Taking the logarithmic derivative of the Legendre's duplication formula is more straightforward, but using infinite series looks much cooler, very nice result indeed.
How timely! This integral is very close to one that I'm working on evaluating, namely integral 0->1 of x^(s-1)/(1+x)^a dx, where 2a is an integer that is at least 4. That integral is related to the area under the curves y^2 = x^n + (1-x)^n.
For the integral I just solved, you could introduce a parameter t in the denominator as a coefficient of x term. Differentiating under the integral sign will give you higher powers of the binomial in the denominator and you'll add -1 to the exponent of the numerator each time you differentiate. The R.H.S will just be a couple of polygamma functions. A pattern should hopefully reveal itself. And for going back to s-1, adjust the parameter in the differentiated integral. That seems to work in my head. Hopefully I haven't missed anything and it works on paper too.
Could you please make a video on how to derive Euler's reflection formula for the Gamma function? I'm also interested in how to derive Euler's integral representation of the Gamma function from his infinite product representation.
@@maths_505 I couldn't find it in the Proofs playlist, what is its name? There are two videos mentioned as not available in that list, though, maybe it's one of them? (I have no clue why they are shown as unavailable to me.)
Usually you mention when you need some sort of convergence or boundedness - don't you need absolute convergence here to do some of the series rearrangement? I'm pretty sure you have it off the cuff but its probably not clear a priori
When he first separates the even and odd values of n in the infinite series, he subtracts two divergent series. It is definitely not correct to say it this way, though corrections can easily be made (perhaps by going to partial sums, or else explaining the result without splitting the series into two series).
How do we know we can split up/rearrange/add "zero" to the infinite series here when they're harmonic series that don't converge absolutely. I don't doubt that this works but surely it's only "correct" if we're more careful with the ordering of the terms inside the sum, right?
that darn 1... if only you were infinite, then I can show off my new found integration techniques papa Gamelin taught me... guess I'm going to just use infinite series now... too bad there isn't a sub we can do to change the bounds of integration...
Taking the logarithmic derivative of the Legendre's duplication formula is more straightforward, but using infinite series looks much cooler, very nice result indeed.
Yeah I know that's faster but I just couldn't stop after those series evaluations
Legendre*
How timely! This integral is very close to one that I'm working on evaluating, namely integral 0->1 of x^(s-1)/(1+x)^a dx, where 2a is an integer that is at least 4. That integral is related to the area under the curves y^2 = x^n + (1-x)^n.
For the integral I just solved, you could introduce a parameter t in the denominator as a coefficient of x term.
Differentiating under the integral sign will give you higher powers of the binomial in the denominator and you'll add -1 to the exponent of the numerator each time you differentiate.
The R.H.S will just be a couple of polygamma functions. A pattern should hopefully reveal itself.
And for going back to s-1, adjust the parameter in the differentiated integral.
That seems to work in my head.
Hopefully I haven't missed anything and it works on paper too.
First time learning of the macaroni constant.
super satisfying at the end:D Keep up the great work!
Could you please make a video on how to derive Euler's reflection formula for the Gamma function?
I'm also interested in how to derive Euler's integral representation of the Gamma function from his infinite product representation.
@@Grecks75 I have a video on the reflection formula that you can find in the proofs playlist
@@maths_505 I couldn't find it in the Proofs playlist, what is its name? There are two videos mentioned as not available in that list, though, maybe it's one of them? (I have no clue why they are shown as unavailable to me.)
@@Grecks75 it's part of the video for int 0 to pi/2 tan^i(x)
Great, thanks! ❤️
Usually you mention when you need some sort of convergence or boundedness - don't you need absolute convergence here to do some of the series rearrangement? I'm pretty sure you have it off the cuff but its probably not clear a priori
Really interesting and smart integral and solution. I like your steps for solution.
Nd it's a new toy for future videos
Great job my friend i love your ideas about inegration
When he first separates the even and odd values of n in the infinite series, he subtracts two divergent series. It is definitely not correct to say it this way, though corrections can easily be made (perhaps by going to partial sums, or else explaining the result without splitting the series into two series).
The partial sums approach is definitely the right way to go.
That's what I had in mind while writing the proof
How do we know we can split up/rearrange/add "zero" to the infinite series here when they're harmonic series that don't converge absolutely. I don't doubt that this works but surely it's only "correct" if we're more careful with the ordering of the terms inside the sum, right?
One day you could do a study of the digamma function, its graph, zeros, poles etc
Is it ok to rearrange the order of terms of an infinite series that is convergent but not absolutely convergent?
What about the Nielsen beta function?
Perhaps I made a mustake. I got PSI(2s) + ln(2). Please Controle it
that darn 1... if only you were infinite, then I can show off my new found integration techniques papa Gamelin taught me... guess I'm going to just use infinite series now... too bad there isn't a sub we can do to change the bounds of integration...
As a regular viewer of the channel my friend you know that I will not shut up about this new toy for the next few weeks😂😂😂
First representation seems more compact ( ψ ( (s+1)/2 ) - ψ ( s/2 ))/2 . Not sure what is gained in the second half.
I guess it's personal preference 👍
Or to show off hehe 😎😎😎
Yes I was definitely showing off there 😂
Exelente
You know the video is fire when one punch man gets excited
Wow
Molto semplicemente è la sommatoria di (-1)^k/s+k
Oh thank you so much no one would ever have guessed that despite being halfway through the video 🤣
@@maths_505 i havent understood well,but i know digamma function not much