Not Reduction Formula

Consider: $I= \displaystyle \int_{0}^{1} \log^{n} xdx$
Set $\log x =t$
So, $I =\displaystyle \int_{-\infty}^{0}t^{n}e^{t}dt$

Also let $t=-p$ So, $I =(-1)^{n}\displaystyle \int_{0}^{\infty}p^{n}e^{-p}dp$
Or, $I= (-1)^{n} \Gamma(n+1) = (-1)^{n}n!$
So, $\boxed{\displaystyle \int_{0}^{1} \log^{10} xdx =10! =3628800}$

Here's my solution

We are required to calculate $\displaystyle I_{10} = \int_{0}^{1} \ln^{10} x dx$

So let us first derive a general formula.

Consider $\displaystyle I_{n} = \int_{0}^{1} \ln^{n} x dx$ .

Put $x = e^t$ and $dx = e^t dt$

Then we have $\displaystyle I_{n} = \int_{-\infty}^{0} t^{n} e^t dt = \lim_{x \rightarrow -\infty} \int_{x}^{0} t^{n} e^t dt$

Take $\displaystyle M_{n} = \int_{x}^{0} t^{n} e^t dt$

Using integration by parts we obtain

$\displaystyle M_{n} = -x^n e^x - n M_{n-1}$

Taking the limit of both sides as $x \rightarrow -\infty$ , we get

$\displaystyle I_{n} = 0 - n I_{n-1} \iff \frac{I_{n}}{I_{n-1}} = -n$

So we also have

$\displaystyle \frac{I_{n-1}}{I_{n-2}} = -(n-1)$

and so on till

$\displaystyle \frac{I_{1}}{I_{0}} = -1$

Multiplying all of these and observing that this is a telescoping product, we have $\displaystyle \frac{I_{n}}{I_{0}} = (-1)^n n!$

We have $\displaystyle I_{0} = \int_{0}^{1} 1 dx = 1$

Therefore $\displaystyle I_{n} = (-1)^n n!$

Finally $I_{10} = (-1)^{10} 10! = 10! = \boxed{3628800}$ is the answer

This exercise led me to calculate a formula for

$\large \displaystyle \int \ln ^{n}( x) \ \mathrm dx$

it is

$\large \displaystyle \int \ln ^{n} (x) \ \mathrm dx = x \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \ln^{n-k}(x)\ \ + c$

I used integration by parts and induction on $n$ to get it, but to prove it works you can simply derive it.

So

$\displaystyle \int_{0}^{1} \ln ^{n}( x) \ \mathrm dx =$ $\displaystyle = \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \ln^{n-k}(1) - \lim_{x \mapsto 0^+} x \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \ln^{n-k}(x) =$ $\displaystyle \sum_{k=0}^{n-1} (-1)^k \frac{n!}{(n-k)!} \ln^{n-k}(1) \ + n! - \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \lim_{x \mapsto 0^+} x\ln^{n-k}(x) =$ $\displaystyle = \sum_{k=0}^{n-1} (-1)^k \frac{n!}{(n-k)!} \cdot 0 \ + n! - \sum_{k=0}^{n} (-1)^k \frac{n!}{(n-k)!} \cdot 0 = n!$ For $n = 10$ we get our result

$\large \displaystyle \int_{0}^{1} \ln ^{n}( x) \ \mathrm dx = 10! = 3628800$