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Could you please tell me how to calculate the limit: $\lim_{x\rightarrow+\infty}\left(\int_0^1\sup_{s>x}\frac{s}{e^{(s\log s)t}}dt\right)$ Thank you so much!

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    how to deal with the case for $t$ near zero?2011-10-10

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I would start by calculating, for fixed $x$ and $t$, the value of the supremum in the integrand. (Note: since the integrand is $e^{-(st-1)\log s}$, this is equivalent to finding where $(st-1)\log s$ is smallest on the interval $(x,\infty)$.)

The answer probably depends on the relationship between $t$ and $x$, and so your integral will split up into two integrals - perhaps one will be a function of $t$, while the other will be some constant depending on $x$. Hopefully then you can evaluate the function in parentheses explicitly as a function of $x$, at which point taking the limit will be easier.

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    I have solve that problem. The limit is $+\infty$.2011-10-13