A beautiful log-trig integral featuring an important constant

Here's a nice log trig integral that evaluates to a famous value of the Reimann zeta function. Solution development using basic trig substitution and a beautiful series expansion of the natural log function.

[manstuckinabox]

It's been a while since I sat my butt down and enjoyed a Maths 505 video, and I see I've got a bunch to enjoy.

[MrWael]

Really impressive solution. Smart substitutions, fascinating integral. Thanks for your selections and your attractive videos.

[pinsonraphael]

Nice video and beautiful result! As someone else pointed out in the comments, substituting u = ln(sin(x)) makes the calculation quicker because cotan(x)dx is simply du, you're left with the integral from -inf to 0 of u.ln(1 - e^2u). Then, series expansion and integration by parts yields the result quite quickly as long as you don't make as many sign errors as I did.

[davidblauyoutube]

Excellent, I didn't expect to see Apery's constant.

[michaellarson]

I love integrals, but this one was a bit too easy for my liking. I even figured this one out, and that’s saying something!

[maalikserebryakov]

Finally an integral on my level

thank you for the confidence boost i now feel like Wolfram Alpha in human form

[devanshsrivastav]

what if you differentiate the beta function twice?

[soup]

I expressed the integral as derivatives of the Beta function and evaluated a really long limit to get the answer

[pluieuwu]

gotta say this one is a bit easier than the other ones 😂 i managed to anticipate every step except for the infinite series expansion which i really should try to use more

[mandlikprajwal]

Can you just solve it by using ln(sinx)=t, all rest terms cancel out leaving just Jonquière's function

[-.-.-.-..-.-.--.]

If you do a substitution by u=sin(x)^2, the ways to solve it is even better

[manthing]

I'm not sure if I saw this somewhere already, but my first idea was to define I(A, B)=integral from 0 to pi/2 sin^A(x)cos^B(x)dx, then differntiate it one time with respect to A, and one time with respect to B, plug A=1 and B=-1, but you have to calculate two derivatives of beta function.
I'm not even sure, if that what would be legal strategy, convergence wise, just my first thought.

[skyethebi]

I would never overcomplicate this for myself by bringing in derivatives of the beta function

[insouciantFox]

Every time you use IBP on the definite int[x^ylnx]dx, I scream internally.
int[x^ylnx]dx = d/dy int[x^y]dx

[lluisteixido]

I'm confused by one fact in many of your videos. You say that, in order to use the infinite series expansion of 1/1-u, |u| must be strictly less than 1. But the limits of integration comprise the case of u being exactly 1. Would not that be a contradiction and a limitation in the use of the expansion?