# Math Help - Nice problem from Johann Bernoulli

1. ## Nice problem from Johann Bernoulli

$\mbox{ Show that }\sum_{n=1}^\infty \frac{1}{n^n} = \int_0^1 \frac{1}{x^x} \: dx\: .$

2. Originally Posted by Bruno J.
$\mbox{ Show that }\sum_{n=1}^\infty \frac{1}{n^n} = \int_0^1 \frac{1}{x^x} \: dx\: .$
$I=\int_0^1 x^{-x} \ dx=\int_0^1 e^{-x \ln x} \ dx=\sum_{n=0}^{\infty} \frac{1}{n!} \int_0^1 (-x)^n (\ln x)^n \ dx.$ now put $x=e^{-t}.$ then: $I=\sum_{n=0}^{\infty} \frac{1}{n!} \int_0^{\infty}t^n e^{-(n+1)t} \ dt=\sum_{n=0}^{\infty}\frac{1}{n!} \cdot \frac{n!}{(n+1)^{n+1}}=\sum_{n=0}^{\infty}\frac{1} {(n+1)^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{n^n}. \ \ \Box$

the fastest way to evaluate $J=\int_0^{\infty}t^n e^{-(n+1)t} \ dt$ is to note that $J$ is the Laplace transform of $t^n$ at $s=n+1$ and we know that $\mathcal{L}(t^n)=\frac{n!}{s^{n+1}}.$