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.
Best Answer
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.