A very cool hyperbolic integral

[@edmundwoolliams1240]

That was a very cool integral! That last substitution was unexpected, but made so much sense after you did it 😮

[@Daniel-yc2ur]

Love seeing integrals that are difficult but don’t require anything fancy

[@noobymaster6980]

Do you think you would be able to find a funky looking improper integral and differential equation which have the same function as solution? Bonus points for funky functions and constants in the solution.

[@maths_505]

Improper integral having a function as solution? Surely you mean a parameterized integral.

[@noobymaster6980]

Might be my english which isnt perfect, just mean an integral without bounds. So your solution is a function and not numerical value. If that makes sense…

[@Unidentifying]

thats a really interesting question, my first thought is that quite often the results of integrals here are trig fn's , though evaluated at limits, and we know also how the trig fn's love to pop up as solutions to diff eqns. Its interesting also because this is definitely not the first integral "solution' ive seen with arctan here, but I dont recall any differential equation has an arctan as solution

[@noobymaster6980]

I did realise after writing this you could just solve a d.e then take that solution and take its derivative. Then ”integrate” that function and solve the d.e and say wow. Altough that kinda feels like cheating, both should come sort of at the same time. Or an interesting rule could be to make the function getting integrated and the d.e ”look similar” / has some of the same functions in them 🤔

[@RandomBurfness]

I know I'm late but I threw this into Wolfram Alpha and apparently this actually has a primitive function that's expressible in elementary functions? It says the primitive is (Sqrt[2] (3 - I Sqrt[3]) ArcTan[(Sqrt[2] Sqrt[-1 + E^x])/Sqrt[1 + I Sqrt[3]]] - 2 Sqrt[6] ArcTanh[(Sqrt[2] Sqrt[-1 + E^x])/Sqrt[-1 + I Sqrt[3]]])/(3 Sqrt[1 + I Sqrt[3]]).

[@MrWael1970]

Thank you for your effort.

[@sandyjr5225]

Amazing solution!!

[@nitzan33]

This might be blasphemy, but I think that at this point it would be very interesting to see a cool integral with no pi in the solution. Correct me if I am wrong, but I don't think we have seen one in a while.

[@maxvangulik1988]

^

[@idjles]

i've called them "cosh" and "shine" since the 80's

[@SuperSilver316]

I hearrd shine recently but it was on a British Quiz show.

[@knivesoutcatchdamouse2137]

I say sink since h kinda looks like a k, and tank for tanh, but obviously I say "coash" for cosh, but with a strong O like the o sound in boat. Some say cawsh, which sounds weird to me but, hey, there's no consensus so you do you.

[@satyam-isical]

Kosh Shine Koth Czeh(sech)🤡

[@cosimo7770]

i've called them "cosh" and "sinch" since the1950's.

[@Unidentifying]

int x^1/2 / (x^1/3 + x^-1) is this an easy one can someone help me ?

[@johnsalkeld1088]

In uk the pronunciation is cosh and shine

[@farfa2937]

I don't particularly like "cosh" and "sinch" because what about the others? tansh? shecant?

[@jackkalver4644]

ds/(s^2-1) past [-1, 1]? This integral could be undefined!

[@cyrex5792]

This is called an improper integral of the second kind. And this integral doesn't converge

[@giuseppemalaguti435]

Basta porre t=√(e^x-1)..INT.2t^2/t^4+t^2+1..a me risulta ,risolvendo tramite integrali razionali,π/√3

[@bandishrupnath3721]

Sir why does the second integral collapse to zero ? the integral from minus to positive infinity of s²- 1

[@Unidentifying]

the bound is actually 2 postivve infinities hence its 0

[@pandavroomvroom]

W videos!

[@theelk801]

>what if you don’t want to go down the complex analytic route I don’t understand

[@alexander_elektronik]

its the residue theorem from complex analysis. it‘s pretty interesting because you can evaluate integrals using their singularities in a specific way.

[@tiagobeaulieu1745]

Wait wait wait wait. So you're telling me that if you simply make a substitution that makes both bounds of integration the same value, you can conclude it indeed is =0 ??? I demand an explanation, Kamal :)

[@Jalina69]

nice

[@hyperboloidofonesheet1036]

That second substitution seems shady.

[@מיכאלקונטרוביץ]

Dear Kamal, I think there is a big problem with the second integral you have introduced through the solution development. This is an improper integral which diverges! Also there is a red alert - when we have the same limit of integration at the bottom and upwards,we may have a problem (not always,but sometimes). In this case you have evaluated the integral only in the Cauchy's principal value sense

[@maths_505]

Hello my friend. In case of that integral, we may consider the upper and lower limits to be the same number a where a > 1. Then it is only a matter of evaluating the integral in the limit as a tends to infinity. We know that the integral from a to a will be zero so the limit evaluates to zero as well. I understand your concerns regarding this technique but I assure you in case any future integral cannot be justified using the reasoning I've given above, I'll will make that clear in the video.

[@מיכאלקונטרוביץ]

@@maths_505 Ya, this is my concern. By your substitution you have created a degeneracy which led us to a big fat zero, especially when a tends to infinity, but when this is the case it may be a big problem with the substitution itself - the function could not be invertible neither differentiable. Also, in order to be pedantic, you Must prove convergence and you haven't done that

[@Ijkbeauty]

this was normal

[@satyam-isical]

So many ex in this video😂

[@txikitofandango]

When I saw u^4 + u^2 + 1, it made me want to multiply it by u^2 - 1 to get u^6 - 1. I wonder if that would've worked...