One of the coolest integrals ever ! and we're gonna solve it using Feynman integration

One of my favourite integrals!
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[maths_]

Apologies for the inteeegral business 😂😂 I was just having a bit of fun since this is the way my high school prof used to say it so I just winged it for the video😂😂. Haven't repeated it since all the comments 😂😂😂

[danigarcia]

Feynman's integration is such a fascinating and beautiful way to solve inteegrals.

[jcantonelli]

"in-TEE-gral" - are we being trolled?

[erichan]

Another interesting property of sin(x)/x is that it's the Fourier transform of a rectangular pulse. I feel invoking FT would be somewhat a generalization of the approach presented in the video. In the conjugated space, the integral would probably quite trivial to evaluate.

[jmcsquared]

Great exercise, but I want to offer one amendment: you can do the integral at 2:15.

If you introduce a regulator - say exp(-tx) (which ironically you do later anyways) - then you can regulate that derivative. The regulation parameter t will be taken to zero when computing the final answer which will indeed be π/2. So, even though the integral is divergent, forcing it to converge using regularization still allows you to use the trick for sin(ax)/x.

As a MathStackExchange commenter once taught me, "convergence is overrated."

[NightWanderer]

Very powerful technique, and wonderfully explained too. I could nitpick about technical details to make it more rigorous but others have already done so. You've got a new sub.

[jez]

Nice video. A slightly technical note: the argument that you give for why the int_0^inf e^(-ax) sin(x)/x dx must tend to zero as a tends to infinity (namely, that for each fixed x the integrand tends to zero as a tends to infinity) isn't quite complete. As a simple example, consider the function f(a, x) which is defined to be 1 if a < x < a + 1 and zero otherwise. Then the integral of f(a, x) dx is always 1 (for a > 0) but for any fixed x, f(a, x) tends to zero as a tends to infinity.

The trick to do this rigorously is to observe that |e^(-ax) sin(x)/x| =< e^(-ax) =< e^(-x) for a > 1 and x > 0, since |sin(x)| =< |x|. Then since int_0^inf e^(-x) dx = 1 we have that the whole sequence is dominated by a integrable function, and so now the argument goes through by the Dominated Convergence Theorem. This is indeed a large part of the reason why negative exponentials work so well for this trick.

Some of the details here are of course a bit technical, but it might be good to mention in spots like this where technical details are being skipped over, to give the viewer somewhere to look if they are interested in the details.

EDIT: Just realised that the Dominated Convergence Theorem is overkill, there is actually a very nice simple argument along the same lines:

| int_0^inf e^(-ax) sin(x)/x dx | =< int_0^inf |e^(-ax) sin(x)/x| dx =< int_0^inf e^(-ax) dx = 1/a

so as a tends to infinity, int_0^inf e^(-ax) sin(x)/x dx must tend to zero.

[dr.osborne]

OMG this is so much easier using contour integration in the plane! Cauchy principal value and you're done. I love integration by differentiation, but this example is a heck of a lot of work!

[adarshhoizal]

Why work so hard? Directly apply Lobachevsky's Dirichlet integral formula where f(x) =1 and get the answer pi/2 in just one step.

[GillAgainsIsland]

I came very close just by looking at the graph. The areas cancel from negative infinity until you get to negative pi, and they cancel from pi to to positive infinity. That leaves an area above the x-axis from -pi to pi. If you draw an isosceles triangle there it will very closely approximate the shape of that area. Then, applying the area of a triangle: 1/2(base)(height) gives you 1/2(2pi)(1) = pi, which surprisingly is the exact answer. Curious.

[manfredwitzany]

This is a shorter path to solve this integral using comlex analysis:
integral(-inf, inf) dx sin(x)/x=integral(-inf, inf) dx Im(exp(ix))/x (using Euler's formula) As 1/x is real and integration is linear we can bring the Imaginary part in front of the integral.
= Im(integral(-inf, inf) dx exp(ix)/x Now let's assume +/- infinity as lim R->inf of R) and add some semicircle with radius R to the path of integration leading to a closed path of integration.
In the limit as R goes to inifinity, the interal along the semicircle vanishes and we have added zero to the original integral. Now using the residue theorem we can easily evaluate this integral as
pi*i*exp(ix)|x=0 (The factor p*i equals the integral along an inifitely small semicircle along the pole at x=0) Plugging x=0 into this expression leads to:
pi*i, from which we have to take the imaginary part being pi.

[maraszynko]

Very impressive, however 10:18 - "0 Times something is zero", fortunately in that case it is indeed, since both sin and cos oscillates. However, it's good to mention that we have to make sure this something does not approach to infinity:)

[DawidEstishort]

Very nice, especially the part with reasoning behind using e^-ax and not another function.

[runneruwu]

I've never heard anyone say integral like that my entire life and I've had math professors from many different countries

[velvetundergrad]

There’s a better way

Improper integrals from 0 to infty where the integrand takes the form f(x)/x simplify to L{f(x)} where L is the Laplacian.

Thus the integrand becomes 1/(1+x^2) which evaluates to atan(infty)-atan(0) which is π/2

[cshsc]

1:28 You cannot just shift differentiation operator inside the integral. You need to check some convergence theorems in order for that to work.

[sid]

Beautiful, no words can define the beauty.

[KingGisInDaHouse]

Great video about differenciating inteegrals.

[goldeer]

when it comes to integrating from 0 to +inf the function sin(x)*e^-ax instead of using integration by parts (or DI method with a table as shown by blackpenredpen which is less prone to error i think) I believe it's easier to compute it as the imaginary part of the integral from 0 to +inf of e^(-a+i)x which happens to be especially nice here becomes you don't have any terms that remain in the exponent (because from 0 to +inf) and it saves quite a lot of time here.

[saumyaadhikari]

I've been looking for this since a long time. Thank you so much ❤