I’m a novice
I want to ask why I can’t find out the limit of input.
[1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=屏幕截图2021-03-19081838.png&userId=2222586
Please post code, not images.
Notice the ( ) are colored red to indicate a syntax error. In the WL arguments to functions are enclosed in [ ], not ( ).
Thank you very much.
I also want to ask why the calculation is so slow?
Does this software need to be connected to the cloud?
Slow compared to what?
Some integrals are easy to compute, some are not, and some do not have an analytic solution. Some limits are easy to compute, some are not.
OK, thank you.
Why does the calculation result just repeat the title, but not come to a specific result?
\(\*UnderscriptBox[\(\[Limit]\), \(n \[Rule] \[Infinity]\)]\)\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(n\)]\(
\*FractionBox[
SqrtBox[\(
\*SuperscriptBox[\(n\), \(2\)] -
\*SuperscriptBox[\(x\), \(2\)]\)], \(2 +
\*SuperscriptBox[\(x\), \(-x\)]\)] \[DifferentialD]x\)\)/n^2
This type integral can’t be found a symbolic solution - closed-form solution. With numerics is easy:
f[n_?NumericQ] :=
NIntegrate[Sqrt[n^2 - x^2]/(2 + x^(-x)), {x, 0, n}]; Table[
f[n]/n^2 /. n -> 10^k, {k, 1, 6}]
(*{0.359439, 0.389349, 0.392364, 0.392699, 0.392699, 0.392699}*)
Looks like limit have a value: 0.392699…
EDITED:
We can use:
AsymptoticLessEqual[x^-x, 1/(x^2 + 1), x -> \[Infinity]]
(*True*)
then:
Limit[Integrate[
Sqrt[n^2 - x^2]/(2 + (1/(x^2 + 1))) // Factor, {x, 0, n},
Assumptions -> {n > 0}]/n^2, n -> Infinity]
(*\[Pi]/8*)
\[Pi]/8 // N
(*0.392699*)
Thank you, friends.