Calculate $\int_0^1 \ln^{2020}(x)dx$ In this video on mark 17:00 gives the tiebreaker question of the integration bee finals. It asks to calculate $$\int_0^1 \ln^{2020} (x)dx$$ We could create a reduction formula for $\ln^n(x)$ which would be $$I_n=x\ln^n(x)-n\int\ln^{n-1}(x)dx$$ Repeatedly doing this we get this sum: $$I_n=n!+x\sum_{k=0}^{n-1}(-1)^k\ln^{n-k}(x)\binom{n}{k}k!$$ Substituting $n=2020$ and then substituting $x=0$, I realized that we could just simply substitute $x=1$ to get the answer, which is $2020!$. This answer is correct, but is my procedure right?
 A: It would be easier to make the substitution $ x = e^{-t}$ which gives $t = - \ln(x)$, limits $t \in (0, \infty)$, and
\begin{align}
I_{n} &= \int_{0}^{1} \ln^{n}(x) \, dx \\
&= \int_{0}^{\infty} e^{-t} \, (-t)^n \, dt \hspace{10mm} x = e^{-t} \\
&= (-1)^n \, \int_{0}^{\infty} e^{-t} \, t^{(n+1)-1} \, dt \\
&= (-1)^n \, \Gamma(n+1) = (-1)^n \, n!.
\end{align}
When $n$ is even the result is
$$ \int_{0}^{1} \ln^{2 n}(x) \, dx = (2 n)!.$$
In view of the proposed problem:
\begin{align}
I_{n} &= \int \ln^{n}(x) \, dx = \int \left( \ln(x) \right)^{n} \, 1 \, dx \\
&= \left[x \cdot \ln^{n}(x) \right] - \int x \cdot \frac{n}{x} \, \ln^{n-1}(x) \, dx = \left[ x \, \ln^{n}(x) \right] - n \, I_{n-1}.
\end{align}
This gives the recurrence relation
$$ I_{n} = x \, \ln^{n}(x) - n \, I_{n-1} $$
with $I_{0} = x$. This relation works without limits. In the case of the limits being $x \in [0,1]$ then, when applied, the recurrence becomes
$$ J_{n} = - n \, J_{n-1}, $$
where $$ J_{n} = \int_{0}^{1} \ln^{n}(x) \, dx$$ and $J_{0} = 1.$ From here it is determined that $J_{n} = (-1)^n \, n!$. This gives
\begin{align}
\int_{0}^{1} \ln^{2 n}(x) \, dx &= (2 n)! \\
\int_{0}^{1} \ln^{2 n+1}(x) \, dx &= - (2 n +1)!.
\end{align}
A: Your solution works. Here is another method which isn't necessarily easier, but is still cool.
Let $$F(a)=\int_0^1 x^adx=\frac1{a+1}\qquad a\ne-1$$
Notice that
$$\begin{align}
F'(a)&=\int_0^1x^a\log(x) dx\\
F''(a)&=\int_0^1x^a\log^2(x) dx\\
F'''(a)&=\int_0^1x^a\log^3(x) dx\\
&...\\
F^{(n)}(a)&=\int_0^1x^a\log^n(x)dx.
\end{align}$$
We can see then that $$\int_0^1\log^n(x)dx=F^{(n)}(0)=\left(\frac{d}{da}\right)^n\frac{1}{a+1}\Bigg|_{a=0}.$$
It is simple to show that
$$\left(\frac{\partial}{\partial z}\right)^nz^\alpha=(-1)^n(-\alpha)_nz^{\alpha-n},$$
where
$$(x)_n=\frac{\Gamma(n+x)}{\Gamma(x)}=(x+n-1)(x+n-2)\cdots(x+1)x,$$
which satisfies the recurrence $$(x)_{n+1}=(x+n)(x)_n,\qquad (x)_0=1.$$
Anyway, we have
$$\left(\frac{d}{da}\right)^n\frac{1}{a+1}\Bigg|_{a=0}=(-1)^n(1)_n(a+1)^{-1-n}\bigg|_{a=0}=(-1)^n\frac{\Gamma(n+1)}{\Gamma(1)}=(-1)^nn!.$$
So,
$$\int_0^1\log^n(x)dx=(-1)^nn!,$$
and thus
$$\int_0^1\log^{2020}(x)dx=2020!$$
