Limit of a integral

1. Aug 8, 2008

dirk_mec1

How can you evaluate this integral?

$$\lim_{n\to\infty}\sqrt[n]{\int_0^1 x^{-nx}\ dx}$$

Are there any rules for taking a square root into an integral?

2. Aug 8, 2008

arildno

No, that is not possible. If you take a root of a sum, can you say that this equals the sum of the roots of the terms? That's basically the same that you're asking!

The above limit seems very nasty, but we can at least roughly estimate a bounding interval containing that limit value:

Note that, for fixed n, the MINIMUM value of the integrand is 1. Therefore, a lower bound for the integral equals 1, and the n'th root of 1, i.e, 1, is a lower bound for the whole expression.

Now, you may readily show that the maximum value for the integrand equals $e^{\frac{n}{e}}$, occurring at $x=\frac{1}{e}$.
Thus, an upper bound for the integral is $e^{\frac{n}{e}}$, and as an upper bound for the whole expression we have:
$$\sqrt[n]{e^{\frac{n}{e}}}=e^{\frac{1}{e}$$

Thus, our limit lies somewhere between 1 and $e^{\frac{1}{e}}$

Last edited: Aug 8, 2008
3. Aug 8, 2008

ice109

may i ask where this integral arises?

4. Aug 8, 2008

nicksauce

Here is a graph I made in maple... looks like the limit is roughly 1.44.

File size:
7.6 KB
Views:
51
5. Aug 8, 2008

snipez90

Can the integral be evaluated through parametric integration or other multivariate calculus methods?

6. Aug 8, 2008

dirk_mec1

I don't know but in contrast to arildno's answer the exact solution can be via integration methods. I just don't which and how to apply them.