I was really expecting Bessel functions to appear in this problem since they have integral representations something like int exp(i*sin(x)-i*x). Your integrals under the summation sign looked somewhat similar. And wikipedia confirms -- series coefficients for Bessel 0 functions ~ 1/(k!)^2. So, maybe this problem can be solved by transforming the integrand into some combination of Bessel functions.
If I am not mistaken, the series in the final result is just the series representation of the modified Bessel function of the first kind I_0(2), or something like that. So yeah, the result IS a Bessel function... I think.
@Maths_505 - a suggestion for your consideration - I think it would be cool to show the Desmos graph of the integrand as part of the videos. Would give at least an approximate sense for the order of magnitude we should expect of the answer.
When i saw the thumbnail, then i proceeds to solve, and solved it. For checking answer, i open the video. I literally did the same steps as done by you I just want to know why you and me used to think same to same !?!😂
what I did is rewrote that integral as intgrl e^(2cosx) dx from 0 to pi/2 + intgrl e^(-2cosx) dx from 0 to pi/2 Then I applied e series. Summation 2^k /k! from k=0 to infinity { intgrl (cosx)^k dx from 0 to pi/2} + Summation 2^k /k! from k=0 to infinity { intgrl (-cosx)^k dx from 0 to pi/2} For k=odd, they will cancel out but will become added into double for k=even, so we take even terms only, so for k=2n we can rewrite it as, 2 summation 2^(2n) /(2n)! from n=0 to infinity { intgrl (cosx)^2n dx from 0 to pi/2} = 2 summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) /(2 T(n+1))} =summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) T(n+1) /(T(n+1))^2 } Apply legendary duplication formula. = summation pi/(n!)^2 from n=0 to infinity.
Pretty cool. I managed to derive the same result using purely real methods by applying e^x maclaurin series as the first step, then splitting the integral as (0 to π/2) + (π/2 to π). I could then make it fit into the form of the beta function. Then after playing around with the terms in the summation i could apply the gamma function duplication formula
You can also use the Taylor series of e^x then you get a beta function with k+1/2, then use Legendre's duplication formula and get the same result. By the way I'd like to mention that this is my number one math channel, no other boi motivates me as much to solve such problems.
I tried using the half limits property and brought the limits to π/2 and now to use the beta function I could use the series expansion of e^x , we have some random constant times the integral of cosⁿx , which is 1/2 B(2n+1/2, 1/2), after converting it into the gamma function I do get 1/n!² but i also get some random stuff in the numerator, is my process wrong?
Utilizzando feyman (e^acosx..)arrivo ad uequazione differenziale del 2 ordine I"(a)+I'(a)/a-I(a)=0..I'(a)=0,I(a)=π..che non so risolvere .ho letto di funzioni di bessel che però non conosco
Tried feynman technique but got a 2nd order differential equation which tbh seemed impossible to "intuit" what the solution would be, did not actually try to solve the differential equation but thought I had hit a roadblock, maybe you could try solving it xf(x)+f'(x)+xf"(x)=0
My friend. I did it myself using summation series of e^x and legendres duplication formula. Isn't the answer pi/(n!)^2, n ranging from 0 to infinity?? Did you do the same?? Or I'll comment the whole process.
@@maths_505 what I did is rewrote that integral as intgrl e^(2cosx) dx from 0 to pi/2 + intgrl e^(-2cosx) dx from 0 to pi/2 Then I applied e series. Summation 2^k /k! from k=0 to infinity { intgrl (cosx)^k dx from 0 to pi/2} + Summation 2^k /k! from k=0 to infinity { intgrl (-cosx)^k dx from 0 to pi/2} For k=odd, they will cancel out but will become added into double for k=even, so we take even terms only, so for k=2n we can rewrite it as, 2 summation 2^(2n) /(2n)! from n=0 to infinity { intgrl (cosx)^2n dx from 0 to pi/2} = 2 summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) /(2 T(n+1))} =summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) T(n+1) /(T(n+1))^2 } Apply legendary duplication formula. = summation pi/(n!)^2 from n=0 to infinity.
this channel has a short about this due to the analytical continuation of the Zeta function at s=-1 it does converge to -1/12 fun fact : this result is used in string theory 😊
@@vascomanteigas9433 Numberphile recently did a video (titled "Does -1/12 Protect Us From Infinity?") where they explained how it made sense even without analytic continuation. Quite a surprising result, based on an approach by Terence Tao and others.
10:03 Well, sure, all that was my first guess. I was a math major at VPI&SU (now Va. Tech), and I don't think I ever got into any really interesting stuff as an undergrad. Alas.
Was bored studying organic chemistry, saw this video posted and decided to take a look Yeah Ima stick to organic chemistry for now, I ain't come this far, for now.
Wouldn’t it be easier to weigh a piece of graph paper and define the unit weight, Then plot the function, cut out the area between the plot and the x-axis and weigh it.🤣
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I was really expecting Bessel functions to appear in this problem since they have integral representations something like int exp(i*sin(x)-i*x). Your integrals under the summation sign looked somewhat similar. And wikipedia confirms -- series coefficients for Bessel 0 functions ~ 1/(k!)^2. So, maybe this problem can be solved by transforming the integrand into some combination of Bessel functions.
If I am not mistaken, the series in the final result is just the series representation of the modified Bessel function of the first kind I_0(2), or something like that. So yeah, the result IS a Bessel function... I think.
Confirm. The integral are given by the modified Bessel function of First Kind:
pi * I[0](2).
The sum are the definition of a Bessel function. I[0](2)
I like how you write your integral signs and keep things neat. The summation symbol is almost there
wolframalpha says that sum is I_0(2) where I_0 is the “modified bessel function of the first kind”
the thumbnail made me immediately think residue theorem, video did not disappoint
@Maths_505 - a suggestion for your consideration - I think it would be cool to show the Desmos graph of the integrand as part of the videos. Would give at least an approximate sense for the order of magnitude we should expect of the answer.
This is my first time seeing a contour integral, amazing job bro as usual. This is light work for you.
When i saw the thumbnail, then i proceeds to solve, and solved it.
For checking answer, i open the video.
I literally did the same steps as done by you
I just want to know why you and me used to think same to same !?!😂
Where did you learn math ?
I guess we can use Cauchy's integral formula directly instead of using residues.
Nice integral!!
what I did is rewrote that integral as
intgrl e^(2cosx) dx from 0 to pi/2 + intgrl e^(-2cosx) dx from 0 to pi/2
Then I applied e series.
Summation 2^k /k! from k=0 to infinity { intgrl (cosx)^k dx from 0 to pi/2} + Summation 2^k /k! from k=0 to infinity { intgrl (-cosx)^k dx from 0 to pi/2}
For k=odd, they will cancel out but will become added into double for k=even, so we take even terms only, so for k=2n we can rewrite it as,
2 summation 2^(2n) /(2n)! from n=0 to infinity { intgrl (cosx)^2n dx from 0 to pi/2}
= 2 summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) /(2 T(n+1))}
=summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) T(n+1) /(T(n+1))^2 }
Apply legendary duplication formula.
= summation pi/(n!)^2 from n=0 to infinity.
Awesome
Whoa! Nice my friend 😊
Pretty cool. I managed to derive the same result using purely real methods by applying e^x maclaurin series as the first step, then splitting the integral as (0 to π/2) + (π/2 to π). I could then make it fit into the form of the beta function. Then after playing around with the terms in the summation i could apply the gamma function duplication formula
That is such a beautiful sum at the end! 😊 I'm surprised it hasn't got a name. You should call it the Kamal constant
Nah bro it turns out it's a Bessel function.
You can also use the Taylor series of e^x then you get a beta function with k+1/2, then use Legendre's duplication formula and get the same result.
By the way I'd like to mention that this is my number one math channel, no other boi motivates me as much to solve such problems.
Very nice! Just glad it didn't result in some special function or constant yet again. I kinda prefer that we have the answer in a raw form.
Hell yeah!
We raw doggin em 🔥🔥
It does result in a special function, see several other comments - you can express the result using a Bessel function.
A beautiful solution.
I tried using the half limits property and brought the limits to π/2 and now to use the beta function I could use the series expansion of e^x , we have some random constant times the integral of cosⁿx , which is 1/2 B(2n+1/2, 1/2), after converting it into the gamma function I do get 1/n!² but i also get some random stuff in the numerator, is my process wrong?
I used I(a)= limits (0,2a) ;f(x) = limits (0 ,a) ; f(x) +f(2a-x) to half the limit
Thank you for this featured solution. I hope to be solved over real domain.
Utilizzando feyman (e^acosx..)arrivo ad uequazione differenziale del 2 ordine I"(a)+I'(a)/a-I(a)=0..I'(a)=0,I(a)=π..che non so risolvere .ho letto di funzioni di bessel che però non conosco
Found the exant same thing using the power series of e^(2cosx) from the beginning
This is Actually Pi times the 0th order modified bessel function of the first kind evaluated at x=2.
You make me happy that I have Sherman K Stein's book Calculus and Analytic Geometry.
Tried feynman technique but got a 2nd order differential equation which tbh seemed impossible to "intuit" what the solution would be, did not actually try to solve the differential equation but thought I had hit a roadblock, maybe you could try solving it
xf(x)+f'(x)+xf"(x)=0
Multiply by x, and you have a special case of Bessel's differential equation (for the Bessel functions of order 0).
@@bjornfeuerbacher5514 oh ye, nicee thanks
My friend. I did it myself using summation series of e^x and legendres duplication formula. Isn't the answer pi/(n!)^2, n ranging from 0 to infinity?? Did you do the same?? Or I'll comment the whole process.
I tried the same but doesn't work
Greetings my friend. Did I not write the same answer?
@@lakshay3745 say didn't not doesn't. 😅 It works
@@Mathematician6124can you please share your solution development
@@maths_505 what I did is rewrote that integral as
intgrl e^(2cosx) dx from 0 to pi/2 + intgrl e^(-2cosx) dx from 0 to pi/2
Then I applied e series.
Summation 2^k /k! from k=0 to infinity { intgrl (cosx)^k dx from 0 to pi/2} + Summation 2^k /k! from k=0 to infinity { intgrl (-cosx)^k dx from 0 to pi/2}
For k=odd, they will cancel out but will become added into double for k=even, so we take even terms only, so for k=2n we can rewrite it as,
2 summation 2^(2n) /(2n)! from n=0 to infinity { intgrl (cosx)^2n dx from 0 to pi/2}
= 2 summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) /(2 T(n+1))}
=summation 2^(2n) /(2n)! from n=0 to infinity { sqrt pi T(n +1/2) T(n+1) /(T(n+1))^2 }
Apply legendary duplication formula.
= summation pi/(n!)^2 from n=0 to infinity.
ramanujan sum said 1+2+3+...=-1/12 what about this sum ? it's true or false ?
this channel has a short about this
due to the analytical continuation of the Zeta function at s=-1 it does converge to -1/12
fun fact : this result is used in string theory 😊
@@aravindakannank.s. interesting 🤔
@@aravindakannank.s. fun fact: string theory is a fallacy
It is only make sense under the Analytical Continuation of Zeta Function, otherwise it is a divergent sum.
@@vascomanteigas9433 Numberphile recently did a video (titled "Does -1/12 Protect Us From Infinity?") where they explained how it made sense even without analytic continuation. Quite a surprising result, based on an approach by Terence Tao and others.
Parseval's identity for x->exp(exp(ix))=sum(exp(ikx)/k!, k=0..infinity) ?
Next do eˆ(cosˆ2(x)) from 0 to pi. (If its even possible, wolfram could do it atleast but im no integral expert)
Since cos²(x) = 0.5 (1 + cos(2x)), that integral would be quite similar to the one in this video.
Do the integrals of bounded functions, integrated over some (bounded) limits, always exist?
Obviously yes
@@maths_505 Such a function may not be Riemann-integrable (it is iff the upper and lower integral are equal).
10:03 Well, sure, all that was my first guess. I was a math major at VPI&SU (now Va. Tech), and I don't think I ever got into any really interesting stuff as an undergrad. Alas.
Before watching the video or even thinking of a solution or anything
I thought: oh this function is a product of two complex functions
Kamal doing kamal fr
Someone said ,"How is not as important as Why", Why did you even get these type of cool Upper esh and their OK cool soln development
Hey Math505 I got the answer as 0.7157
Was bored studying organic chemistry, saw this video posted and decided to take a look
Yeah Ima stick to organic chemistry for now, I ain't come this far, for now.
😂😂😂
Same 🫠 and then some complex analysis and learn it and then rewatch this to understand something
organic chemistry is also cool unless or until the question asks me about the 10-20 step mechanism 😅
@@aravindakannank.s. i like the subject its just that i like math more so comparatively speaking, it gets boring.
Wouldn’t it be easier to weigh a piece of graph paper and define the unit weight,
Then plot the function, cut out the area between the plot and the x-axis and weigh it.🤣
Don't talk to stranded intergrals.
BesselJ(α,z)=
∫_{0}^{pi} dt/pi e^(-i*(αt-z*sin(t)))
So, I(0,2)
Too messy.
Put 2cosx = t, and apply ILATE