# Nice problem from Johann Bernoulli

• Jun 30th 2009, 06:45 PM
Bruno J.
Nice problem from Johann Bernoulli
$\displaystyle \mbox{ Show that }\sum_{n=1}^\infty \frac{1}{n^n} = \int_0^1 \frac{1}{x^x} \: dx\: .$
• Jun 30th 2009, 09:26 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
$\displaystyle \mbox{ Show that }\sum_{n=1}^\infty \frac{1}{n^n} = \int_0^1 \frac{1}{x^x} \: dx\: .$

$\displaystyle I=\int_0^1 x^{-x} \ dx=\int_0^1 e^{-x \ln x} \ dx=\sum_{n=0}^{\infty} \frac{1}{n!} \int_0^1 (-x)^n (\ln x)^n \ dx.$ now put $\displaystyle x=e^{-t}.$ then: $\displaystyle I=\sum_{n=0}^{\infty} \frac{1}{n!} \int_0^{\infty}t^n e^{-(n+1)t} \ dt=\sum_{n=0}^{\infty}\frac{1}{n!} \cdot \frac{n!}{(n+1)^{n+1}}=\sum_{n=0}^{\infty}\frac{1} {(n+1)^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{n^n}. \ \ \Box$

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