$int_0^1 (lnx)^n "dx"=(-1)^n*n! \quad\quad nin NN$

[Steven11]

Nice and not too hard (the knowledge of Liceo is sufficient)

Show that
$int_0^1 (lnx)^n "dx"=(-1)^n*n! \quad\quad nin\mathbb{N}$

Bye!

[Steven11]

"WiZaRd":
The proof in finished.
It's OK?

Of course.

This is the mine:
$int_0^1 (lnx)^n "dx"=(-1)^n*n!$ (1)
Let's suppose (1) is true (inductive hypothesis).
You just verified the relation is true if $n=0$
So, if we have $n+1$
$int_0^1 (lnx)^(n+1) "dx"=$
I make the following sobstitution: $lnx=t=>x=e^t$ and I get
$=int_(-oo)^0 t^(n+1)e^t "dt"=$
With an integration by parts
$=[t^(n+1)e^t]_(-oo)^0-(n+1)int_(-oo)^0t^n e^t "dt"$
On the other hand,
$[t^(n+1)e^t]_(-oo)^0=0$ and $int_(-oo)^0t^n e^t "dt"=int_0^1 (lnx)^n "dx"=(-1)^n*n!$
so the final result is
$=-(n+1)(-1)^{n}\cdotn! =(-1)^(n+1)\cdot(n+1)!$

[G.D.5]

I'll try to prove the statement by Induction, but I don't know if it's correct or not.

Since $f(x)=lnx$ is not defined in $x=0$, the statement given become

$lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}(lnx)^{x} dx = (-1)^{n} \cdot n!, \ \forall n \in \mathbb{N}$.

If $n=0$ we have:

$lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}(ln)^{0}dx=lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}dx=\lim_{\alpha ->0^{+}}(1-\alpha)=1=(-1)^{0}\cdot0!$

and the statement is true.

Now we assume that the statement is true for $n$: $lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}(lnx)^{n}dx=(-1)^{n}\cdotn!$.
For $n+1$ we have:

$\lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}(lnx)^{n+1}dx=\lim_{\alpha ->0^{+}}\int_{\alpha}^{1}(lnx)^{n+1}\cdot1dx=\lim_{\alpha -> 0^{+}}\int_{\alpha}^{1}(lnx)^{n+1}\cdot(x)'dx=\lim_{\alpha ->0^{+}}[(ln1)^{n+1}\cdot(1)-(ln\alpha)^{n+1}\cdot(\alpha)-\int_{\alpha}^{1}(n+1)(lnx)^{n}\frac{1}{x}\cdot(x)dx]=$
$=lim_{\alpha -> 0^{+}}[-\alpha(ln\alpha)^{n+1}-(n+1)\int_{\alpha}^{1}(lnx)^{n}dx]=-(n+1)(-1)^{n}\cdotn! \ =(-1)(-1)^{n}(n+1)n! \=(-1)^{n+1}\cdot(n+1)!$

The proof in finished.


It's OK?