UK Integration Bee 2024 #9

[@doronezri1043]

Perfecto👏👏👏 Loved the didactic step-by-step approach🍻

[@owl3math]

and thank you for the idea! :) 👍👍👍

[@DihinAmarasigha-up5hf]

Interestingly enough...you could also use feynman's technique by letting I (t) = the integral given but with e^(-xt) replacing e^(-x)( where t>0).....then by differentiating twice and integrating twice and letting t =1 we get the answer 4ln4 -3 = ln 256 -3

[@owl3math]

Nice! That was the way I did the other one I mentioned at the beginning of the video. I think it was this one: kzbin.info/www/bejne/amq1aqihrZV1ZpIsi=7mrWFybP69ScedNn

[@DihinAmarasigha-up5hf]

@owl3math yup exactly....

[@DihinAmarasigha-up5hf]

@@owl3math for your next integral video can you solve the integral of (ln (cos x))^2 from zero to pi/2...try it using complex analysis the contour and the final result is really nice...

[@owl3math]

@@DihinAmarasigha-up5hf - sorry I generally can't take requests but I will keep it in mind for some day in the future.

[@slavinojunepri7648]

Fantastic

[@owl3math]

thanks Slavino!

[@Vengeance-yb8lm]

do you know if the integral of 0 to infinity of e^-(x+1/x)^2 is possible, or have you done a vid on it ?

[@owl3math]

Good one! Not too hard. It's a nice variation on the Gaussian integral. I think i want to do a video.

[@MathCastt]

Where do you find these integrals? I sat the UK int bee 2024 this wasn't on it or is it a different competition?

[@owl3math]

Hi. I got these for UK 2024 here: www.scribd.com/document/703490462/UK-Integration-Bee-Practice-DUTIS-and-Double-Integrals

[@toadjiang7626]

Or, you can just use the Gamma Function. I=lim(s to 0) int_(0 to inf) e^(-x)*x^(s-2)*(3x-1+e^(-3x))dx, int_(0 to inf) e^(-x)*x^(s-2)*3x dx= 3*Gamma(s), int_(0 to inf) e^(-x)*x^(s-2)dx=Gamma(s)/(s-1), int_(0 to inf) e^(-x)*x^(s-2)*e^(-3x) dx= 4^(1-s)*Gamma(s)/(s-1), so I=lim(s to 0) (3+1/(1-s)-4^(1-s)/(1-s))*Gamma(s), when s approachs to 0, Gamma(s)=1/s, 1/(1-s)=1+s, 4^(1-s)/(1-s)=4+4*(1-ln4)s, so I=4ln4-3. You can also use the same method to solve the other integral you mentioned.

[@owl3math]

Hi. I follow you up to this point: I=lim(s to 0) (3+1/(1-s)-4^(1-s)/(1-s))*Gamma(s) but then this is an indeterminate form. How did you evaluate this to 4ln4-3?

[@toadjiang7626]

@owl3math That indeterminate form is easy to evaluate using series expansion. First, you need to know the series expansion of Gamma(s) at s=0, which is Gamma(s)=1/s-r+O(s), where r is Euler-Mascheroni constant, but here we only need the first term, i.e Gamma(s)=1/s, because all the other terms will result to 0 anyway. And Gamma(s)=1/s when s approachs to 0 is easy to prove using s*Gamma(s)=Gamma(s+1), since s approachs to 0, s+1 approachs to 1, Gamma(s+1) also approachs to 1, so Gamma(s)=1/s. Now all you need is the Taylor expansion of 3+1/(1-s)-4^(1-s)/(1-s), and since all the s^2 terms and higher terms will result to 0, you'll only need the constant term and s term, which is very easy to calculate. which means 1/(1-s)=1+s, 4^(-s)=1-s*ln4, 4^(1-s)/(1-s)=4*(1-s*ln4)*(1+s)=4+4*(1-ln4)s, 3+1/(1-s)-4^(1-s)/(1-s)=(4ln4-3)*s, so the indeterminate form is equal to 4ln4-3.

[@owl3math]

@@toadjiang7626 nice! Thanks for the explanation :) 👍