant and you unfortunately didn't, but, now that you did it, you are one of my favourite integrals/series/limits solver youtubers! (great content as always, just kidding ❤)
This is a gorgeous result for a mouthwatering integral. Thanks for sharing.
Not to mention great for party tricks
I think I_2 could be found more easily by using the fact that the original integral is even wrt x, so u can swap to -infty, 0 which when applying the same substitutions leads to the integral from 0 to 1
Yes you're right
If you integrate from -infinity to infinity and just halve the result you get after a little simplification you the Beta-function with the normal bounds and that little trick changes the argument of the Beta-function, so that you can use the reflection formula. The endresult is (π/4)×csc(πs/2).
Very smart solution. Thank you.
GODDANG IT I WAS DOING JUST FINE UNTIL I FORGOT BETA MALE FUNCTION EXISTED, meh I should make a new rule of thumb as to when to use invoke it:
so our methods are quite similar, I immediately multiplied up and down by e^sx and did an u = e^x sub, but I did not like my boundaries so I immediatly invoked the u = 1/v sub and wrote it as half instead of invoking the beta function before, here I got stuck because I forgot beta existed, so after I watched the video I used the v^2 = t sub and got the result.
ok so when ever we have "polynomial/(1+x^(someparameter)^someparameter" try using beta function.
which reminded me of a recent problem posted by Micheal Penn.
1/4B(s/2,s/2)
I(1) = pi/4 = pi*2^-2 = pi*2^-(2^1)
I(3) = pi/32 = pi*2^-32 = pi*2^-(2^5)
I(2k+1) = pi/2^(2^(4k+1)) ?
I think you could have saved quite some time if you had gone to the 1/t-world already at 3:20.
Nah she likes it when I finish later
Oh the integrals yeah I'm pretty sure they enjoy the journey too
please inform of: indefinite integral of
( x^6 / (x^13-1) ) dx without using infinite series expansion...of course, without partial fraction , not involving big expression
It doesn't have an elementary integral. Perhaps you have a typo?
x⁶/(x¹⁴-1) is pretty simple, it's just 1/7 arcoth x⁷
x⁶/(x¹²-1) is integrable but REALLY messy. Lots of partial fractions.
Nice
Can someone tell me that as the Riemann integrable function graphically represnt the area bounded by that function then what does Riemann Stieljes integral represent graphically?
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Can i know the name of that pen ?
Isnt Г(1/2) actually (-1/2)! because Г(x) = (x-1)! when x is a non negative integer ?
1/2 isn't an integer
@@frijoless22Yeah but from what I understand, if we want x! we need to compute Г(x+1), which means to get 1/2! we need Г(3/2) not Г(1/2).
doesn't the factorial equivalence only hold for the values the factorial itself is defined on? (aka nonnegative integers) you can't say the factorial of any rational number without using some kind of extension to the factorial
hi math 505: how to evaluate the following integral: from 0 to infinity [ cos(lnx)/ (e^x) dx]?
Laplace transform
Though you won't have a nice closed form
Can this be done through contour integration?
Yes ofcourse. I think qncubed3 did some similar integrals last year.
@@maths_505 can you link the video?
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