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This question comes from the MIT integration Bee 2022 Final Round.

As $10^{-x^3} = e^{-x^3\ln10}$, and by substitution $u=x^3\ln10$, the integral becomes $$\int^\infty_{2022^3\ln10}\frac{1}{3(\ln10)^{1/3}}u^{-2/3}e^{-u}du$$ However, I don't know how to tackle this incomplete gamma function and get the final answer. Thank you for your help.

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    $\begingroup$ You definitely are not meant to compute the integral. If $I$ is the integral, you need to show: $$10^{-2022^3-8}\leq I <10^{-2022^3-7}.$$ You can easily show $I<10^{-2022^3}.$ But not sure how to proceed. $\endgroup$
    – Thomas Andrews
    Aug 5 at 16:06
  • $\begingroup$ The integral is dominated by the integral $\int_{2022}^{2023}.$ The other part is very small, even relative to that interval. In particular, $$\int_{2023}^{\infty} 10^{-x^3}\,dx<10^{-2023^3}\int_{0}^{\infty}10^{-3\cdot 2023 x}=\frac{10^{-2023^3}}{3\cdot 2023\cdot \ln 10}$$ While $\int_{2022}^{2023} 10^{-x^3}dx>10^{-2022.5^3}/2.$ And $2023^3-2022.5^3>6000000.$ So the remainder term is very tiny compared to the start. $\endgroup$
    – Thomas Andrews
    Aug 5 at 16:19

1 Answer 1

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Use that for $x \geq 2022$ the following bounds hold: $$10^{-x^3} \leq 10^{-2022^3 - 3\cdot 2022^2\cdot (x-2022)}$$ and $$10^{-x^3} \geq 10^{-2022^3}(1 - 3\cdot 2022^2\cdot \log(10) \cdot(x-2022)).$$ (The lower bound is the tangent line at $x=2022$.) Integrate the upper bound over $[2022, \infty)$ and the lower bound over the interval $[2022, 2022 + (3\cdot 2022^2 \cdot \log(10))^{-1}]$ (where it is non-negative) to get good enough bounds for the original integral.

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  • $\begingroup$ Thanks for the suggestion. However, when integrating the lower bound over 2022 to infinity, $$[x-\frac{(3\ln10)(2022)^2(x-2022)^2}{2}]^\infty_{2022}$$ How to treat the infinity? $\endgroup$
    – HeyFan
    Aug 6 at 3:47
  • $\begingroup$ Integrate only over the interval (starting at $2022$) where the tangent is positive to get a lower bound. $\endgroup$
    – WimC
    Aug 6 at 5:29
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    $\begingroup$ Finally, I got the inequality $$-2022^3-\log_{10}(6\ln10)(2022)^2<\log_{10}\int^\infty_{2022}10^{-x^3}dx\leq-2022^3-\log_{10}(3\ln10)(2022)^2$$ so the final answer for this question is $-2022^3 - 8$. Thanks WimC. $\endgroup$
    – HeyFan
    Aug 8 at 8:59

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