Complex Analysis: Integral of sin(x)/x using Contour Integration
qncubed3
144,740 views
Today, we use complex analysis to evaluate the improper integral of sin(x)/x, also known as the Dirichlet Integral.
Laplace Transform Method:
Laplace Transforms: Integral of sin(x)/x f...
20 Comments
@AnibalRiba 1 month ago
You can skip the indentation entirely: subtract 1 from the numerator and integrate (e^(iz) − 1)/z over a plain semicircle.
The −1 isn't arbitrary — subtracting 1 up top is the same as subtracting 1/z, which is exactly the principal part at the origin. Strip that off a simple pole and the singularity becomes removable, so the integrand is entire and plain Cauchy gives zero. No residue theorem, no notch, no principal values, and one limit instead of two.
It costs you nothing on the real axis, because the subtraction only hits the real part: cos x over x becomes (cos x − 1) over x, which is fine at the origin, while the imaginary part is still sin x over x.
The catch: the large arc no longer vanishes. Split it — the e^(iz)/z piece dies by Jordan, and the 1/z piece is exactly iπ for any radius. That's where your π comes from. Same iπ as in your version, just billed on the large arc instead of the small one.
Not strictly better, mind you: you trade the pole at the origin for a tail at infinity that only cancels symmetrically, and indentation still handles cases where this trick has nothing nice to leave behind.
The −1 isn't arbitrary — subtracting 1 up top is the same as subtracting 1/z, which is exactly the principal part at the origin. Strip that off a simple pole and the singularity becomes removable, so the integrand is entire and plain Cauchy gives zero. No residue theorem, no notch, no principal values, and one limit instead of two.
It costs you nothing on the real axis, because the subtraction only hits the real part: cos x over x becomes (cos x − 1) over x, which is fine at the origin, while the imaginary part is still sin x over x.
The catch: the large arc no longer vanishes. Split it — the e^(iz)/z piece dies by Jordan, and the 1/z piece is exactly iπ for any radius. That's where your π comes from. Same iπ as in your version, just billed on the large arc instead of the small one.
Not strictly better, mind you: you trade the pole at the origin for a tail at infinity that only cancels symmetrically, and indentation still handles cases where this trick has nothing nice to leave behind.
@kayleeweatherspoon6526 3 years ago
I might actually pass my qualifying exam because of your videos--thank you so much!
@icee562 6 years ago
noice All this just to prove inverse fourier transform of Fourier transform is the function itself .
@alicesmith5361 6 years ago
This is so cool! Thank you so much for this example; I was quite lost when my professor brought this kind of process up as a point tangential to the lecture.
@abcdef2069 2 years ago
lets say epsilon be r to reduce words
at 10:35 integ [ exp(i r exp (i * t)) ] ~ 1/r (1 - exp(- r * constant)) when r goes to 0, this results in 0/0 and, this was the same as the jordan thing. or he could have done the same just to apply R going to infinity BEFORE the integration, but he didnt. but ok the limit goes to ~ r const / 1 = 0
at 10:35 integ [ exp(i r exp (i * t)) ] ~ 1/r (1 - exp(- r * constant)) when r goes to 0, this results in 0/0 and, this was the same as the jordan thing. or he could have done the same just to apply R going to infinity BEFORE the integration, but he didnt. but ok the limit goes to ~ r const / 1 = 0
@jperez7893 2 years ago
this is how to show a solution, you showed every step and why. it infuriates me to follow a proof and they skip several steps
@barryzeeberg3672 7 months ago
11:02 - it is interesting that the integral over the little half circle is -i * pi. If we had used a full circle instead of half circle, by Cauchy we know that the integral would be -2 * i * pi . I wonder if we could have avoided the previous several lines of computation by just recognizing that the integral for the half circle is 1/2 * (-2 * i * pi)?
Complex Analysis: Integral of sin(x)/x using Contour Integration
qncubed3
144,740 views
Today, we use complex analysis to evaluate the improper integral of sin(x)/x, also known as the Dirichlet Integral.
Laplace Transform Method:
Laplace Transforms: Integral of sin(x)/x f...
https://www.youtube.com/watch?v=yDiC_tZdEp8