# Show that $\int_0^1 (\ln x)^n dx =(-1)^n n!$

How can I prove the following: $$\int_0^1 (\ln x)^n dx =(-1)^n n!$$ where $n$ is an integer and $n>0$?

By using partial integration, I started by finding a reduction formula

\begin{align*} I_n &= \int (\ln x)^n dx \\ &= x(\ln x)^n - nI_{n-1} \end{align*}

however the bounds 0 and 1 complicate things seeing as $\ln x \to -\infty \quad \mathrm{as} \quad x \to 0^+$

Note that

\begin{align*} \lim_{x \to 0} x (\ln x)^n &= \lim_{x \to 0} \frac{(\ln x)^n}{\frac 1 x} \\ &= \lim_{x \to 0} \frac{n (\ln x)^{n - 1} \frac 1 x}{-\frac 1 {x^2}} \\ &= -n \lim_{x \to 0} x (\ln x)^{n - 1} \end{align*}

as an application of L'Hospital's rule. Repeat as needed to reduce the exponent to zero, and one sees that the limit is zero. It follows that, in your notation,

$$I_n = -n I_{n - 1}$$

and the result follows.

• Very neat and to the point. +1 Dec 26 '15 at 9:14

You can use the $\; \Gamma\;$function to evaluate the integral. Substituting $x=e^t \;$ $$\ \int_0^1 {(lnx)^ndx}=\int_{- \infty}^0 {t^n}e^tdt$$ Substituting $\, p=-t \,$; $$\int_{- \infty}^0 {t^n}e^tdt=(-1)^n\int_{0}^{\infty}{t^ne^{-p}dp}=(-1)^n n!$$

We have $$I_n = \left[x \ln^n x\right]_0^1 - nI_{n-1}$$

But $$\left[x \ln^n x\right]_0^1 = \ln^n 1 - \lim_{x \to 0} x \ln^n x = 0$$

So $$I_n = -nI_{n-1} = n(n-1)I_{n-2} = -n(n-1)(n-2)I_{n-3} = \cdots$$