Integral ln tan from 0 to pi/2

[@eliyasne9695]

You could easily find the first integral by using the fact that tan(x)=sin(x)/cos(x) and therefore showing that ln(tan(x))=ln(sin(x))-ln(cos(x)) Because of the nature of the integral the could cancel out.

[@avdrago7170]

Before watching, I’m guessing the answer is 0 because you turn it into ln(sinx)-ln(cosx) and with the usub of u=pi/2-x the 2 integrals are equal so the whole integral is 0.

[@holyshit922]

I calculated it in mind also before watching and in the same way as you had done it

[@nournote]

2:55 even if I is infinity [...] we still have I=0 How is that possible?

[@drpeyam]

The point is that I can’t be infinity. At the beginning there might be a possibility that the integral diverges, but I’m ruling that out

[@nournote]

@@drpeyam I get your point. I am still wondering if that is enough to justify the existence of such integral. I would calculate the integral from alpha to pi/2-alpha, which would be 0. Then take the limit. Even proceeding likewise I am not totally sure that is rigorous enough.

[@fabiotiburzi]

exponential is like god, their ways diverge from our ways

[@nournote]

@@angelmendez-rivera351 Why not (your 1st comment)? I meant from ALPHA to PI/2 - ALPHA, not from 0.

[@nournote]

@@CHector1997 thanks for sharing

[@connorshea9085]

Is it possible to find the integral from 0 to pi/4, to avoid things cancelling out?

[@fanyfan7466]

Connor Shea it is actually and the solution is very elegant, especially if you know what Catalan’s Constant is

[@factsheet4930]

Wolfram glitches and adds a small real number in the solution of the integral from 0 to Pi 🤔

[@fNktn]

That's probably due to rounding errors, as any approximation (and wolfram alpha has to use approximations) is only accurate to some degree and if you then have to also convert your calculations from binary to decimal these small deviations can happen

[@alannrosas2543]

Brilliant solution!

[@newtonnewtonnewton1587]

U r wonderful D peyam السلام عليكم

[@sergioh5515]

Can u do it using riemann definition?

[@sergioh5515]

@@angelmendez-rivera351 but it converged? It just wasnt in the set of R

[@sergioh5515]

@@angelmendez-rivera351 for the bonus one I mean

[@ΦοιβοςΒασαλιος]

If this integral is 0,then the area of the function ln((tanx)) between 0 and pi/2 is 0 is this fact makes a contradiction with the graph itself ;

[@healthandfitnessglobal9838]

Brilliant you are

[@romualdaszapolskasromualda4249]

Very good

[@999bigsmoke]

Cool!

[@fabiotiburzi]

3:06 WTF? 3:19 ah, ok, let's go

[@sea34101]

I don't like the second part of the video. Over [pi/2,pi] tan(x) is negative, so calculating ln(tan(x)) doesn't really make sense. Using that "main branch" without any proper explanation looks like a magical trick to me.

[@drpeyam]

It’s a standard convention in complex analysis