Okay now this was definitely one of the hardest problems I've ever solved and one of the most unorthodox solution developments I've ever come up with. But WOW this was awesome!!!
For anyone wondering the integral inside the log with base 10 the value is => (1/{3•[ (ln10)^1/3 ] } )•Γ( 1/3 , (2022^3)ln10 ) Where Γ(s,a) is the incomplete upper gamma function To find the solution i transformed the 10^(-x^3) into e^(-ln10•x^3) and then did u=(x^3)•ln10 and the gamma function appears The only problem is the bounds thats why we have to use the incomplete gamma function.
Since the integrand decays super fast I tried estimating the integral much more sloppily, and in the end I got -2022³-6, so I'm a little disappointed that I was 2 steps off, but still, not too shabby. btw my argument was (and I replaced 10 with b, 2022 with n, until the final steps) b^-(n+1)³ ~ b^-n³ × b^-3n² The second factor is so small that we really could settle with estimating the integral from n to n+1 and the rest of the integral is many orders of magnitude smaller, so ignore it. Then set x = n + t, with 0
This was part of the 2021 marathon I did a few days back. While solving it I had to make a cut scene where I exclaimed "what the f**k am I doing with my life"😂😂😂 Yeah this was horrible...and there's not even much integration in it anyway 🤣
It doesn't. Instead, we can argue that as f(x) is positive, int_{2022}^{infty}f(x) dx > int_{2022}^{2022+1/(3*2022^2*log(10))}f(x) dx, and now this is greater than the integral of l(x) over that same interval since f(x) > l(x).
Hi We can use polar coordinates . first we use double integral of 10^-(x^3+y^3) which is the answer of this integral^2 .for the upper bound we can use double integral of square root of x^2+y^2 times 10^-(x^3+y^3) and this integral can be solved because we have the 10 ^-r^3 times r^2 . and for the lower bound we can write double integral of 1/((x^2+y^2)times x) times 10^-(x^3+y^3) and this can be solved too but a little bit different than first one we have different variable transformation in this one than the first one sounds good right?
