int 0 to infinity (cos(x))/(π^2 - 4x^2): the best thing you'll see next week

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[@yoav613]

Nice. If you are lazy like me you can also use: cosx/(pi^2-4x^2)=1/4pi (cosx/(pi/2-x)+cosx/(pi/2+x)) now by the subs pi/2-x =u for the first and pi/2+x=u for the second you get: integral 1/4pi (sinu/u) from -inf to inf which is 1/4😂

[@maths_505]

You literally destroyed this integral in seconds 😂😂😂

[@yoav613]

@@maths_505 what to do,i'm lazy🤣🤣

[@adamding3873]

My first reflection is also the real analysis method as you described. But I guess the author just wanted to demonstrate complex analysis skill. 😂

[@Calcprof]

2πi times the residue at (say) π/2 is 1/2 (but you are going the wrong way around), so you pick up half the value, which is what you would expect.

[@zunaidparker]

I'm loving the graph in the thumbnail! 😍

[@maths_505]

You often requested me to include graphs so I asked Myers to design the recent thumbnails.

[@zunaidparker]

@@maths_505 indeed, I like how it turned out 👍🏾 Next step is to introduce it in the first 30s of the video talking about the area we are calculating to give people a visual picture, then diving into the calculus 👍🏾

[@anasharere]

i like the new thumbnails style 🔥

[@KramRemin]

4:30: I would save a variable, by letting the radius be 1/epsilon

[@MrWael1970]

Very interesting and smart solution. Thank you

[@ashotdjrbashian9606]

Interesting fact, which may or may not have anything to do with this integration. An approximation to cosx known since the 7th century (by Indians) is (pi^2-4x^2)/(pi^2+x^2)

[@ericthegreat7805]

If you divide cosx/(pi^2 - 4x^2) in the integral, the approximation is 1/(pi^2 + x^2) which is an arctan integral, which probably sorts out to 1 in this limit.

[@Calcufast001]

Dear Mat505,
I enjoy every bit of your videos.
I am a first class graduate of mathematics and I could tell you are extremely good with integrals
Can you recommend great books to master this?
Regards.

[@maths_505]

Books won't help as much as the internet especially math stackexchange.

[@sachacloot1925]

hi i am a bit young but love integrals and stuff like that, i can recommande uyou a books call "(Almost) Impossible Integrals, Sums, and Serie" by Paul J Nahin aand publiched by Springer it's an amazing book with some stuff i had never seen and detailed explication :)

[@KramRemin]

You are literally integrating on a KILROY contour!

[@manstuckinabox3679]

I'd love to reference a certain book... But it has become a cliche at this point, I'm loving the new thumbnail designs btw!
Fine I'll reference it, so in gamlin's book there's an entire chapter on the "angular theorem of residues" [not the actual name but too lazy to open the book rn...] which is the same thing as our closed contour residue theorem, the difference being that it's theta*i*the residue at the point. where theta is the change in angle along the path...
I think it's way more detailed in the book...

[@nahbro1321]

I searched on Google gamlins book doesn't exist

[@bartekabuz855]

Leg days are very relaxing, agree

[@maths_505]

Ah hell nawh bro
It's not the relaxation
It's the satisfaction of having destroyed your legs by working each muscle group to failure and struggling up the stairs 🔥🔥🔥

[@agrajyadav2951]

​@@maths_505 Exactly 🔥🔥🔥

[@giuseppemalaguti435]

Feyman...I(a)=[....e^(-ax)...],con I(0)=I...risulta una eq.differenziale -4I"(a)+π^2I(a)=a/(1+a^2)...omogenea ok,ma la particolare???

[@amritlohia8240]

For the particular integral, you end up having to use the exponential integral (Ei) function (en.wikipedia.org/wiki/Exponential_integral), so it becomes intractable.

[@agrajyadav2951]

I like leg day too

[@danielrosado3213]

COMPLEX MAN

[@agrajyadav2951]

But chest is my favorite, sir

[@maths_505]

I had a killer chest workout on Monday. Gonna hit arms today🔥

[@sergten]

Jordan lemma... What are they gonna teach you in that grad school, as it seems like you already know everything and then some?

[@amritlohia8240]

Jordan's lemma is a standard result taught in undergraduate complex analysis courses - it's nowhere near graduate-level mathematics!

[@milanashadrina611]

О лемма Жордана