A (literally) complex integral

Рет қаралды 9,598

Күн бұрын

Пікірлер: 41

@MyOldHandleWasWorse 8 ай бұрын

As soon as you got to the step in which you only had a trig in the exponent, and a composite of two trig functions, I realized (with the help of the title) that this is like one of my favorite qualifying exam problems, just slightly more complicated. With that in mind, you could have used Cauchy's Integral Theorem after one more change of variables after that same step, say letting z=e^{ix} and the problem becomes a contour integral. Of course, the two methods are equivalent by the same theorem - many such cases in complex analysis!

@SuperSilver316 8 ай бұрын

Yeah I was thinking about this substitution as well!

@jeromemalenfant6622 8 ай бұрын

Or you could use Euler's formula at the getgo and write the cosine as the real part of the exponential: cos (1/(1+x^2) = Re exp [ i /(1+x^2) ], and bring it into the exponential of x/(1+x^2): I = Re integral from (- inf) to (+ inf) exp [ (x+i)/ (1+x^2) dx/(1+x^2) Since 1+ x^2 = (x+i)(x-i), the exponential becomes exp [ 1/(x-i) ]. The other factor becomes 1/(x+i)(x-i), with poles at z= +- i. Closing the contour below the x-axis to enclose the pole at z=-I, and to avoid the exponential pole at z= +I, you get from Cauchy's Theorem I = Re (-2 pi i) /(-2i) exp [ 1/(-2i) ] = Re pi exp [ I/2 ] = pi cos (1/2)

@CM63_France 8 ай бұрын

Hi, "ok, cool" : 3:17 , 6:33 , 9:27 , 10:48 , "terribly sorry about that" : 3:42 , 5:25 , 9:13 .

@srikanthtupurani6316 7 ай бұрын

If we consider regions like semicricle with centre at origin and use cauchy residue formula. This entire thing will get simplified. Many of the integrals can be evaluated using cauchy residue formula. Feynmans techniques, fubini theorem can sometimes be tedious. When it comes to complex residue theorem we are hundred percent sure we will get the final answer. We have to choose the region properly in a clever way

@euler1 8 ай бұрын

@10:45 why does the exponential term simplify to 1?

@maths_505 8 ай бұрын

Use Euler's formula and you'll see why.

@ArtcticKnightX 8 ай бұрын

Bro you are literally Euler😂

@GeraldPreston1 8 ай бұрын

our boy Kamal bringing out another BANGER of a video!!!

@vladimir10 8 ай бұрын

Wow Really satisfying indeed!!

@MrWael1970 8 ай бұрын

Thank you for this pretty solution.

@gregwochlik9233 8 ай бұрын

Stopped the video at 0:16 entered into cheat mode: Off to Desmos to graph the thing to see what it looks like. Performing the absolutely obvious u-substitution. Restarted the video, and it is the wrong u-sub. I had u = x^2 + 1

@srikanthtupurani6316 7 ай бұрын

We can use complex analysis contour integrals.

@tioulioulatv9332 8 ай бұрын

الله يرحم والديك

@tomasstride9590 8 ай бұрын

I like to try these having looked only at the thumbnail. I also only like to try theorem if I can use contour integration. This is what I did and I did in fact get your solution. The integrand simplifies to Re exp(1/(z+i))/(z**2+1). Close the contour in the lower half plane and use the pole at -i. I think this is all valid as the integrand vanishes like 1/z**2 on the big semicircle. What do you think?

@maths_505 8 ай бұрын

Reasoning is on point and I like your solution development.

@Mathematician6124 8 ай бұрын

@@maths_505hey friend have you seen my solution?? 😊

@maths_505 8 ай бұрын

I believe I have

@maths_505 8 ай бұрын

It was excellent

@Mathematician6124 8 ай бұрын

@@maths_505 thank you so much friend. 😊 May the almighty bless you profusely.

@aymenstorm3087 8 ай бұрын

Can we use the SE-METHOD ?

@Mathematician6124 8 ай бұрын

Hey Friend 😊. I found something terrible. Go to 2:24 Int - pi/2 to pi/2 e^(sinxcosx) cos(cos^2 x) dx Int - pi/2 to pi/2, e^(1/2 sin(2x)) cos(1/2 + 1/2 cos2x) dx Let, I(a) = int - pi/2 to pi/2, e^(a/2 sin(2x)) cos(1/2 + a/2 cos2x) dx I'(a) = 1/2 int - pi/2 to pi/2, e^( a/2 sin(2x)) sin( 2x - (1+acos2x)/2 ) Convert sin(***) into polar form, I'(a) = 1/2 int -pi/2 to pi/2 e^(a/2 sin2x) img{ e^(2ix) * e^-i( (1+acos2x)/2 )} dx =1/2 img int - pi/2 to pi/2, e^(a/2 sin2x) e^(2ix) e^-i(( 1 +acos2x) /2) =1/2a img int - pi/2 to pi/2, e^-i/2 e^{-ia/2 ( e^2ix)} (-ia/2) (2i) e^2ix dx Take - ia/2 e^(2ix)= z we shall find it's just intgrl 1/2a img (e^z dz) = 1/2a img [ e^{ - ia/2 (e^2ix)}] put limits (pi/2, -pi/2) = 1/2a img [ e^{ - ia/2 e^ipi} - e^{ - ia/2 e^-ipi}] =0 Cuz, We know e^ipi=-1, and e^-ipi=-1 too So, I'(a) =1/2 img(0) =0 I(a) =C =I(0)=pi cos(1/2)=I(1) =I(💙)=I(kamaal) For every real a, I(a) has a fixed value of pi cos(1/2)

@satyam-isical 8 ай бұрын

Can you tell me a book which contains these types of integrals along with solutions?

@Anonymous-Indian..2003 8 ай бұрын

Who else used contour integration ?

@MZ-em1rw 8 ай бұрын

Amazing ❤

@BladimirRemon 8 ай бұрын

Wow 😮

@Jalina69 8 ай бұрын

I am lost at 5:40 for a moment. Where did isin go? Why only real? I need complex analysis(

@xizar0rg 8 ай бұрын

It's nothing to do with complex analysis. There are two functions, Re(arg) and Im(arg) that are essentially "take the Real part" and "take the imaginary part". Since both cos and sin are real valued functions, the only real part of (cos(x) + i * sin(x) ) is just the part attached to the cosine. Since the function in Integral_One (I hate the lack of serifs on the web) is a cosine function, we can sneak in the i * sin so that we can talk about the exponential function (which has a much simpler Taylor expansion) and say "we're just going to be needing the real part in the end". He backtracks it at the end when he extracts the real part to get the cosine out. If the function had been Sin(f(x)), he would have used Im() on exp(f(x)) instead.

@emanuellandeholm5657 8 ай бұрын

Oh look: It's the "Tangent double half-angle substitution". :D

@Chris_387 8 ай бұрын

Make the denominator another integral in terms of e^x take real part to make cosx in terms of e^x and proceed

@dogukanbirinci2099 8 ай бұрын

AMAZİNG

@giuseppemalaguti435 8 ай бұрын

Ho fatto come te,ma poi mi sono bloccato su I1,perché avevo indici di integrazioni diversi..ho sistemato gli indici di integrazione e risulta πcos(1/2)

@xizar0rg 8 ай бұрын

While I am not so froggy as to seek out integrals to handle on my own, I do appreciate the ability to see a lot of the techniques used just by eyeballing the function in the thumbnail. Some of your substitutions are still wild, but I've experienced far fewer "how the hell did he come up with that" moments. I guess having watched your channel from the start has done something. (well, minus the physics stuff... with apologies)