2
$\begingroup$

"Sophomore's Dream" says $\sum_{n=1}^{\infty}n^{-n}=\int_0^1x^{-x}$

Can you replace the $x$ and $n$ with $2x$ or $x^3$ (and $2n$ or $n^3$) or something? I would guess not, because replacing $x$ with the constant $1$ ($x^0$) doesn't work. And if this doesn't work, why does it not work? I understand its probably a basic math operation that prevents you from doing this, so thanks for any input!

$\endgroup$
2
$\begingroup$

The standard way of proving this identity is to write $x^{-x} = e^{-x\log x}$ and then expand $e^{-x\log x} = \sum_{n=0}^{\infty} \frac{(-x\log x)^{n}}{n!}.$ Now the integral is $$\int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-x\log x)^n}{n!} \mathrm{d}x = \sum_{n=0}^{\infty} \int_{0}^{1}\frac{(-x\log x)^{n}}{n!}\mathrm{d}x= \sum_{n=0}^{\infty} \frac{(n+1)^{-n-1}}{n!}\int_{0}^{\infty} u^{n}e^{-u}\mathrm{d}u = \sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{n!(n+1)^{n+1}} = \sum_{n=0}^{\infty} \frac{1}{(n+1)^{n+1}}.$$

The important steps here are interchanging the order of summation and integration (which is okay here since the power series uniformly converges), and the consequent substitution $x=e^{-u/(n+1)}.$ Replacing $x$ by $f(x),$ we would instead have $$\sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-f(x)\log f(x))^n}{n!} \mathrm{d}x.$$ Generally this integral would be impossible to work out by substitution as before, but to address one of your examples:

$$\sum_{n=0}^{\infty} \int_{0}^{1} \frac{(-3x^3 \log x)^n}{n!}\mathrm{d}x = \sum_{n=0}^{\infty} \frac{(-3)^n}{n!}\int_{0}^{1}(x^3 \log x)^n\mathrm{d}x \\= \sum_{n=0}^{\infty} \frac{3^n}{n!(n+1)^{n+1}}\int_{0}^{\infty} u^{n} e^{-(3n+1)u/(n+1)}\mathrm{d}u = \sum_{n=0}^{\infty}\frac{3^n}{n!(n+1)^{n+1}}\cdot \left(\frac{n+1}{3n+1}\right)^{n+1} \int_{0}^{\infty} v^{n}e^{-v}\mathrm{d}v\\ =\sum_{n=0}^{\infty}\frac{3^n}{(3n+1)^{n+1}},$$ so as it turns out, you see that the term $n^{-n}$ doesn't directly correspond to $x$ itself, rather it varies with the exponent of $x$. Of course, this isn't precise, but you see the point.

Edit: I should be careful in saying that it varies with the exponent of $x,$ because as you can see, in the limit the term behaves like $\frac{1}{3}n^{-n},$ so it's more of that in the limit you end up with an extra factor of $1/3.$ I didn't check this for other exponents, but this derivation should work for any $x^{k},$ with a similar result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.