Formula for integral of $x^a \ln^n(x) dx$

Question

Can you explain this formula, I don't get how it works. Given $a > 0$, the integral is $$ \int_{0}^{1} x^a \, \ln^n(x) \, dx = \frac{(-1)^nn!}{(a+1)^{n+1}}. $$ I started integrating it (by parts) and understood that the first part is always $0$. So we are left with $$-\frac{n}{a+1} \, \int_{0}^{1} x^a \, \ln^{n-1}(x) \, dx $$ What happens next?

Answer 1

Let $\displaystyle I(a) = \int_0^{1} x^a \ \mathrm{d}x$. It is straightforward to note that $I(a) = 1/(a+1)$. Using differentiation under the integral, we have, \begin{align*} I'(a) = \frac{\mathrm{d}}{\mathrm{d}a} \left(\int_0^{1} x^a \ \mathrm{d}x\right) = \int_0^{1} x^a \ \ln(x) \ \mathrm{d}x. \end{align*} However, since $I(a) = \dfrac{1}{a+1}$, $I'(a) = -\dfrac{1}{(a+1)^2}$. Repeating this process for a total of $n$ times yields the answer.

Answer 2

What happens next?

For $a\ne -1$, using integration by parts gives $$ \begin{aligned} \int_0^1 x^a \ln ^n x d x&=\frac{1}{a+1} \int_0^1 \ln ^n x d\left(x^{a+1}\right) \\ & \quad= \underbrace{ \left[\frac{x^{a+1} \ln ^n x}{a+1}\right]_0^1}_{=0} -\frac{n}{a+1} \int_0^1 x^a \ln ^{n-1} x d x \end{aligned} $$ So repeatedly using the reduction formula yields $$ \begin{aligned} I_n & =-\frac{x}{a+1} \cdot I_{n-1} \\ & =-\frac{n}{a+1}\left(-\frac{n-1}{a+1}\right)\left(-\frac{n-2}{a+1}\right) \cdots\left(-\frac{1}{a+1}\right) I_0 \\ & =\frac{(-1)^n n !}{(a+1)^n} \int_0^1 x^a d x \\ & =\frac{(-1)^n n !}{(a+1)^{n+1}} \end{aligned} $$

Answer 3

I was astonished to not see anyone use the gamma function which seems to me the most straightforward route to derive this result, so here goes.

$$I=\int_{0}^{1}x^a\ln^n(x)\,dx$$

To involve the gamma function lets make the following substitution

Let $$\ln(x)=t$$ so $$x=e^t$$ $$dx=e^tdt$$

$$I=\int_{-\infty}^{0}t^ne^{t(a+1)}\,dt$$

Now let $$t(a+1)\longrightarrow{-t}$$

$$I=\frac{(-1)^n}{(a+1)^{n+1}}\int_{0}^{\infty}t^ne^{-t}\,dt$$

$$I=\frac{(-1)^n\Gamma(n+1)}{(a+1)^{n+1}}$$

As you can see, this was pretty straightforward and we even have a general formula for when n is not an integer! Be careful though, you need to make sure that I converges, because it doesn’t for all non-integer n.

Answer 4

As mentioned in the comments and proposed problem a recurrence can be obtained by using integration by parts. In another answer differentiation of a parameter can be used. In this solution the use of series will be made to obtain the same result.

Consider the integral $$ \int_{0}^{1} x^{a} \, dx = \frac{1}{1+a} $$ and letting $a \to a-t$ which gives \begin{align} \int_{0}^{1} x^{a-t} \, dx &= \frac{1}{1+a-t} \\ \int_{0}^{1} x^{a} \, e^{-t \, \ln x} \, dx &= \frac{1}{1+a} \, \frac{1+a}{1+a - t} \\ \int_{0}^{1} x^{a} \, \sum_{n=0}^{\infty} \frac{(- t \, \ln x)^{n}}{n!} \, dx &= \frac{1}{1+a} \, \sum_{n=0}^{\infty} \frac{t^n}{(1+a)^n} \\ \sum_{n=0}^{\infty} \frac{(-1)^n \, t^n}{n!} \, \int_{0}^{1} x^a \, \ln^{n}(x) \, dx &= \sum_{n=0}^{\infty} t^n \, \frac{1}{(1+a)^{n+1}}. \end{align} Equating coefficients gives $$ \int_{0}^{1} x^{a} \, \ln^{n}(x) \, dx = \frac{(-1)^n \, n!}{(1+a)^{n+1}}. $$

Answer 5

Just another way using EGF.

Write $I(a,n)$ for the integral and consider the exponential generating function $G(t)$, $$G(t)=\sum_{n=0}^\infty\frac{I(a,n)}{n!}t^n=\int_0^1x^a\sum_{n=0}^\infty\frac{(t\log(x))^n}{n!}\ dx=\int_0^1x^{t+a}\ dx=\frac{1}{t+a+1}$$ expanding the result as a geometric series in the same region, $$\frac{1}{t+a+1}=\frac{1}{a+1}\frac{1}{1+t/(a+1)}=\sum_{n=0}^\infty\frac{(-1)^n}{(a+1)^{n+1}}t^n=\sum_{n=0}^\infty\frac{(-1)^nn!}{(a+1)^{n+1}}\frac{t^n}{n!}$$ comparing coefficients, $$I(a,n)=[t^n/n!]G(t)=\frac{(-1)^nn!}{(a+1)^{n+1}}.$$