Sophomore's Dream

Unfortunately, I am no longer a sophomore, but the sophomore's dream is anyone's dream! Find out why down below.

Sophomore's Dream

The sophomore's dream is a pair of identities discovered in 1697 by Johann Bernoulli. $$\begin{aligned} \int_0^1{x^{-x}} &=\sum_{n=1}^{\infty}{n^{-n}} \\ \int_0^1{x^{x}} &= -\sum_{n=1}^{\infty}{(-n)^{-n}} \end{aligned}$$ The two sums converge to an approximate numerical value of $1.291285997$ and $0.7834305107$, respectively.

Note:If you had guessed that a freshman's dream might exist, then you would be correct... except in that case, it really is just a dream. Freshman's dream refers to the erroneous identity, $(x+y)^n=x^n+y^n$. Ah, how good it would be if that were true!

A similar, related problem

Before we begin, however, we will differentiate $x^x$ instead of integrating (which is always harder than differentiation). The initial technique we use will be sort of similar.

To find $\frac{d}{dx}x^x$, we will use logarithmic differentiation. $$\begin{aligned} \text{Let } y&=x^x \\ \ln{y}&=\ln{x^x} \\ &=x\ln{x} \end{aligned}$$ Implicitly differentiate both sides: $$\begin{aligned} \\ (\ln{y})' &= (x\ln{x})' \\ \frac1y \cdot y' &= x'\ln{x}+x(\ln{x})' \\ \frac {y'}{y} &=\ln{x}+x\cdot \frac{1}{x} \\ y' &= y(\ln{x}+1) \\ y' &= x^x(\ln{x}+1) \end{aligned}$$ This was just a simple demonstration of logarithmic differentiation. By taking the natural logarithm of this complicated expression and using the log rules to simplify from an exponentiation to a multiplication, we were able to use the product rule to implicitly differentiate. When integrating $x^x$, we will do something sort of similar.

Proof

Now, we are ready to tackle the Sophomore's Dream! We begin by expressing $x^x$ as a sum: $$ \begin{aligned} x^x&=e^{\ln{x^x}}=e^{x\ln{x}} \\ &= \sum_{n=0}^{\infty}\frac{(x\ln{x})^n}{n!} \\ &=\sum_{n=0}^{\infty}\frac{x^n(\ln{x})^n}{n!} \end{aligned} $$ ,using the Maclaurin series expansion for $e^x$. Let's now integrate: $$ \begin{aligned} \int_0^1{x^x}&=\int_0^1{\sum_{n=0}^{\infty}\frac{x^n(\ln{x})^n}{n!}}dx \\&={\sum_{n=0}^{\infty}\int_0^1\frac{x^n(\ln{x})^n}{n!}}dx \\&={\sum_{n=0}^{\infty}\frac{1}{n!} \int_0^1{x^n(\ln{x})^n}dx} \end{aligned} $$ Now, we introduce a u-substitution, $u=\ln(x)$ $$\begin{aligned} du&=\frac{1}{x}dx \\ &=\frac{1}{e^u}dx \\e^u du&=dx \end{aligned} $$ Now we want to find the new limits of integration in terms of $u$ by substituting the existing limits in terms of $x$:

• At $x=0$, $u=-\ln{0}=\infty$ (since $\ln{0}$ approaches $-\infty$)
• At $x=1$, $u=-\ln{1}=0$

Putting everything together, $$={\sum_{n=0}^{\infty}\frac{1}{n!} \int_{\infty}^0{{(e^{-u})}^n (-u)^n}(-e)^{-u}du}$$ We will now flip the endpoints of the definite integral by changing the sign of the integral, and this allows us to get rid of the negative sign in $(-e)^{-u}.$ Then, we will factorise the $e^{-u}$ terms, and factorise $(-u)^n$ as $(-1)^nu^n$. Note, $(-1)^n$ is just a constant, so we can then take that out of the integral: $$={\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!} \int^{\infty}_0{{e^{-u(n+1)}}u^n}du}$$ This looks very similar to the Gamma function, which is given by $$\Gamma(z)=\int_0^\infty{e^{-t}t^{z-1}}dt=(z-1)!$$ In our case, $t=u$ and $z=n$ so that: $$\Gamma(n+1)=\int_0^\infty{e^{-u}u^{n}}dt=n!$$ We can now introduce a second substitution, $v=u(n+1)$, in order to manipulate the current integral into the form of the gamma function. Then, $u=\frac{v}{n+1}$ and $dv=(n+1)du$, and the limits of the integrals do not change. $$\begin{aligned} &={\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!} \int^{\infty}_0{{e^{-v}}(\frac{v}{n+1})^n}\frac{dv}{n+1}} \\ &={\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!} \int^{\infty}_0{e^{-v}v^n\frac{1}{(n+1)^{n+1}}dv}} \\ &={\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!} \frac{1}{(n+1)^{n+1}}\int^{\infty}_0{e^{-v}v^ndv}} \\ &={\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!} \frac{1}{(n+1)^{n+1}}n!} \\ &={\sum_{n=0}^{\infty}(-1)^n \frac{1}{(n+1)^{n+1}}} \end{aligned} $$ Finally, we can increase the index of the summation by $1$ so that we go from $n=1$ to $n=\infty$. $$={\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n^{n}}} \\ ={-\sum_{n=1}^{\infty} (-n)^{-n}}$$

Historical Proof

When Bernoulli discovered this identity, the gamma function was not yet known. Therefore, instead of using substitutions to manipulate the integral in to the gamma function, he used integration by parts to integrate term by term. I will not show it here, because it is just heavy computation, but if you want to learn more about it, click on the Wikipedia page right below!

Wake Up!

I took all of this information from this Wikipedia page. On the article, it says to take the substitution $x=e^{-\frac{u}{n+1}}$ which results in $u=-\ln{x}(n+1)$. This is exactly what I did above, except I separated it into two substitutions that actually made sense!

In this post, I wanted to share about this fascinating identity and offer a more detailed explanation of the proofs, as I found that the proofs on the Wikipedia page were quite difficult to understand for the average high school sophomore!