Cool! I first took lnx=t and then applied (1-t)/(1+t)=z and it popped out like a popcorn. Easy problem it was compared to others. A Secular hatsoff from this Secular guy 😎
Love From india, Your videos made me more passionate abt integrals😅 Jee advanced(India)exam is on 24th may Very much excited to learn these type of integrals in college😊
Yes, but then the bounds of integration become (0,inf) so the negative's can be used to make the bounds (inf,0). He did make a mistake in not writing it though
Always nice to see a Mellin transform pop out of nowhere.
By the way, shouldn't we have alpha > 0 rather than -1 for the zeta series?
Yes you're right.
@@maths_505 the result would not be the one we're looking for if we used the functional equation for zeta(s)/zeta(1-s) for -1 < s < 0 ?
I think this was the first time I managed to (successfully) solve one of your integrals without hints from the video. Was a nice one
Welcome to the maths505 channel, the only place on earth an integral would be called “cute” and “hospitable”
Pretty hardcore Integral there.
Cool! I first took lnx=t and then applied (1-t)/(1+t)=z and it popped out like a popcorn. Easy problem it was compared to others.
A Secular hatsoff from this Secular guy 😎
Thank you for your effort.
Fantastic
d/dx e^-x = -e^-x and not e^-x right?
Solved today's integral 🔥🔥
Where do you bring these beautiful integrals from? This is breathtaking! Thanks a great deal, I enjoyed it a lot!
This one I made up and others I come across by accident as a result of incorrectly solving other integrals 😂
Love From india,
Your videos made me more passionate abt integrals😅
Jee advanced(India)exam is on 24th may
Very much excited to learn these type of integrals in college😊
It's on 26 may bro
@@arnavmeena525 @7yamkr Whoops!
Oo sry for typo 🙏😂
💀 last time me mocks pyq ke alawa kya hi kr skte math 505 best for entertainment
You almost gave me heart attack when you said 24th MAY
Although doesn't really matter its just a 2 day span
@ 5:37 shouldn't it be dt = -e^-x dx ?
Yes, but then the bounds of integration become (0,inf) so the negative's can be used to make the bounds (inf,0). He did make a mistake in not writing it though
well this is probably the first time there is a constraint which hasnt come while solving the integral
asnwer=-1x
domain expansion: geometric series
name of the app? ty!
Nice thumbnail
Ans (-1)^a a! (2^(a+1) - 1) z(a+1) /2^(a)
Isnt it??
Intanto t=lnx...I=INTln((1-t)/(1+t))^a/t..t=0..1,poi vedremo....ecco I=2((-1)^a)Γ(a+1)(ξ(α+1)(1-(1/2)^(a+1)).. ovviamente devo ricontrollare