Відео

Sweet

i'd like to know how you evaluate the digamma(1/2) and digamma prime of 1/2

I'm not studying on such a high level of calculus but l can't understand what the B function that you use on 3.30.Can you please explain?

Beautiful! I am going to read up on the beta function now!

how much time do you spend daily doing all this?

great

Excellent

Nice, but you should also have shown the conditions for them to apply.

Could you do video on proving beta function?

That's a great formula you proved ❤

Keep up the good work bro , subscribed :)

Hang on! This is a JEE integral? Isn't it?

Nice! 😃

Excellent

Hello... I have a question and I hope you can help me. My question is that I would like to learn mathematics in order to study classical physics and quantum physics. Do you have courses, UA-cam channels or resources where you learned math from algebra to calculus? This was my question and I hope you can help me ❤.

👍

What software do you use for presenting these?

Excellent

Mental integral

π / √φ Φ is golden ratio.

Excellent

Excellent

How about Int_0^pi/4 (1+tan^2 t) d(tan t)= ( tan t + tan^3 t /3)_0^(pi/4) = 1+1/3=4/3

Excellent

Good stuff

A nice video would be: Prove arcsin a + arcsin b= ; and arccos a +arccos b =.

It can be quite long but it can be calculated with Ostrogradski's method of isolation rational part of integral

Quick question if you change the bounds from 0 to infinity to 0 to 1 does this technique still work? You can’t add the two integrals when there bounds are different?

Hello... How can I contact you because I have some questions about learning mathematics and I hope you can help me?

Nice question

How can i send an integral question for you? Its national contests

I could do that one

My first Olympiad problem was of this type lol.

I remember struggling with this kind of Diophantine equations and then i discovered that they're in a class called "Mordell's equations". The way to prove there are no solutions using the technique I saw in the file i downloaded is to find a way to factor the expression and get a contradiction by quadratic reriprocy and I'm pretty sure this problem was in that file y²=x³+23 y²+4²=x³+3³ y²+4=(x+3)(x²-3x+9) it's easy to see that x must be odd and x=1(mod 4) and to get those claims just look at the original equation modulo 4. With this we get that x²-3x+9=3(mod 4) so x²-3x+9 must contain a prime factor p=3(mod 4) so looking at the equation modulo p you get y²+4=0(mod p) so y²+1=(mod p) but p=3(mod 4) and it is well known that this case has no solutions by quadratic reciprocy.

If a is even then 3a²-1 is going to be -1 modulo 4 which is a contradiction. If a is odd then 3a²-1 is going to be -1 modulo 3, contradiction.

Wait why do we get n|x+1?

The solution makes sense and i even get the feeling that everything is obvious but coming up with the idea is the hard part :(.

I still have a long waaaay to go in NT huh...

Yeah my aprooach was the second solution. x³ is only congruent to 0,1 or -1 modulo 9 and if none of the numbers a,b or c is divisible by 9 we'll get a contradiction because a³+b³+c³ will only be congruent to 3,1,-1 and -3 modulo 9.

Fantastic evaluation. Beta and Gamma functions make everything so interesting.

Nice

I solved more simplify. On one of start step we can subs 1/x+arctgx =t ===> dt = 1/(1+x^2) - 1/x^2 = (-1) /(x^2*(x^2+1)) and interval from 1+π/4 to π/2. At result we have int dt/t = ln t = ln (2π/4+π) 🎉

Fantastic

This sum is very sneaky, and it only takes the McLaurin's expansion of exponential to compute it. This is so cool!

At 0:57 B is missing. It should be B(1/4, 3/4) and B(1/4, 1-1/4). Sorry for this mistake

yess

Thanks a lot for rendering premium content to free of cost . May God bless you

What's the name of background music?

How did you thought of that substitution

Fantastic evaluation. Please discuss the other methods of evaluating this integration.