Very good 👍

Very clever! But G isn't one of my favourite constants. I like the Euler-Mascheroni constant and Chaitin's omega. 🙂

mindblowing question sir and ur approach was fire..

Why did you calculate to G??

interesting. Nicely done. However, you could also take f(n) = 1/(n+5)(n+4)(n+3) and show that your sum is -1/3 sum (f(n+1)-f(n)). Then, using telescopic relation, you get the sum is 1/180.

Incrrdible

3.14 de verdiğiniz formülün kanıtını nerde bulabilirim

Helpful

A much Easier method would to convert them into integrals and then solve it .

Thank you sir, useful for JEE advanced 2022

What's the name of the transformation ?

😤😤😤समझाए तो हो नही वो कैसे आया समाकल्न

How did you get a double factorial

I=Γ(2)π^2/12(ln2)^2

There's a problem with this proof! You are trying to prove the Euler's Reflection Formula. You used the prior result that the integral_0^infinity of x^{m-1}/(1 + x^n) = (1/n)*pi / sin(m*pi/n). However, you forgot that this prior result was proved by utilizing the Euler's Reflection Formula itself. You cannot use this result if you are trying to prove it. Perhaps you forgot, it happens to everyone once in a while. I have the proof of this formula by leveraging both the Gamma function and Residue theorem from Complex Analysis.

Noice

Very tidy.

Hola Informàtic 23' We didnt get the Beta function yet, but we need to solve this problem, is it possible ?

Thanks for sharing this problem! At 5 minutes into the video, when you changed the summation index from 0 to 1, in order to get e^n/n!, you effectively set (n - 1) to a new variable, say, t. And when n = 0, t = -1. If so, shouldn't the index of the new summation go from -1 to (x - 1) instead of 1 to x ?

This is some strange coincidence. Today , in our final paper of Integral Transforms, this question came. I think even our professor watches your videos

Thanks for sharing! I understand your proof but what I don't understand is where you got your version of the Beta function which doesn't look like anything I have seen. The Beta function, by definition, is the integral defined from 0 to 1 but yours goes from 0 to infinity. In addition, the integrand doesn't look familiar. Can you please point me to a resource where you got the formula B(a, b) = int_0^infinity x^(a - 1)/(1 + x)^(a + b) ? Many thanks!

Pls share link where you have solved it by Laplace transform ...

Thanks for this video! I wonder how you got the Beta function B((1 + i)/2, (1 - i)/2) based on the definition of the Beta function which can be found on Wikipedia at en.wikipedia.org/wiki/Beta_function. Can't make that connection. I'll appreciate your clarification!

This could be calculated very quickly if you remember the derivative of Γ(1/2) Γ ' (1/2)= - √π(γ+2ln2) and then ψ(1/2)=Γ ' (1/2) / Γ(1/2)

👌💯

I would use substitution x=(1-u)/(1+u) but x=tan(t) is still OK, not as great as x=(1-u)/(1+u) but OK

Wow, you made solving that integral seem like a sinh!

That was definitely feyn, man! 👏

You definitely went off on a tangent, but it was great! 👍

Amazing

Nice,elegant solution.

Your content is unique❤

Posto coslnx=Re(e^ilnx)=Re(x^i),,si arriva facilmente a 1/2-3/10+5/26-7/50+9/82-11/122...ah,ah,deve vedere come sintetizzare la serie...goodnight

This is a nice one! I think you could also compute it by writing it as a sum of two geometric series and integrating term-by-term.

Repeat?

I have a suggestion for video for this integral , int 0 to pi/2 x²cotxln(1-sinx) dx , Try it.

Please, give some explanation on exp(-x)/(exp(-x)+1) expansion, because it dosn't look like Taylor series and this final stop for me 😅

Con le sostituzioni diventa un integrale gaussiano

Ho usato feyman, ma non sono sicuro della costante Di integrazione nel passaggio da I'(a) a I(a).. Comunque a me risulta pi^2/12...io ho fatto come te,ma al 4:30 non potevi integrare direttamente la Σ?passi subito da s^(-3)--->s^(-2)/(-2)

Very nice😊🙌🙌

Sir can we apply L-hospital rule in this sum by limit as a sum?

And the only integration you had to do was ∫dx. Awesome.

Nice😊

Intanto è funzione pari..I=2..poi uso feyman con I(a)=..arctg a/coshx..I(0)=0..I=2I(2024)...risulta ,dopo aver svolto i calcoli,I=2*(π/2)*arcsh(2024)=πarcsh2024

Simply (eˣ +e⁻ˣ +eⁱˣ +e⁻ⁱˣ)/4 coz 1, -1, i, -i are the roots of x⁴ - 1 = 0 Just those IIT-JEE things 🗿

Good one

Great example

Very nice! I wasn't aware of that representation of the beta function before.

Is posible resolve: int 0 to infinite of (cos(ax)-cos(bx))/x dx

Questa è facile..l=ln3