# Sophomore's Dream

Using IBP or otherwise prove that:

\displaystyle \int_0^\infty x^n e^{-x} \mathrm dx = n!

for n \ge 0.

Hence evaluate:

\displaystyle \int_0^1 x^n (\log x)^n \mathrm dx

Hence show that:

\displaystyle \int_0^1 x^{-x} \mathrm dx = \sum_{n = 1}^\infty \frac 1 {n^n}

Also evaluate:

\displaystyle \int_0^1 x^x \mathrm dx

leaving your answer in a similar form.

You can ignore complications exchanging sums and integrals.

We will need this simple lemma for the solution. It is quite obvious, but I will still provide a proof :

Lemma: Given a function f: \Bbb Z^+_0 \to \Bbb R, if some recurrence relation u_n satisfies, for non-negative integer n, u_n=f(n)u_{n-1}, u_0=k \in \Bbb R then \displaystyle u_n=k\prod^n_{r=1}f(r)

Proof: By induction on n.

base case: When n=1, u_1=f(1)u_0=kf(0)

Suppose the statement is true for some n=k. Consider n=k+1:
\displaystyle u_{k+1}=f(k+1)u_k=f(k+1)k\prod^k_{r=1}f(r)=k\prod^{k+1}_{r=1}f(r)
By induction, the lemma holds. \blacksquare

## Solution:

Let \displaystyle u=x^n, v'=e^{-x} then u'=nx^{n-1}, v=-e^{-x}
Let I_n=\displaystyle \int_0^\infty x^n e^{-x} \mathrm dx
\displaystyle I_n=[-x^ne^{-x}]^\infty_0-n\int^\infty_0-x^{n-1} e^{-x} \mathrm dx
\displaystyle I_n=nI_{n-1}, I_0=1
By the lemma, we have I_n=n!.

For the next part I have appended an alternative method that does not need the usage of the first result for anyone interested. The intended method is included here, due to the usage of â€śhenceâ€ť, rather than â€śhence or otherwiseâ€ť:

use the substitution y=-(n+1)\log x \iff e^{-\frac y{n+1}}=x so dx=-\frac1{n+1}e^{-\frac y{n+1}}dy

\displaystyle \therefore\int^1_0x^n(\log x)^n \mathrm dx=-\frac1{n+1}\int^0_\infty\left(-\frac y{n+1}\right)^n(e^{-\frac {ny}{n+1}})e^{-\frac y{n+1}}\mathrm dy

\displaystyle=\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty y^ne^{-y}\mathrm dy

\displaystyle=\frac{(-1)^nn!}{(n+1)^{n+1}}

Let x \in \Bbb R^+ then:

x^{-x}=e^{-x\log x}

\displaystyle e^x=\sum^\infty_0\frac{x^n}{n!}, \ x \in \Bbb R
\displaystyle \therefore e^{-x\log x}=x^{-x}=\sum^\infty_{n=0} (-1)^n\frac{(x\log x)^n}{n!}, \ x \in \Bbb R^+

\displaystyle \int^1_0x^{-x} \ \mathrm dx=\int^1_0\sum^\infty_{n=0} (-1)^n\frac{(x\log x)^n}{n!} \mathrm dx

\displaystyle =\sum^\infty_{n=0} \frac{(-1)^n(-1)^nn!}{(n+1)^{n+1}n!}

\displaystyle=\sum^\infty_{n=0}\frac{1}{(n+1)^{n+1}}

\displaystyle=\sum^\infty_{n=1}\frac{1}{n^n}

x^x=e^{x\log x}
\displaystyle \therefore x^x=\sum^\infty_{n=0} \frac{(x\log x)^n}{n!}, \ x \in \Bbb R^+

\displaystyle \int^1_0x^{x} \ \mathrm dx=\int^1_0\sum^\infty_{n=0} \frac{(x\log x)^n}{n!} \mathrm dx

\displaystyle=\sum^\infty_{n=0}\frac{(-1)^n}{(n+1)^{n+1}}

\displaystyle=\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n^n}

## Alternative Method for the Second Integral:

We can derive the second result directly by being a little clever with the recurrence relation. However, do note that the question uses â€śhenceâ€ť rather than â€śhence or otherwiseâ€ť so this method would (strictly speaking) not be valid for the purposes of the question. Or perhaps Iâ€™m being pedantic due to STEP.

Let \displaystyle J_r=\int^1_0x^n(\log x)^r \mathrm dx, r \in \Bbb Z^+
Consider integration by parts on J_r with u=(\log x)^r, v'=x^n.
\displaystyle u'=\frac rx(\log x)^{r-1}, v=\frac{x^{n+1}}{n+1}

\displaystyle\therefore J_r=\left[\frac{x^{n+1}(\log x)^r}{n+1}\right]^1_0-\int^1_0\frac{rx^n(\log x)^{r-1}}{n+1}\mathrm dx

\displaystyle J_r=-\frac r{n+1}J_{r-1}, \ J_0=\frac1{n+1}

\displaystyle J_r=\frac {(-1)^rr!}{(n+1)^{r+1}}

\displaystyle\therefore \ J_n=\frac {(-1)^nn!}{(n+1)^{n+1}}

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