#Integral with sinx / x^2
pliant violetBOT
jaunty spoke
Unsure how to proceed with this integral. I've managed to find a closed form for the infinite sum within the integral : e^(cos x) sin(sin x).
After that, I'm lost
Things I've tried:
- by parts with v' = e^(cosx) sin x
- t-substitution: unsure how to evaluate the resulting integral
frigid granite
Unsure how to proceed with this integral. I've managed to find a closed form for...
Is that the original integral ?
jaunty spoke
Is that the original integral ?
yes
frigid granite
yes
This part :
$$ \int_0^\infty \frac{sin(x)}{x}\frac{sin(x/n)}{x/n}dx$$
Looks like a borwein integral so it should be pi/2
You can use laplace's transform to show it
So the result must be (e-1)pi/2 if i dont mess up
summer oxideBOT
Ævan
jaunty spoke
This part :
$$ \int_0^\infty \frac{sin(x)}{x}\frac{sin(x/n)}{x/n}dx$$
Hey, thanks for responding.
I'm not seeing how you get the integrand into the form of a borwein integral tho
nimble monolith
to clarify, after rewriting the infinite sum the integral becomes from 0 to inf of sinx/x² * e^cosx * sin(sinx)?
jaunty spoke
to clarify, after rewriting the infinite sum the integral becomes from 0 to inf ...
Assuming my work is correct, yes
jaunty spoke
yeah
nimble monolith
do you have reason to believe one exists?
jaunty spoke
It's a problem on one of the berkeley maths challenge so I kinda just assumed from there
nimble monolith
ah they don't have a closed form answer?
jaunty spoke
The solutions aren't out yet
jaunty spoke
I don't know the former and I'm in the process of teaching myself the latter
nimble monolith
if you can write this as a nice and equivalent complex function
let me try something
jaunty spoke
alright
nimble monolith
your answer is π/2 * (e-1)
jaunty spoke
Holy shit, you did it
jaunty spoke
your answer is π/2 * (e-1)
How did you get write it as a complex function?
I made some attempts but nothing really worked out
nimble monolith
i tried for hours using a complex analytic technique or feynman’s but couldn’t make any good progress
so the trick was actually not rewriting the summation at all
the thing inside the integral clearly uniformly converges so there’s no issue with switching the integration and summation operators, so you can just pull the sum and the 1/n! outside of the integral and just integrate sinx*sin(nx) / x² from 0 to infinity
the integral ends up always equalling pi/2 for the values that n takes on
you can prove that either by residue theorem or integration by parts
so you just get the sum from n = 1 to infinity of 1/n! * pi/2, and if you recognize that the sum of 1/n! is just another way to represent e if the sum starts at 0 (think the taylor series of e^x with x = 1), then you get (e-1) * pi/2
jaunty spoke
the thing inside the integral clearly uniformly converges so there’s no issue wi...
I spent the past hour trying to wrap my head around this but I think a proper understanding of it is a bit advanced for my level anyway.
I did manage to achieve the same result for "sinx*sin(nx) / x² from 0 to infinity" with by parts
Anyway, thanks for your help! After 3 days on this problem, I think I'm somewhat satisfied with the outcome(for now)
.close
pliant violetBOT
Solved