Given$\displaystyle \int_0^\infty x^n dx$ find a value for n, which is a real number, such that the limit exists. (Such that the limit converges to some, presumably, real number.)

I'm omitting the work/algebra I've done to save time/space and because this all seems pretty clear to me.

It seems pretty obvious, for the case of $\displaystyle n\geq1$ that it doesn't converge, just by looking at the end behavior of any polynomial function with a positive exponent greater then 1, say 2 and the fact that an integral is the area under a curve. While x^1 is just a constant function and doesn't converge either.

For 0<n<1, we have some root function, which also doesn't converge. Again I'm just looking at the graph.

For -1<n<0, this is the inverse of a root. Which looks similar to the inverse of a polynomial. This range appears to converge on the interval (0,1] but diverge on [1,+$\displaystyle \infty$] (I'm omitting work here.)

Since $\displaystyle \int \frac{1}{x} dx= ln|x|+c$* Yet again I'll just reference the graph and conclude that this diverges over the given interval.

For n<-1, the function is just an inverse of a polynomial. They converge over the interval [1, +$\displaystyle \infty$). But all diverge over (0,1].

So assuming this is as clear as I think it is, a big assumption, and I haven't made any errors . There is no n satisfying the conditions.

So if i'm correct, and there is no n such that the limit exists/converges, could someone give me some hints as to how I can write this out in a way that isn't so bulky?

Thanks!