1. ## integration with factorials

im having a lot of trouble showing that the

definite integral from 0 to 1 (ln x)^n dx = n!(-1)^n
if someone could help get me started cause i have no idea how to get the factorial in the answer ... all i get is this infinite loop of integration by parts which keeps going on and on

2. Put $u=-\ln x$ and your integral becomes ${{(-1)}^{n}}\int_{0}^{\infty }{{{u}^{n}}{{e}^{-u}}\,du}={{(-1)}^{n}}n!.\quad\blacksquare$

See some of

Gamma function - Wikipedia, the free encyclopedia

3. i am not familiar with gamma functions ... is there an easier way of solving this problem ... maybe this will help

there was a previous part to this question which i was able to solve

I(subscript)n = integral (ln x)^n dx implies that
I(subscript)n = x (ln x)^n - n*(I(subscript)n-1)

maybe i have to somehow use this to solve the second question ... but i cannot see anyway to use it ...

4. The way Kriz showed is about the easiest. If you don't know the Gamma, then get to know it. It's a great tool to have in the arsenal.

Here's another subtle difference in the sub Kriz mentioned.

$\int_{0}^{1}(ln(x))^{n}dx$

Let $x=e^{-t}, \;\ dx=-e^{-t}dt$

Make the subs and change the limits of integration.

When x=1, then t=0. When x=0, then $t\to {\infty}$

$-\int_{\infty}^{0}(-t)^{n}e^{-t}dt$

Factor out the $(-1)^{n}$ and reverse the limits of integration to shed the negative sign.

$\int_{0}^{\infty}(-1)^{n}t^{n}e^{-t}dt$

Now, since ${\Gamma}(n)=\int_{0}^{\infty}t^{n-1}e^{-t}dt$

Then we have for ours, ${\Gamma}(n+1)$ which equals n!

So, we get $(-1)^{n}n!$

5. Now, if you make an iterative integration by parts for your integral, you'll get the following:

\begin{aligned}
\int_{0}^{1}{{{\ln }^{n}}(x)\,dx}&=\int_{0}^{1}{(x)'{{\ln }^{n}}(x)\,dx} \\
& =-n\int_{0}^{1}{{{\ln }^{n-1}}(x)\,dx} \\
& =n(n-1)\int_{0}^{1}{{{\ln }^{n-2}}(x)\,dx} \\
&~\,\vdots \\
& ={{(-1)}^{n}}n!.
\end{aligned}

6. and can u show me how to get that (-1) in your answer ... as when i replace x with 1 and 0 they end up cancelling out and giving me 0

7. Induction, I assume since you put that you have $n!$ and not $\Gamma(n+1)$ that $n\in\mathbb{N}$.

Proposition: $\int_0^1 \ln^n(x)~dx=n!(-1)^n$

Base case: $\int_0^1 \ln(x)~dx=\left[x\ln(x)-x\right]\bigg|_{x=0}^{x=1}=-1=1!\cdot (-1)^1$

Inductive hypothesis: $\int_0^1 \ln^n(x)~dx=n!(-1)^n$

Inductive step:

\begin{aligned}\int_0^1 \ln^{n+1}(x)~dx&=\left[x\ln^{n+1}(x)\right]\bigg|_{x=0}^{x=n}-(n+1)\int_0^1 \ln^n(x)~dx{\color{red}\star}\\
&=0-(n+1)\overbrace{n!(-1)^n}^{IH}\\
$\color{red}\star$: This is gotten by parts letting $u=\ln^{n+1}(x)\implies du=(n+1)\frac{\ln^n(x)}{x}~dx$ and $dv=dx\implies v=x$