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 😂😂😂
It's a lot older than Feynman though. The general rule for differentiating an integral with respect to a parameter, where both the integrand and the endpoints vary, is referred to as the Leibniz integral rule. It includes both the fundamental theorem of calculus and differentiating under the integral sign as special cases. Feynman just made the latter method popular by discussing it in one of his autobiographical pieces. The most general form of the rule is used e.g. in the theory of weak solutions of quasilinear PDEs, for deriving the jump conditions (Rankine-Hugoniot) from conservation laws integrated across a shock discontinuity.
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.
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."
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.
Thanks I'm actually more inclined towards physics and hence the tendency to skip the rigor in favor of "yeah that seems about right " or "that seems trivial"😂😂 But I really do love teaching math especially on RU-vid. I'm gonna keep uploading too so I'm looking forward to your feedback as well as anyone else.
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!
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.
Very clever intuition, Sin(x)/x approaches to 1 as x approaches to 0, so the triangle kinda works. As for areas beyond +/- pi, I don't have good intuition to confirm that the areas would cancel out. They indeed are a bounded sequence of decreasing quantities with alternating positive and negative signs, but I wouldn't be so sure intuitively as to whether they actually cancel out or leave some residue. Any intuition around that that I am missing perhaps? Great observation, and a great way to visualize what's going on instead of blindly following a technique.
@@sitosopti nope, what you are saying is correct. They don't actually cancel out, to convince yourself, just try a numerical integral from -pi to a very large negative value and keep making it bigger to guess the limit behavior.. it is not vanishing/ zero. What happens is the residue you refer to exactly matches the deviation of the [-pi pi] interval integral from the isosceles triangle area OP mentioned
The signed areas outside the interval I = [-pi, pi] don't cancel. If you draw your beforementioned isoceles triangle, you will see that its' area is smaller than the area between the graph and the x-axis from -pi to pi. Hence the total integral outside of this interval on the entire real axis will have to be negative in order to result in a total value of pi for the generalized integral from -inf to inf. Even if you don't know the result pi beforehand, you can also see that the areas don't cancel from noting that the function f(x) = sin(x)/x is asymptotically approaching zero when |x| --> inf. So the crests and troughs of the graph become damped in amplitude, while the period remains unaltered. The biggest of these crests and troughs are throughs on the adjacent sides of the central peak and the interval I (with negative values). Hence they will give the biggest contribution (negative value) to the original integral aside from the positive contribution from the central peak that is slightly larger than pi in area. (Integral f(x) from -2pi to 2pi would have a value less than pi, but integrating "outwards" from the origo the area will approach pi as the limit, from altenating sides for every extra period of the function)
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.
Loved the explanation and yes I agree that there are areas where the explanation can be more rigorous. However being primarily a physics teacher I tend to skip over a bit of rigor in favour of "yeah that seems about right"😂 In my head it went like sinx is a bounded function anyway so it all comes down to a "race" between an exp function and a polynomial (x) and the exp function would get to zero much faster than the x in the denominator would....yeah not much rigor but seems to work😂
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.
Why do I get answer 2pi instead of pi following your way. Closed integral, couchy: int (f(z)/(z-0))=2pi*i*f(0)=2pi*i On the other hand, closed integral is a sum of: 0 (R->inf) + integral of e(ix)/x along the x axis. Infinitizemal smal semi circle path integral around 0 is zero, right? What am I missing? Small semi circle goes around and BELOW zero, right? Tnx for reply
@@meeehc The correct integration path is the following: Suppose R being a large radius and r being a small readius. Integrate from -R to -r, make a semi circle with radius r around pole z=0, integrate from r to R and make a big semi circle with radius R back to -R. As there is no pole within the path of integration, this path integral will be equal zero. In the limit R-> infinity and r -> 0 the straght paths conicide with the goal integral and the big circle vanishes. Consequently the goal integral + the integral along the small semi circle must be equal 0. It is important to distinguish between real and complex poles. If the pole is real as in the present problem, you need a semi circle to complete the closed path. On the other hand, if you have a complex pole, you have to make the following path: Assume the pole is x + iy: Integrate from -R to x, integrate from x to x+ i(y-r), integratge a full circle with radius r surrounding pole, integrate from x+i(y-r) to x, integrate from x to R... You will see that both integrals in the imaginary direction are eqal with opposite signs and therefore cancel each other. The remaining difference is that the path around the pole will be a full circle. I think your problem is that you do not start with a correct Cauchy path. Make a drawing of the desired path using finite values for r and R nearly comprising the goal integral and correctly surrounding all poles. This path consits of straight lines, semi cicles and full circles only. If you do so, you will see that you will need a semicircle instead of a full circle. In a second step you can apply the limits to it.
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:)
there is still a problem in this moment: he says that with x-->inf exp(-ax) goes to 0, but later he uses result he gats from this statement while integrating with a-->0, so basically he multiplies 0 by inf at some point
The error is at 12:00 when he use the continuity of I(a) at the point a=0. That needs at least 10 to 12 lines of justifications. And I have the proof right close to me.
That would have been cool to point out for sure! I know there is some people not used to these arguments that would really need some clarification on the rigorousity in integration. It's not widely spread on the internet, but it's so important in order to get to higher mathematics 😬
@Jasper Antonelli how about the fact that it gets you the correct answer? That’s enough for physics Edit: I feel like I overstated the point a bit too much here, but nevertheless I think it's safe and pedagogically correct to sometimes separate the physics from the more rigorous parts of the mathematics. It's the same reason we don't have to derive everything from scratch every time we learn something new.
@Jasper Antonelli you do need to check that but not by checking the convergences and all that as mathematicians would do. That’s how physicists can do many integrals that mathematicians can’t because those are ill defined mathematically. Physicists check the the results make sense by checking it against a set of rules but they are all physics motivated, causality for example is typically used. Ideally, they should spend some time to make their formulae more rigorously defined mathematically but that is not happening. Nor is it clear if it’s entirely possible. That’s the nasty reality about QFT these days.
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
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.
when u have taken the factor exp(-ax) then it became as Laplace integral transform. So, there may be another type of integral transform if u take log(+ or - ax) , sin(ax), ax , etc. but on the condition that the integral should exist when we do so ?
When reduced to e^(-ax)*sinx you can convert sinx to Im(e^ix), integrate with respect to dx and be left with (i-a) as a denominator,multiply by its conjugate (i+a) both top and bottom, separate the real and imaginary parts and bingo!!
To avoid integration by parts, write the integral as the imaginary part of exp(-(a-i)x) integrated from 0 to infinity. So that gives you Im (a-i)^{-1} = 1/(1+a^2) much more quickly. The other way to do this integral elegantly is by contour integration, but Feynman's claim was that he could always avoid contour integrals by finding an alternative trick.
the fact that he had the urge to do this is real ( the mathness screaming inside his head) but it's too early for the general yt audience to get a second existential crisis in the same video which they have thought they know to solve. 😊
an interesting formula I have come across recently that could work is called the Dirichlet Lobachevsky formula, where then the integral from 0 to inf of sinx/x dx becomes the integral from 0 to pi/2 dx which yields pi/2, multiply that by two since we are integrating an even function from -inf to inf, you get pi
The integral over e^(-ax) sin x is easier to solve: instead of twice IBP, use 2 sin x = e^(ix) - e^(-ix) and you get a simple exponential to integrate.
Very cool technique, thanks! Two things I noticed could be more rigorous: sinx/x is not defined at 0, but we can replace it with its limit at 0 to make a new function to integrate, and to evaluate I(inf) at the end I believe you also have to check when x goes to inf as well, whether that changes anything (it doesn't).
@@tupoiu not that, u have to verify that after differentiating under the integral sign with respect to the dummy variable, that the domain for that variable is valid when u sub back in at the very end, and if not u have to use limits to justify it. For example, if u set a change of variables involving a partial fraction 1/a-1 as the intermediary step, and then substitute a = 1 at the final stage to determine your integral, this is a prod pen because of the hole at a = 1. We instead have to deduce that I(a) is continuous and the limit as x->1 I(x) exists. Same story with functions going off to +_ infinity under the new variable
I think the IBP part of the calculation can be simplified by using sin(x) = Im exp(ix), which reduces the integral to that of an exponantial function with complex argument. But the given proof also works if the students are not familiar with complex numbers yet.
could you please explain why at the end, I(infinity)-I(0)=-pi/2, we didn't accept the answer -pi/2, but had to find what is I(infinity) and get to the answer pi/2
I might like to make math videos--or tutor math online--but I find myself stymied by an inability to draw mathematical symbols on a keyboard. How are you allowing your own drawing to be turned into a video?
Richard Feynman was a genious in Physics. As for Maths, his tricks allow to solve interesting problems but not this one. The integral of the sinc function was deciphered 2 centuries before him: Laplace, Fourier, Residues, etc.
@@MSloCvideos "When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration." From the wiki article you linked. No one likes a pedantic ass.
Residues ? There's no circulation path on which to integrate here ....Just minus infinite , plus infinite 1D .... Sure you could integrate on half an infinite circle and find a way to find the value of the curved part to retract it ...But still ,very cumbersome way of solving that ( i've done a few of those a while back ) , a change of variable , or some kind of substitution trick is always welcome ...
You do not need to have infinity in integration limit of the integral with respect of a. It can arbitrary, so just as well make it the most convenient.
@@maths_505 Yes indeed. Point taken. Feynman saves the day for many such tricky ones. It is just that many videos at this part move to IBP (incl BPRP one) and i always wondered if i was missing a nuance at that point. Thank you!
Yeah bro I just watched it... you used complex analysis which I found really cool. Keep up the good work! Just do one thing....don't use the typed stuff in your videos Just write everything yourself cuz that makes it more engaging. Handwriting doesn't matter cuz it's perfectly legible.
If we take 1/ x as a log x and another sinx fun as it is and then if we solve this by integration by parts the ans will be right ...but why are u doing this extra ?
Why is this called Feynman's technique? It's a classic technique for integrals like this, and with some grain of salt in exchanging limits it can be tought in high school. However, it's very beautiful when you encounter this the first time. Feynman's technique usually refers to calculate path integrals by regularized determinants and zeta functions.
Well this method was actually popularised by Feynman. However its actually the Leibniz integral rule. Its normally called Feynman's trick rather than technique.
@@maths_505 Referring this integral to Feynman would be appropriation (and Feynman would like not claim this). However, Wikipedia seems to wrongly refer to this as Feynman's Trick, too). This integral is historically called Dirichlet-Integral, who solved this by Laplace transformation which is very close to your presentation ("Sur la convergence des séries trigonométriques qui servent à représenter une fonction arbitraire entre des limites données", 1829), without dirctly spelling out the ODE for J. However, that might we a wrong referrel already, as first consideration already go back to Euler. Hardy ("Mathematical Gazette 5, p98-103)", 1909) calls it a classical proof.
Well You can integrate indefinitely but the definite integral treatment seems more elegant and I didn't have to worry about figuring out the value of C
@@jshadow1988 well I(0) is what I wanted to find and I knew the behaviour of I(a) in the limit a approaches infinity...so it made the calculation convenient That's what the Feynman technique is all about...its more about getting creative and making things work to your liking rather than just some systematic algorithm.
@@maths_505 yes, that's right now! But in the video seems like you integrate alpha from zero to infinity just because the first integral in x is from zero to infinity. Maybe my english is not really good, but thats what I've understood.
@@jshadow1988 yeah that's sort of an accidental motivation stemming from my choice of the alpha function The original integral prompted my choice of the function and so it provided me with information that came in handy doing the last integration w.r.t a
The sine limits the numerator to +-1, and the 1/x limits to zero. The function is odd because sin(x)/x = sin(-x)/(-x). Therefore the answer is a non zero positive value.
That's actually quite a nice question... To say that something is approaching infinity means that its growing arbitrarily large. So if something approaches negative infinity it means that its growing arbitrarily small.
@@maths_505 Thank you for your answer. Sorry about my math level, I may be wrong. Is it means there is no line in reality? Line should be part of circle, elipse or connected curve in space with particals. This is like a why any object can not move in same direction in space if there are more than one object in space. They will effect on other objects, even if they are infinitely far from each other. Because of they can not stay in same point and it means if an object is f(x)=x^(x ∞). It should keep take a part from negative infinity to exist. When it takes it it means there is a (1/infinity) point of 1/-∞ plus amount of its on loan - taken in its centre. x^x/ ∞. So it can act like a small black hole with around the hole it will want to attract different particles that want to get to the other side but get stuck. or The particles that make up the matter in the positive field will form an internal spiral curve at the tangent point to the boundary of the negative field. (Like the shape formed by the friction of gases of different densities on Jupiter). In this case, if we take x=1 singular basis. 1 to the +infinity, 1 to 0 - infinity. In this case, 1 unit in a + infinity field; when increasing the square multiplier. - at infinity it will increase by an equal amount like a 1 unit ruler. Therefore, there will be a flow from the Positive field to the other field. I do not know enough math to explain. It is like just a question.
@@maths_505 I think it's safer, in avoiding misinterpretation, to think of "approaching" negative infinity as getting big negatively without bound--from -1 to -1000 to -1,000,000 to negative googol to negative googolplex, and so on. "Arbitrarily small" is right in the sense that negative 1000 is less than negative 2, but "arbitrarily small" sounds like getting closer and closer to zero, which of course isn't what is intended.
