3
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I use the Limit and DiscreteLimit to solve it but failed. How to solve it by Mathematica? $$\underset{n\to \infty }{\text{lim}}n \left(\int_0^{\frac{\pi }{4}} \tan ^n\left(\frac{x}{n}\right) \, dx\right)^{1/n}$$

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  • $\begingroup$ If you try Integrate[Tan[x/n]^n, {x, 0, Pi/4}] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist? $\endgroup$ – Ulrich Neumann Sep 10 '18 at 7:29
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    $\begingroup$ @UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4 $\endgroup$ – LLlAMnYP Sep 10 '18 at 7:32
  • $\begingroup$ @LLlAMnYP The limit would be 1/4 \[Pi]^((4 + \[Pi])/\[Pi]) (4 + \[Pi])^(-4/\[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2 $\endgroup$ – Ulrich Neumann Sep 10 '18 at 7:39
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    $\begingroup$ @King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure. $\endgroup$ – LLlAMnYP Sep 10 '18 at 7:45
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    $\begingroup$ Try: Limit[n*Integrate[Tan[x/n]^n, {x, 0, Pi/4}, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity] $\endgroup$ – Mariusz Iwaniuk Sep 10 '18 at 9:13
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Integrate returns an anti-derivative which seems reasonable for large n:

antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, {x, 0, π/4}, PlotRange -> All]

And then

Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
π/4
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  • $\begingroup$ how simple a tricky solution can be... $\endgroup$ – Ulrich Neumann Sep 10 '18 at 7:55

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