# Reduction formulae

• Jun 14th 2011, 12:33 AM
BabyMilo
Reduction formulae
i) Given that $\displaystyle In= \int_{1}^{0} (lnx)^n dx$, where n is an integer, prove that

$\displaystyle In = -nI(n-1)$

ii) Hence find In in terms of n.

I can do the first part.
How do i do the second part?

thank you.
• Jun 14th 2011, 12:54 AM
emakarov
Re: Reduction formulae
Note that $\displaystyle I_0=-1$, write several first values of $\displaystyle I_n$ and try to guess the general formula for $\displaystyle I_n$.
• Jun 14th 2011, 01:06 AM
BabyMilo
Re: Reduction formulae
In= -n*-(n-1)*-(n-2)....
is this correct?
• Jun 14th 2011, 01:11 AM
emakarov
Re: Reduction formulae
$\displaystyle I_n=(-1)^{n+1}n!$
• Jun 14th 2011, 01:16 AM
BabyMilo
Re: Reduction formulae
can you explain step by step?
im feeling stupid :)

thank you.
• Jun 14th 2011, 01:30 AM
emakarov
Re: Reduction formulae
Starting from $\displaystyle I_0=-1$ and calculating subsequent values of $\displaystyle I_n$ using the recurrence equation $\displaystyle I_n=-nI_{n-1}$, we get

$\displaystyle \begin{array}{c|c|c|c|c|c|c}n & 0 & 1 & 2 & 3 & 4 & 5\\\hline I_n & -1 & 1 & -2 & 6 & -24 & 120\end{array}$

As you noted, $\displaystyle |I_n|=n(n-1)(n-2)\cdot\ldots\cdot1$, which, by definition, is $\displaystyle n!$. Further, we note that $\displaystyle I_n>0$ for odd n and $\displaystyle I_n<0$ for even n. The sequence $\displaystyle (-1)^{n+1}$ shows a similar behavior with respect to alternating signs. Altogether, $\displaystyle I_n=(-1)^{n+1}n!$.