what a nice integral!

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[@goodplacetostop2973]

0:04 Bless you 0:07 Actual start

[@mtaur4113]

4 seconds of hard work, not gonna throw that away, we have come too far.

[@Cloud88Skywalker]

Help, please. I couldn't find a good place to stop and now I'm lost.

[@jay_13875]

If we introduce I(b) = int_{-infty}^0 e^y/y * sin(b*y) dy, we get dI(b)/db = int_{-infty}^0 e^y * cos(b*y) dy, which is known to be 1/(1+b^2). (basically the Laplace transform of cos(b*t) evaluated at s=1 after substituting t=-y) Therefore I(b) = arctan(b) + C. And since I(0) = 0, we have C = 0, giving us I(1) = arctan(1) = pi/4.

[@rotoboravtov4354]

4:15 After getting the dz integral we can use exponential formula for sin and immideately get the last integral using the fact \int_0^inf e^(az)=1/a

[@maurobraunstein9497]

Clever! I did this integral less cleverly but I still got the answer, which doesn't usually happen with these integral videos. I also did the substitution, which gives the integral of sin(y)e^(y)/y dy from -∞ to 0, but then I got a bit stuck (because I never think of differentiating under the integral sign) so I just... turned sin(y)/y into an infinite series, the sum from n = 0 to ∞ of (-1)^n · y^(2n)/(2n + 1)!. Then it's just integration by parts. If f(n) = ∫y^(n)·e^y dy from -∞ to 0, you can easily see that f(0) = 1, and by applying integration by parts, you get that f(n) = -n·f(n - 1), so f(n) = (-1)^n · n!. This eventually gets you to the integral as 1 - 1/3 + 1/5 - 1/7 + ..., which is a series that converges to π/4 *famously* slowly. I feel like I got lucky that I remembered that series. Not sure I would have been able to get π/4 from that otherwise.

[@dalek1099]

I did it by writing sin(alnx)/lnx with feynman's trick the derivative becomes cos(alnx)=Re(e^(ialnx))=Re(x^(ia)). Then integrate, to get Re(1/(1+ia))=Re((1-ia)/(a^2+1)=1/(a^2+1). Integrate to give arctana+C. Setting a=0 gives the integral=0 because sin0=0. Therefore, we get arctana and setting a=1 we get arctan1=pi/4.

[@SuperSilver316]

Feynman’s trick or something

[@TheEternalVortex42]

Feynman's trick is pretty much the same thing as what Michael did.

[@AndyBaiduc-iloveu]

Feymanns trick or contour integration seems good

[@karlyohe6379]

Okay. That was like a f*cking magic trick :D

[@barryzeeberg3672]

Need a good place to start :)

[@demenion3521]

i think writing sin(y) in exponential form gives the same result a bit quicker.

[@BikeArea]

0:04 You should do more backflips again to relax your muscles. 😮 😏

[@tm8539]

I don't understand, it's a beautiful feeling, I wanna understand so much!

[@mr.soundguy968]

And that's a good place to stop

[@holyshit922]

t = -ln(x) L(f(t)/t) where f(t) = sin(t) then plug in s = 1

[@sqigger]

Another way to do it: sin(log(x)) = (x^i - x^(-i))/2i I(p) = int from 0 to 1 (x^(p*i) - x^(-i))/log(x) dx And there you go.

[@solcarzemog5232]

Why this? Can you elaborate?

[@rotoboravtov4354]

@@solcarzemog5232 sin(x) = (e^(ix) - e^(-ix))/(2i). This is a bit loose but correct application of complex analysis.

[@sqigger]

@@rotoboravtov4354 yea, my bad 😅

[@gghelis]

B-b-but where's the good place to stop?

[@robblerouser5657]

What did you do in the third step when you introduced "z" into the integral?

[@mtaur4113]

Looks like sin(u)e^u/u du to me. That u denominator looks rough though. Maybe there is a definite-integrals-only trick too. 🤔

[@phatguardian]

Why didn’t he just restart filming 😂

[@stevenpurtee5062]

Why do you always skip over the details of if/how to change the order of integration. That's an interesting and important step. You have at least said "by Fubini's Theorem" before, but it think it would be important and interesting to go through the steps.

[@EliavShalev]

Lol you way overcomplicated it. Could have easily solved it using the properties of the laplace transform.

[@AndyBaiduc-iloveu]

Can you teach me how to do that? I only have learned a little about it when we use it to solve differential equations

[@EliavShalev]

@ sure! First you do the substitution that he did and convert the integral to that of -(sin(x)/x)e^(-x) from 0 to inf this is the laplace transform of sinx/x and from the properties of the laplace transform we know that L(f(t)/t) = L(f(t)) integrated from s to inf thus the result is the integral of 1/(x^2+1) from s to inf with s=1

[@AndyBaiduc-iloveu]

Ok , I kinda get it ! Thanks!