Відео

quality

I think the sum has to be finite

how did you 'make' the last one

ua-cam.com/video/rKQt0O66D0I/v-deo.htmlsi=ADPpPhRE_ViuJZyp I unable to understand it, can you kindly help to make it clear. Thank you sir.

Cool

I'd like to see a generalized integral kernel video from you

Cho so dai mien man dung ngay thang noi xem nhe

in the last integral du/dx= -u2 . What happens with the -1?

Can u explain the integral involving (x^2-2lnx-1)/(lnx)^2 separately in other video 🫥

arcosh(1)=0 because (e^0+e^-0)/2=(1+1)/2=1

ooo finally

its about time lol

thought bro said feynman was too slow lmao

Hey can u make a strategy vedio which will help us to identify where we could apply the feynman trick because it's very confusing for me to where i can actually apply it

first I know yall been hella waitin for this xD

7:00 the argument of that second gamma should be negated

Gonna binge this whole series :)

the goat

excellent content, probably the most extensive and best content creator on integration

My God bruh have some rest

hey i thought it would be a good idea if we use trignometric substitution in the 6th question

hey in the integral dx/cos(x) sqrt(sin(x)) after u substituted u= sin(x) then in the resulting integral in terms of 'u' can we also just substitiute z= 1/u to solve it further

These names are getting crazy and crazy

Why it is called MAZ identity???

6:06 How do you sub in the bounds? ln(infinity) Idk if the integral formula works in this example bc the bounds are from infinity to x, not 0 to x. So I did it by writing out the Laplace expression, doing IBP like you did when making the formula, but then the first term doesn't cancel out completely when you sub x = 0. Then you get some Frullani integral and end up with the same answer of -ln(s+1)/s.

Do I have to know what Laplacian is and its basics before watching the Laplacian videos or have you explained it in the videos itself?

L{ln(t)}=int[0,♾️](e^-st•ln(t))dt u=st du=sdt L{ln(t)}=1/s•int[0,♾️](e^-u•(ln(u)-ln(s)))du L{ln(t)}=(Ř'(1)-ln(s))/s L{ln(t)}=(-ř-ln(s))/s where Ř is the gamma function and ř is the euler-masceroni constant.

tan(1+log x)/x + C

8:24 lol that evaluation scared me for a sec

i love your videos bro, so intelligent and super underrated

6:46 ; shouldn't it be 'du 'in plce of 'dt' ?

cool man, just don't stop uploading , loved it

I went through the integral without worrying about the bounds and it worked 19:00. I'm not sure I learned my lesson. It feels like it works as long as I split the integral. However, it was much easier starting with the knowledge of periodicity.

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Yooo fuck yeah

Thanks man that's was very helpful

this guy is cooking

The last one is a great example to use the weisstrass theorem, since if you had a power of x instead of e^x you would have gotten 2 and you know that e^x = sum x^n/n! so the int would be 2 * sum 1/n! = 2e

🐐

could we have substituted nx = u and then taken the limit, the answer comes out the same in the second one

go to sleep brother goodness gracious

Nice

The first one is wrong. It is infinity. The integral is constant with respect to limit, which diverges. And limit is also divergent. Infinity times infinity is infinity and you can not parametrize integral when you have already limit of n. The rest is ok, because it is already paramterized before limit takes action.

great vid man, keep up the good work!

2:15 There's the mistake. You forgor 💀 to change the bounds of the integral after the substitution.

I just wanna mention because you haven't until now that IBP is discustingly helpfull sometimes and also for people to look up the weisstrass approx theorem because it helps a lot in the problems of the type int g(x,n)*f(x) over smt compact since you need only to solve the problem for f(x)=x^n for some natural n.

It's time for multivariable stuff now...

I am fairly convinced that the example at 16:56 is completely wrong. You assumed that the limit exists, that cos^n is >0, and that its valid to interchange limit and integrand. In fact, I am pretty sure the limit does not exist. Moreover, if you take the limit over even integers, it should equal infinity. The sqrt(pi/2) is equal to the limit when the integration is over (0, sqrt(n)). Of course, I understand this is not meant to be an analysis video, but it's risky business doing these calculus tricks without justification!