Went with the classic rectangular contour, you can take the integral you have over the whole real line and just take half of the result. Then it’s just residue theorem at the poles. I quite like that the result is (pi/2)/cosh(pi/2), very neat!
Forgot to add, perform contour integration on e^(i*x)/cosh(x), then just take the real part at the end. Choose the appropriate corners for the rectangles such that the denominator doesn’t change, and you are good to go!!
I've done this another way and get a result SUM n=0 to infinity (-1)^n(4n+2)/(2n+1)^2+1 Does that means it equal to Pi/2 Sech (Pi/2) Can someone shows me about that ?
3:17 From here on out, you can use two things, 1. the Mellin transform of 1/1+x^2 or 2. Gamelin's generalized integral exercise that states that... let me find it... ah! so, both methods (since one is the proof of the other, Gamelin you've done it again) lead to the integral from zero to one of x^(a-1)/1+x^b = pi/b*csc(pia/b) => pi/2*csc(pi(i-1)/2) which is pi/2*sec(i*pi/2) which well... we all know what hyperbolics are to trigonometrics.
All good. But I think we should be careful with things like x^i as it has multiple values: x^i = (x*e^(2*k*pi*i))^i = x^i * e^(-2k*pi) or in other terms: x = exp(ln(x) + 2k*pi*i) => x^i = exp(i*ln(x) - 2k*pi) It doesn't mean that we can't use x^i at all, but we should be careful, knowing which of its infinite number of values we use. As of now this video shows that that integral _could_ be equal to that result.
@@moeberry8226 alhamdullilah everything's going great. Just the usual work routine for the channel plus working on complex analysis lectures coming out in September. The rest of my time is spent with football and the occasional movie. How about you? How's work and how's the gym routine paying of?
Very smart approach
Thanks
Went with the classic rectangular contour, you can take the integral you have over the whole real line and just take half of the result. Then it’s just residue theorem at the poles. I quite like that the result is (pi/2)/cosh(pi/2), very neat!
Forgot to add, perform contour integration on e^(i*x)/cosh(x), then just take the real part at the end. Choose the appropriate corners for the rectangles such that the denominator doesn’t change, and you are good to go!!
@@SuperSilver316 bruh I made a video doing exactly that 6 months ago😂😂😂
Oh wow I didn’t realize, I just really like contour integrals I guess!!
how do you learn how to use contours to integrate? any books you recommend?
I've done this another way and get a result
SUM n=0 to infinity (-1)^n(4n+2)/(2n+1)^2+1
Does that means it equal to Pi/2 Sech (Pi/2)
Can someone shows me about that ?
3:17 From here on out, you can use two things, 1. the Mellin transform of 1/1+x^2 or 2. Gamelin's generalized integral exercise that states that... let me find it... ah! so, both methods (since one is the proof of the other, Gamelin you've done it again) lead to the integral from zero to one of x^(a-1)/1+x^b = pi/b*csc(pia/b) => pi/2*csc(pi(i-1)/2) which is pi/2*sec(i*pi/2) which well... we all know what hyperbolics are to trigonometrics.
All good. But I think we should be careful with things like x^i as it has multiple values:
x^i = (x*e^(2*k*pi*i))^i = x^i * e^(-2k*pi)
or in other terms:
x = exp(ln(x) + 2k*pi*i) => x^i = exp(i*ln(x) - 2k*pi)
It doesn't mean that we can't use x^i at all, but we should be careful, knowing which of its infinite number of values we use.
As of now this video shows that that integral _could_ be equal to that result.
Fun to watch these. Thanks
awesome result , keep it up , I am enjoying ❤❤❤
Beautiful
Very nice result
Awesome Solution. Thanks.
Very cool result Indeed
So many vids lately I can keep up!!! :D
AND I AIN'T STOPPING!!!
2nd upload dropping in 1 hour
No geometric series this time :/ JK, beautifully done mate :)
Thanks bro
You know I can never resist a gamma function
@@maths_505 Lol!
nice video
Thanks mate
Phi the constant??
Plz tell me that was a joke
So the integral of sin(x)/sinh(x) has pi/2 with pi/2 on the inside except one is sech(x) and the other is tanh(x).
Yeah I found that pretty cool
@@maths_505 how you been bro what’s new
@@moeberry8226 alhamdullilah everything's going great. Just the usual work routine for the channel plus working on complex analysis lectures coming out in September. The rest of my time is spent with football and the occasional movie. How about you?
How's work and how's the gym routine paying of?
@@maths_505 Al Hamdiallah bro. The gym is amazing and also watching football keep up the videos bro you always got the best.
@@moeberry8226 its written Al Hamdullilah bro😂 points for the effort though.
Thanks mate....get those gains to complement the math brains😎
video has 100pi likes at the moment
And I thought I had bad handwriting.