Okay I did the extra digging You can either get a difference of two trigamma functions or the result I’m most fond of, which is in terms of the Lerch Transcendent 1/(2*pi)*Lerchphi[-1,2, (s+pi)/(2pi)] I think s being positive is a sufficient enough criterion (it might even work for some values of negative s if I’m being honest) but I think that’s a really nice result for this problem.
1:21 tear drop in my eye, I miss complex analysis :'( SHEESH That was totally not expected! definitely saving this technique in my mental tool box. this involved literally every single one of my favorite identities. it was one heck of a 15 minutes.
Aliter: contour integration Use f(z) =1 / {(lnz)(z-1)²√z} Now use the keyhole contour from (-π to π) with residue at z=1 (of 3rd order) If you forgot me, I'm that indian one 😂 By the way, great solution, respect for you 📈
With a little bit of manipulation from the form in 11:26 (represent cosh in exponential form, assume s is real and take the imaginary part of the expression) you can bring this to the form of pi * the laplace transform of t sech (pi t). Wolfram Alpha tells me that that's a linear combo of trigamma functions. I would love to know the steps inbetween those two... because then you get a general closed form expression for I(s).
Excellent video as always, though I think the final result for all s greater than pi using the trigamma function also deserves some credit ! Keep up the great work man !
Yes through some further digging you can get some expressions in terms of trigamma functions, and one other expression in terms of the Lerch Transcendent, it’s actually a super cool result!
As a highschool student preparing for college entrance exams, I always struggle with integrals but videos like these revive my curiosity for maths once again.Thanks for making videos like this ❤
I gotta say this one had me thinking for quite a bit. I approach this in a similar manner to you, but I used the substitution x = exp(-u). This changes the problem significantly, though not necessarily one where it makes the result easier to work with. My goal right now is to generalize this for any parameter, but the value pi actually makes this integral sing. You need it for this to work. There were contour integrals involved and in the end I only needed to solve the following integral pi/8*Int((2t-1)(2t-3)) from 0 to 1 The pi made things very easy to work out somehow, it’s actually crazy that even a problem like this can exhibit some kind of symmetry.
another alt: change of variable x=y^2. I=-4*integral_0^inf{dy*ln(y)/((1+y^2)^2*(4*ln^2(y)+pi^2))}=-integral_0^inf{dy/(1+y^2)^2*(1/(2*ln(y)+i*pi)+1/(2*ln(y)-i*pi))} =-integral_-inf^inf{dy/((1+y^2)^2*(2*ln(y)+i*pi))}, with the log branch cut taken along any direction above the real axis, which is also the integration path. Then the contour integral can be taken for the bottom half plane, with just one third-order-pole at -i.
Awesome video and beautiful development! But you noted at the start that the parameter s is characterized by the fact that its real part is greater than zero. However, the stage of invoking complex numbers assumes s is purely real. Just worth noting that.
@@maths_505Sure! Still, your development is valid for real s only. Which does not make it less breathtaking, obviously. In general, the original expression does not guarantee a purely real result for s with the non-zero imaginary part.
Heavy Duty! RE: 3:05 "Complicatedness" is not a word. Alternatively, you could have said, "Let's introduce more complication." or "Let's increase the complication." or "Let's increase the difficulty."
The beta function is defined for both real and complex numbers technically, when you consider that the real numbers are a subset of the complex numbers, just that the imaginary part is equal to 0. For example, you could plug in u=1 and v=2, which is the same as saying u= 1+0i, and v = 2+0i.
Both of them appears frequently enough in various places to be widely accepted as natural extensions of elementary functions, much like the logarithm came historically as the result of an integral and only latter got the implicit definition of the inverse of exponentials. For an expression to be or not considered a closed form, depends mostly on your definitions only or sometimes the context. If your not doing number theory or something of that sorts I highly doubt anyone is gonna consider a Clausen function an elementary object or if you brute force through a transcendental equation with the Lagrange inversion theorem instead of a more direct numerical approach you will probably get weird stares from programmers.
That being said both of them, specially the gamma function have wonderful properties that would be harder to see from their original form alone and both have been studied extensively due to their usefulness or appearances in seemingly unexpected places. The gamma function specifically is one of the most reoccurring said "transcendental functions" in physics and surprisingly we haven't ever found a simple differential equation whose family of solutions it would be a part of, which is kind unusual since a lot of exotic functions from physics come about this way, e.g. Bessel functions, Elliptical Integrals...
I tried for like twenty minutes to solve the integral for any S than I realized that I can only solve it if the integration is -infinity to infinity and not 0 to infinity.
@@SuperSilver316 Yes, but s only shows up as s^2. So any number s with Re(s)>0 has a corresponding -s which has Re(s) < 0. But s^2 is the same for both values.
Yeah that’s correct, this is more in relation to the identity he uses in the later step 1/(s+i*ln(x)) = int(exp(-(s+i*ln(x))t) from 0 to inf This is only works if the Re(s) > 0
Okay I did the extra digging
You can either get a difference of two trigamma functions or the result I’m most fond of, which is in terms of the Lerch Transcendent
1/(2*pi)*Lerchphi[-1,2, (s+pi)/(2pi)]
I think s being positive is a sufficient enough criterion (it might even work for some values of negative s if I’m being honest) but I think that’s a really nice result for this problem.
1:21 tear drop in my eye, I miss complex analysis :'( SHEESH That was totally not expected! definitely saving this technique in my mental tool box.
this involved literally every single one of my favorite identities. it was one heck of a 15 minutes.
Hell yeah 🔥🔥🔥
I love when you stop talking for a moment and i listen to you on earbuds
Aliter: contour integration
Use f(z) =1 / {(lnz)(z-1)²√z}
Now use the keyhole contour from (-π to π) with residue at z=1 (of 3rd order)
If you forgot me, I'm that indian one 😂
By the way, great solution,
respect for you 📈
With a little bit of manipulation from the form in 11:26 (represent cosh in exponential form, assume s is real and take the imaginary part of the expression) you can bring this to the form of pi * the laplace transform of t sech (pi t). Wolfram Alpha tells me that that's a linear combo of trigamma functions. I would love to know the steps inbetween those two... because then you get a general closed form expression for I(s).
Excellent video as always, though I think the final result for all s greater than pi using the trigamma function also deserves some credit ! Keep up the great work man !
Yes through some further digging you can get some expressions in terms of trigamma functions, and one other expression in terms of the Lerch Transcendent, it’s actually a super cool result!
Hi everyone, please do take a look into the channel named "The Hidden library of Mathematics". You won't regret it.
Very nice solution. Try to solve in general at s. Thank you.
As a highschool student preparing for college entrance exams, I always struggle with integrals but videos like these revive my curiosity for maths once again.Thanks for making videos like this ❤
I gotta say this one had me thinking for quite a bit. I approach this in a similar manner to you, but I used the substitution x = exp(-u). This changes the problem significantly, though not necessarily one where it makes the result easier to work with. My goal right now is to generalize this for any parameter, but the value pi actually makes this integral sing. You need it for this to work. There were contour integrals involved and in the end I only needed to solve the following integral
pi/8*Int((2t-1)(2t-3)) from 0 to 1
The pi made things very easy to work out somehow, it’s actually crazy that even a problem like this can exhibit some kind of symmetry.
Exactly what I thought while solving it 😂
Hi everyone, please do take a look into the channel named "The Hidden library of Mathematics". You won't regret it.
another alt: change of variable x=y^2.
I=-4*integral_0^inf{dy*ln(y)/((1+y^2)^2*(4*ln^2(y)+pi^2))}=-integral_0^inf{dy/(1+y^2)^2*(1/(2*ln(y)+i*pi)+1/(2*ln(y)-i*pi))}
=-integral_-inf^inf{dy/((1+y^2)^2*(2*ln(y)+i*pi))}, with the log branch cut taken along any direction above the real axis, which is also the integration path.
Then the contour integral can be taken for the bottom half plane, with just one third-order-pole at -i.
Awesome video and beautiful development! But you noted at the start that the parameter s is characterized by the fact that its real part is greater than zero. However, the stage of invoking complex numbers assumes s is purely real. Just worth noting that.
The dissection still works during the expanding with conjugate step so it's cool overall.
@@maths_505Sure!
Still, your development is valid for real s only. Which does not make it less breathtaking, obviously.
In general, the original expression does not guarantee a purely real result for s with the non-zero imaginary part.
10:33 sin(pi/2+x)=cosx
sin(pi/2+i pi t)=cos(i pi t)=cosh(pi t)
Very interesting integral and pretty solution. I hope to solve the integral in general form, not at s = pi. Thank you
i think a very good solution is with inverse laplace transform!
Was gearing up for you to do the whole thing with the fattest partial fraction decomp in history
*""it didn't do anything but it was unsettling so why not""*
Ah yes. Math is all about making ugly looking expression beautiful.
Heavy Duty! RE: 3:05 "Complicatedness" is not a word. Alternatively, you could have said, "Let's introduce more complication." or "Let's increase the complication." or "Let's increase the difficulty."
Oops, complicatedness has a dictionary entry. My bad. YMMV.
6:30 only for complex u and v? This formula aint for real u and v?
The beta function is defined for both real and complex numbers technically, when you consider that the real numbers are a subset of the complex numbers, just that the imaginary part is equal to 0. For example, you could plug in u=1 and v=2, which is the same as saying u= 1+0i, and v = 2+0i.
Eta and gamma are functions for what reason? Just simplification reasons?
instagram.com/p/CubW6kJtAcc/?igsh=MW8wY3VxZGZlbzlieg==
Both of them appears frequently enough in various places to be widely accepted as natural extensions of elementary functions, much like the logarithm came historically as the result of an integral and only latter got the implicit definition of the inverse of exponentials.
For an expression to be or not considered a closed form, depends mostly on your definitions only or sometimes the context. If your not doing number theory or something of that sorts I highly doubt anyone is gonna consider a Clausen function an elementary object or if you brute force through a transcendental equation with the Lagrange inversion theorem instead of a more direct numerical approach you will probably get weird stares from programmers.
That being said both of them, specially the gamma function have wonderful properties that would be harder to see from their original form alone and both have been studied extensively due to their usefulness or appearances in seemingly unexpected places.
The gamma function specifically is one of the most reoccurring said "transcendental functions" in physics and surprisingly we haven't ever found a simple differential equation whose family of solutions it would be a part of, which is kind unusual since a lot of exotic functions from physics come about this way, e.g. Bessel functions, Elliptical Integrals...
Thank you!
@@lynxrose3005You're welcome ^-^
I tried for like twenty minutes to solve the integral for any S than I realized that I can only solve it if the integration is -infinity to infinity and not 0 to infinity.
Very nice
The only thing I need to criticize is that this is *one* exotic function. I want TEN! 👍
I think 3628800 exotic functions is a bit excessive
@@mitri4939 Why?
hey there is a challenge for you sir can you solve this equation x^6-x^5+x^2+x+1=0
Why does Re(s) have to be > 0?
More than likely it’s the only way for some of these integrals to converge properly.
@@SuperSilver316 Yes, but s only shows up as s^2. So any number s with Re(s)>0 has a corresponding -s which has Re(s) < 0. But s^2 is the same for both values.
Yeah that’s correct, this is more in relation to the identity he uses in the later step
1/(s+i*ln(x)) =
int(exp(-(s+i*ln(x))t) from 0 to inf
This is only works if the
Re(s) > 0
Hi everyone, please do take a look into the channel named "The Hidden library of Mathematics". You won't regret it.
asnwer=1dx isit
I think I found a way to do this integral with Laplace convolutions, but boy is it ugly.
FU