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750

Part of 2011 Today's Calculation Of Integral

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Today's calculation of Integral 750

Source: 2011 Tsukuba University entrance exam

8/31/2011
Let an (n≥1)a_n\ (n\geq 1) be the value for which ∫x2xe−tndt (x≥0)\int_x^{2x} e^{-t^n}dt\ (x\geq 0) is maximal. Find lim⁡n→∞ln⁡an.\lim_{n\to\infty} \ln a_n.
calculusintegrationlimitlogarithmscalculus computations