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sauravshakyaBest ResponseYou've already chosen the best response.0
dw:1356849576104:dw
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
seems it doesn't have any closed form... http://www.wolframalpha.com/input/?i=integral+%28x%5E%281%2Fn%291%2Fx%5En%29%5En
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
are we assuming n is a constant?
 one year ago

malevolence19Best ResponseYou've already chosen the best response.1
\[=\int\limits \left( \frac{x^{\frac{n^2+1}{n}}1}{x^n} \right)^ndx\]
 one year ago

malevolence19Best ResponseYou've already chosen the best response.1
I see 2 approaches. Either do something like: u=x^(n^2+1)/n  1 and solve for x in terms of u; then find dx, etc. Or try integration by parts by letting u=(...)^(n1), dv=(...)^1dx; from there you could try to establish a pattern for n. But it is more than likely unable to be solved in closed form.
 one year ago

wioBest ResponseYou've already chosen the best response.0
@abb0t We know \(n\) is constant... but it would help to know if \(n\) is just a natural number or any real number.
 one year ago

wioBest ResponseYou've already chosen the best response.0
For some reason things like this want me to try binomial theorem, lol I'm just a bit crazy I guess.
 one year ago

wioBest ResponseYou've already chosen the best response.0
If you assume \(n \in \mathbb{N}\) you can just use binomial theorem and get an answer in summation notation.
 one year ago

wioBest ResponseYou've already chosen the best response.0
Then if you're lucky, you might be able to simplify it from there, into something algebraic, but there is no guarantee it will happen.
 one year ago

wioBest ResponseYou've already chosen the best response.0
There is no reason to be afraid of summations when it comes to certain integrals... there is no guarantee the function isn't transcendental.
 one year ago

malevolence19Best ResponseYou've already chosen the best response.1
=\[\int\limits \sum_{k=0}^{\infty} \left(\begin{matrix}n \\ k\end{matrix}\right)(x^{\frac{1}{n}})^{nk}x^{nk}dx\] My only question with doing this is that in the binomial theorem the don't make mention of things of the form: \[(\alpha(n)+\beta(n))^n\] Where alpha and beta are numbers that depend on n.
 one year ago

malevolence19Best ResponseYou've already chosen the best response.1
I would assume it doesn't matter.
 one year ago

malevolence19Best ResponseYou've already chosen the best response.1
And that last x should have an (1)^k also.
 one year ago
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