A generalized Fresnel integral: int from zero to infinity of sin(x^n)

[@maths_505]

Note: At the 11:27 mark, v=1/2 - 1/2n and the result for u changes accordingly; the reason the mistake didn't affect the final result is the symmetry of the Beta function in its arguments u and v so you get the exact same result as in the video

[@rajendramisir3530]

Besides research and rigorous trial & error, I think creativity and logical reasoning play a huge part in arriving at these integral solutions as worked out by Maths 505. He works tirelessly to solve these integrals and differential equations. He educates us by sharing his work. I suggest he compiles all his impressive work into a book for easy reference.

[@lexinwonderland5741]

i agree! I think his problems perfectly walk the line of clever and accessible, and every time his videos end i catch myself just saying "huh. neat!". I'm glad to see him on youtube!

[@danielrosado3213]

Use ramanujans master theorem!! So much faster and easier though you do have to show the integral converges which is a bit difficult.

[@joaomatos6598]

You can prove this integral converges by the Dirichlet’s convergence theorem.

[@lexinwonderland5741]

highly recommend @Michael Penn video on this!!

[@Calcprof]

The inside exponential/trig integral (about 7:50 or so) is just the Laplace transform of sin(x)

[@riadsouissi]

I used to know of only two methods to solve the Fresnel integrals, Laplace integration transform and complex contour integration. I like this one where 1/x^a is replaced by a convenient integral. This reminds me of a similar method to find the sum of 1/(n^2+a^2) which can be replaced by integral of sum of sin(nx)e^(-ax)/n.

[@alainleclerc233]

From around 8 to 9:00, you could have applied this beautiful Integration method to the two other Fresnel integrals i.e. cos(xˆn) and exp(ixˆn) with suitable minor adjustment. Thanks a lot for such different approach. More please!

[@chrisbarrington108]

Thanks Kamal… Loved this one… Your explanations are really clear… I considered the integral of exp(-ix^n), then used a slightly different subst u^(1/n)=i^(1/n)*x, which gives dx=(1/n)(i)^(-1/n)(u)^((1/n)-1)du… The integral immediately becomes (i)^(-1/n)*(1/n)*gamma(1/n) without any further substitutions or integration. The term (i)^(-1/n) is cos(pi/2n)-i*sin(pi/2n) and consequently you get the result for both the integral of sin(x^n) and cos(x^n) in a slightly more streamlined form: Integral of sin(x^n) dx = (1/n)*gamma(1/n)*sin(pi/2n). And Integral of cos(x^n)dx = (1/n)*gamma(1/n)*cos(pi/2n) The solution for sin is the same as yours, just apply the reflection formula for the gamma function and the double angle formula for sin to your solution… What I love about your videos is that they inspire us to look at these problems in different ways… Thank You!!!

[@illumexhisoka6181]

The moment you used that substation I knew that you are going to you the integral representation of x to some power

[@manstuckinabox3679]

The Golden question is, where in the world did you learn all these integration techniques? do you recommend any books or is it just creative thinking at it's finest. P.S: You do an awesome job at explaining the intuition behind them, but I'm becoming increasingly jelous of your magic tricks magician....

[@maths_505]

I got the parametrization of 1/x^n from an exercise in advanced calculus by Woods As far as everything else is concerned it's mostly hit and trial through self learning and lots of searching on the internet....alot of stuff is available on the stack so do check it out from time to time....and you can find lots of pdfs on the internet with examples on applying Feynman's technique, contour integration, the beta and gamma function, the digamma functions, polylogarithms etc

[@manstuckinabox3679]

@@maths_505 Ahhh sweet, the ol' exchange. Oh, and I'll definitely check out Woods' book.

[@elibrahimi1169]

7:31 i thought you'd recognize this as the laplace transform of sinx , except the s is replaced with t, and the t is replaced with x however , the way you handled it was intresting aswell

[@lexinwonderland5741]

my guy i love your content (ive been on a commenting spree on your videos lately lol) but the handwriting is KILLING me, especially how handwritten n looks like u and such, i get writing on a phone/tablet is hard but i would definitely appreciate if the handwriting was a little more careful. i want to appreciate your kickass content more easily!! i'm thrilled seeing your videos, you have a very casual and friendly yet nerdy demeanor and you address all sorts of decently advanced topics in a very accessible way. i watch you and Michael Penn's videos the way my parents did the crossword in the morning paper, it's just the right level of challenge for casual entertainment. all in all great vid and keep it up!!

[@usernameisamyth]

thanks man... keep making these

[@TheHellBoy05]

I obtained 1/n{Gamma(1/n)}sin(π/2n). And I'm very curious to see why my result is wrong, because at n=1, the answer is invalid

[@cameronspalding9792]

Whenever I do problems like this I always like using contour integration

[@lucadituri]

At 11:27 isn't v supposed to be equal to 1/2 - 1/2n ?

[@maths_505]

Yup I just pinned a comment

[@aranaedgar3]

Excelente video, excelente explicación, lo felicito por el nivel matemático y la forma de tratar cada caso ❤❤❤

[@realdebil.]

this gameplay fire

[@MrWael1970]

v = 1/2(1-1/n) there is a mistake in this step. Thank you very much for this video.

[@pestopoppa]

After the initial variable transofrm, why not expand SIN(x) via taylor and write is as a residue integral in Mellin form? The original integral then becomes the Mellin transform of an inverse mellin transform and everything pops out as desired.

[@jonathan3372]

I noticed that the integral is the Mellin transform of sinx evaluated at a=1/n. Could you elaborate on your method, sounds interesting?

[@harovar7]

{integral( sin(xⁿ) ), x=0 to inf }= (1/n)! sin(pi /(2n)) n>1

[@hoangquang7153]

If int from 0->pi/2 ??

[@cameronspalding9792]

I did my own method using the contour integration and I got a term involving sin(pi/2n)

[@TheHellBoy05]

Did you obtain 1/n{gamma(1/n)}sin(π/2n)? Because I obtained the same result. However at n=1, the result is invalid. But I'm curious to see where the error arose from.

[@holyshit922]

It looks like Gamma function

[@giuseppemalaguti435]

A me risulta 1/nG(1/n)sinpi/2n.. G funzione gamma

[@Decrupt]

oh damn, more content today. nice

[@povijarrro]

Hello nice video. But you should not use n and u in same problem solving.

[@renesperb]

Mathematica gives the result I[n]= Gamma[1+1/n]*sin[π/2n] . Can you check if this is your result ?

[@maths_505]

Myers correctly pointed out that the results are equivalent. This definitely a nicer result though

[@renesperb]

I just checked that it is the same .

[@paulboutemy8991]

@@maths_505 Wonderful, keep it up but how can we prove this equivalence ?

[@davidblauyoutube]

Incredible result!

[@maths_505]

Indeed

[@digxx]

Well.... While interchanging the x- and t-integral may work, the Fubini condition for the absolute value integral is not fulfilled...