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If I Go To another city. I have to Know someone in Bradford that Has a Cousin on The Outskirt of that city. Then I Can Go to that Outskirt. If I Want to got to The Centre of that city: I Have to Know Someone that is a High class Ba5tard of That Cities King. The Outskirt has given a B1tch to the King of The city.

Beautiful! Got to love Feynman's😊

I like this one. It occurs to me that a nice generalisation is x^(a-1) (ln x)^b where b >0 is a non integer. I think that a fractional derivative (the Weyl version, I suspect) will let that work. I'll have a play with it.

Converting to a contour integral and solving in the complex plane can be very helpful with some stubborn integrals, but that is not as widely applicable as Leibniz rule/ Feynman's trick. I guess Feynman wasn't really joking after all. LOL

The square root has been put on purpose to take the problem solver by surprise. A memory fog can cause someone to forget that a square root is equivalent to 1/2 power.

Personally I think it is a shame that "Feynman's trick" is not a part of the standard curriculum, not just because it is useful by itself, but because it helps you to gain deeper insight into how integrals work.

Excellent 👌

Super powerful result

The change of variable must describe the interval of integration to be correct. A bijection is an ideal choice of course.

bruh just remember zeta(s)=1/gamma(s)(integral 0 to+ ♾️ u^(s-1)/(e^u-1)

It should be clear that the result is zero , because of the symmetry properties of cos x .

😂 this was a question in uk my school teachers and friends will laugh after seeing these type of question

I tried a different approach which seemed to work. First multiplied both the numerator and denominator by e^-x then chose 1-e^-x as u and then found that x^2 as (-ln(u+1))^2 which then I used IBP wherein I was going in circles thus I made use of the solution I and equated it with that like in the e^xsinx function integral. I don’t know if my approach is correct but that’s what I did. Also one thing to note is that I did it for the indefinite case since I don’t care for potentially transforming limits

u=½x du=½dx e^(½x)=e^u ½*integral of e^udu=½e^u+C=½e^(½x)+C

LETS GOOOO

f(x)=sin(e^x)-sin[e^(-x)] f(-x)=sin[e^(-x)]-sin(e^x) f(-x)=-{sin(e^x)-sin[e^(-x)]} f(-x)=-f(x) a=ln(pi) b=-a=-ln(pi) An integral of an odd function defined on a symmetrical interval is equal to 0.

Beautiful! Gamma function??? I was hoping you would use Laplace Transform 😊

My first thought was the gamma and zeta connection and the integral is ζ(3)Γ(3)

funny enough i asked chatgpt for an integral a few days ago due to boredom and it gave me this same problem but with an x^3 in the numerator instead. Equally shocking is that it got the correct solution. We also encounter ζ(4) which has a known value of π^4 / 90 which is pretty nice in that integral.

I don't think that zeta(3) is any more or less unsatisfying than ln(3), sin(3), sqrt(3) or e^3. In every case it's a number that has to be approximated by some value if we want to use it to calculate a real world answer.

We can also just integrate √tanx itself easily by just substituting tanx as u²

Beautiful! Loved the geometric series and the Laplace Transform🍻

I=int[0,♾️](x^2•e^-x/(1-e^-x))dx 1/(1-e^-x)=sum[k=0,♾️](e^-kx) I=int[0,♾️](x^2•sum[k=1,♾️](e^-kx))dx I=sum[k=1,♾️](int[0,♾️](x^2•e^-kx)dx) t=kx dt=kdx I=sum[k=1,♾️](k^-3•int[0,♾️](t^2•e^-t)dt) I=sum[k=1,♾️](2!/k^3) I=2•zeta(3)

I had fun with this. Started with e^-x/e^-x and then made a sub u = 1-e^-x to get ln^2(1-u)/u. I then went down the feynmann rabbit hole and it got messy fast. However, I'm going to keep playing. Currently at I'(a) = integral of -2 ln(1-au)/u(1-au) which I can partially decompose and get some Li(x) function.

Can u do a video on the inverse tangent from -infinity to infinity. I know the answer is zero do too the function is odd but if you work out the improper integral it gives a value of pi. It seems when working the integral you be adding limits which isn’t the area under the curve.

The ζ is just a Z with a little curl between the first two strokes, and a little hook at the end.

Could have done all the same tricks with x on the top instead of x squared and had an actual solution

according to the engineering rule integranatomatics pi=e

This is so satisfying

I broke it down further to ln(a+ix)(a-ix)-ln(b+ix)(b-ix) = ln(a+ix) + ln(a-ix) - ln(b+ix) - ln (b-ix). From there, integrate and realise that that using the same trick for 1/(1+x^2) is a partial fraction of 1/(1+ix)(1-ix) the above integration pops out with the same answer. There is some jazz hands for the ♾️ limit of the integral.

the straussian integral

I just used Taylor series.

Great job!

That sec^2/sec^2 trick is powerful - and one I rarely remember to try!

Did you know it was leibniz who first came up with the rule

Surely you're joking??

MATHEMATICA gives the answer Coth[π/2]/π , numerical value 0.347 ...,whereas your answer gives 0.904... There might be a little error in your caculation.

I used lobacheskiy's formula Integral of f(x)*sin(x)/x from 0 to infinity=integral of f(x) from 0 to pi/2 where f(x) is both continuous and has a period of pi

if you wanna make f(infinity)=0 more rigorous i think you can show that the sequence of functions e^(-nx) sin^3(x)/x converges uniformly to 0 on any closed interval [epsilon, t] for 0 < epsilon < t so the limit of the integral as s -> infty is 0 by the integral limit theorem

Great video and great evaluation. Please do some videos on the Lobachevsky Dirichlet Integral Formula.

We can use the triple angle formula for sin and Dirichlet's integral to arrive quickly at the answer. The exponential parametrization in this video was used to derive the Dirichlet integral in the first place, but repeating it provides good review into Laplace Transform for rusty viewers. Good stuff 👌

it’s probably faster to use power reduction here since you already had the previous value

It's interesting that differentiating under the integral sign is given only very brief coverage in Woods' Advanced Calculus, but Feynman used it to great effect and made it quite popular. Feynman also had a differentiation technique he called the "dispatch method", but that is not nearly so well known.

This question came in ny exam and i did cause i know the way

Pretty easy

Writing the integral as \int^{\infty}_{0} d(\pi x) \sin (\pi x)/(\pi x) - \int^{\infty}_{0} d(e x) \sin (e x)/(e x) = \int^{\infty}_{0} dx \sin (x)/(x) - \int^{\infty}_{0} dx \sin (x)/(x) by suitable rescaling, we see that the integral is zero.

This is a good one. Just for fun (?!) I decided to try a variant with x^2 in the denominator. (Because there are two sine functions upstairs and we've already done the ( sin x)/x problem.) Introducing the Feynman e^(-ax) again and differentiate twice with respect to a. When we do the integral, we end up with an expression for I"(a) = a/(a^2 + pi^2) - a/(a^2 + 9 pi^2), which we have to integrate twice. We get two constants, but I'(infinity) = I(infinity) = 0 so they are both zero. We finally get I(a) = 1/2 [ a ln(a^2+x^2) - a ln(a^2+9 pi^2) + 6 pi arctan{3 pi/a) - 2 pi arctan(pi/a) ] and then it's easy to let a --> infinity to get I(0) = pi^2 as the answer. (Verified in Mathematica) Feel free to use this beautiful result and thanks for prompting it.

of course only Florida is capable of making such as crazy looking integral. Everywhere else on earth is too sane to do such a thing.

I=1/2•int[3,♾️](e^-(3+4/(x^2-6x+9))/(x^2-6x+9))dx t=x-3 t^2=x^2-6x+9 dt=dx I=1/2e^3•int[0,♾️](t^-2•e^-(4/t^2))dt s=t^-2 ds=-2t^-3•dt I=1/4e^3•int[0,♾️](s^(-1/2)•e^-4s)ds b=4s db=4ds I=1/8e^3•int[0,♾️](b^(-1/2)•e^-b)db I=sqrt(pi)/8e^3

I really enjoyed that one.