I like your clips and this one is very nice again. There is, however, a small mistake after 12:42 where you state the supposed reason for the integral along Gamma going to zero. While it is correct that the expression exp(-R sin(phi)) approaches zero as R become large, this and the fact that sin(phi)>0 for 0
Oh yes indeed That's exactly the reason why the integral evaluates to zero. What I meant by "dominating discourse" meant that that term actually relieved our concerns about convergence. It seemed quite trivial but now I do realise that I should be more clear in my wording rather than just brushing such things off as trivial. I'm really sorry for that and I'm actually quite grateful that you pointed this out. And thank you so much for the appreciation.
@@maths_505 You could have also just replaced this expression by the largest possible value happening at phi=0 or phi=Pi (with R s.t. the cos() becomes 1), which gives the upper bound e. The rest still goes to zero as R->infinity.
13:47 Im(e^e^ix)=Im(e^cosx e^isinx) =e^cosx×Im(e^isinx) =e^cosx×sin(sinx) This is an odd function Hence Imaginary part of int(e^e^ix)/(x²+1)dx from(-inf,inf)=0
On the contrary! I am absolutely new to complex analysis and contour integration but I am fascinated by it! I am still trying to make heads and tails of its basics though. For me, it seems to be the key to solve ANY integral. Without it, I have the feeling that we are skipping (being blind to) a whole world, which is as REAL. Without contour integration we will be seeing only the tip of the iceberg...
I like your clips and this one is very nice again.
There is, however, a small mistake after 12:42 where you state the supposed reason for the integral along Gamma going to zero. While it is correct that the expression exp(-R sin(phi)) approaches zero as R become large, this and the fact that sin(phi)>0 for 0
Oh yes indeed
That's exactly the reason why the integral evaluates to zero. What I meant by "dominating discourse" meant that that term actually relieved our concerns about convergence. It seemed quite trivial but now I do realise that I should be more clear in my wording rather than just brushing such things off as trivial. I'm really sorry for that and I'm actually quite grateful that you pointed this out. And thank you so much for the appreciation.
@@maths_505 You could have also just replaced this expression by the largest possible value happening at phi=0 or phi=Pi (with R s.t. the cos() becomes 1), which gives the upper bound e. The rest still goes to zero as R->infinity.
13:47
Im(e^e^ix)=Im(e^cosx e^isinx)
=e^cosx×Im(e^isinx)
=e^cosx×sin(sinx)
This is an odd function
Hence Imaginary part of int(e^e^ix)/(x²+1)dx from(-inf,inf)=0
So the sin(sin(x)) version is the imaginary part which just equals 0?
This is Problem 2116 from the Mathematics Magazine.The solution is in vol 95(2), 2022, p. 158.
do u have any pdfs of those magazines or any book that has countour integration problems?
Sssooooo satisfaying...DAMMN
HEY IT'S MY FAV FUNCTION DEVIDED BY MY FAV DENOMINATOR!!
Constructing integrals 101
@@maths_505 and then when you can't solve them... we never ran out of greek letters anyways...
@@manstuckinabox3679 there are few things as true as this statement 😂😂😂
Epic 🔥
app check :)?
❤️
Everything can be contour integrated if you are brave enough
Well said
Contour integration is an absolutely broken method. Thankfully Mathematics can’t be nerfed.
i'll be honest, i hate contour integration 😢
On the contrary! I am absolutely new to complex analysis and contour integration but I am fascinated by it! I am still trying to make heads and tails of its basics though. For me, it seems to be the key to solve ANY integral. Without it, I have the feeling that we are skipping (being blind to) a whole world, which is as REAL. Without contour integration we will be seeing only the tip of the iceberg...