A limit of a trigonometric integral.

Building on our last integral video, we prove a nice result involving the limit of a certain trigonometric integral.
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[@jegasi8082]

This is very nice, thanks for sharing!

[@adityabharadwaj7665]

hey can you please bring questions that also involve a bit of physics cause they rack your brain a lot. btw love your questions and your solutions

[@goodplacetostop2973]

13:28 Daily stuff... Find an equation for the parabola x = y² + 1 rotated counter-clockwise by an angle of π/4 around the origin

[@elengul]

Did you mean parabola, or did you mean to square the x term?

[@leastsignificantbit5069]

Thank you for another fun exercise My answer: Using formula for point (x,y) rotated around origin by an angle of p: (x',y')=(x*cos(p)-y*sin(p),y*sin(p)+x*cos(p)) and parametrization of the parabola: x(t)=t^2+1 y(t)=t (t ranges from -infty to +infty) We get: x'(t)=(t^2+1)sqrt(1/2)-t*sqrt(1/2) y'(t)=(t^2+1)sqrt(1/2)+t*sqrt(1/2) Solving for y in terms of x' and y' (sadly, I can't find any way to solve for y' in terms of x' without loss of generality), we get: y'=(y'-x')/2+((y'-x')^2/2+1)*sqrt(1/2) Derivation of formula used before: Consider point (a,b). We can express the same point as (r*cos(q),r*sin(q)), where r is distance from origin to this point and q is the angle beetween line connecting origin and (a,b) and OX axis. Then (a,b) rotated by some angle w around axis is given by (r*cos(q+w),r*sin(q+w)), which is the same as: (rcos(q)cos(w)-rsin(q)sin(w),rsin(q)cos(w)+rcos(q)sin(w)) Substituting a=rcos(q), b=rsin(q): (a*cos(w)-b*sin(w),b*cos(w)+a*sin(w)) Which is the formula used before.

[@blazedinfernape886]

You mean hyperbola x^2=y^2+1 The ans is xy=1 btw

[@goodplacetostop2973]

@@elengul Parabola, the equation is fine

[@elengul]

@@goodplacetostop2973 Cool, thanks for the clarification! Replacing x with (x - y)/sqrt(2) and y with (x + y)/sqrt(2), gets: x^2 + 2xy + y^2 - x*sqrt(2) + y*sqrt(2) + 2 = 0.

[@riitk69]

Love from india grt job sir

[@arvindsrinivasan424]

If you rewrite the sin as complex exponential, this problem is relatively easy (obv not trivial) using the residue theorem

[@riadsouissi]

What contour did you use ? I tried a semi circle but the circle part of the contour integral diverges.

[@arvindsrinivasan424]

@@riadsouissi once you re write the sin as complex exponentials, you use the geometric sum definition to express as a sum of simple complex exponentials (see Dirichlet Kernel). Then if you replace the order of the sum and integration you get something that when you use the semicircle contour vanishes as the radius goes to infinity. You get the finite sum that Michael ends with. Then you can take the limit easily as was done in the video.

[@rafael7696]

Beautiful result : pi and e in a limit

[@NoamK24]

Isn't that the Dirichlet Kernel? I took the course on Furrier a few years ago

Indeed, the best place to stop is at a pretty nice result.

[@hopegarden7636]

Nice result but I'm gonna go and overkill this with invoking gamma functions relation to sine function...will be fun!

[@stewartcopeland4950]

How can this integral converge if the sine takes an indeterminate form when n tends to infinity (oscillates between -1 and 1)?

[@MrRyanroberson1]

the x^2+1 in the denominator makes it trend to 0 on either infinity

[@leif1075]

Wait question if you don't know the first tool is it not possible to solve this?? Seems like it.

[@lost3834]

👍🏻👍🏻👍🏻👍🏻👍🏻

[@sakinarajpoot7577]

I hv one question why we calculate argumnt of z in tan inverse neither in cot or sin or cosine inverse in complex plzz answr mee

[@coderdemo9169]

Iover of integration

[@riitk69]

Am in class 11 preparation for iit and watching this 😁😁

[@adityabharadwaj7665]

been there done it jee 2020

[@riitk69]

@@adityabharadwaj7665 hope so but who are u

[@chhabisarkar9057]

@@riitk69 why always relate maths and calculus to iit ? Iit isnt everything in your life , just enjoy maths and thats it .