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Math Help - Nice problem from Johann Bernoulli

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    MHF Contributor Bruno J.'s Avatar
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    Nice problem from Johann Bernoulli

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