# Proof of the relation $\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$

I had the thought that by introducing some parameters into simple integrals and taking derivatives we can get exact values for infinitely many 'complicated' integrals.

$$\int_0^1 x^a dx = \frac{1}{a+1}$$

$$\int_0^1 x^a \log x dx = -\frac{1}{(a+1)^2}$$

$$\int_0^1 x^a \log^n x dx = \frac{(-1)^n~ n!}{(a+1)^{n+1}}$$

Since $|x|<1$ the following sum has a closed form:

$$\sum_{a=0}^{ \infty } x^a=\frac{1}{1-x}$$

And by definition:

$$\sum_{a=0}^{ \infty } \frac{1}{(a+1)^{n+1}}=\zeta(n+1)$$

So we have:

$$\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$$

Is this proof correct? Can we use this method to find more complicated integrals (assuming the derivatives exist of course)?

For example, if we use two more parameters in the original integral:

$$\int_0^1 (b+cx)^a dx = \frac{(b+c)^{a+1}-b^{a+1}}{c(a+1)}$$

We can get much more complicated expressions.

• Looks good to me. Mar 1, 2016 at 20:07
• well done (+1) ! Mar 1, 2016 at 21:17
• I think one has to be really careful that by introducing the extra parameters, the derivatives w.r.t to $a$ become not to messy by chain rule application. even in the case $b=0$ things becomes complicated it is not obvious how to obtain a closed form for the $n$-th derivative. Mar 2, 2016 at 8:26

\begin{aligned} \int_0^1 \frac{(\ln x)^n}{1-x} & \stackrel{x\mapsto e^{-x}}{=} \int_0^{\infty} \frac{(-x)^n}{1-e^{-x}} d x \\ &=(-1)^n \sum_{k=0}^{\infty} \int_0^{\infty} x^n e^{-k x} d x \\ & \stackrel{kx\mapsto x}{=} (-1)^n \sum_{k=0}^n \frac{1}{k^{n+1}} \int_0^{\infty} x^n e^{-x} d x \\ &= (-1)^n \sum_{k=0}^{\infty} \frac{\Gamma(n+1)}{k ^{n+1}} \\ & =(-1)^n \Gamma(n+1)\zeta(n+1) \end{aligned}

Seems to work with a Mellin Transform under the integral sign. We can introduce a parameter $a$ $$\mathcal{M}_a\left[\frac{\log(x)^n}{a-x}\right](s) = \pi\csc(\pi s) (-x)^{s-1}\log(x)^n$$ where $\mathcal{M}_a[f](s)$ is a Mellin transform over parameter $a$, then $$\pi\csc(\pi s)\int_0^1 (-x)^{s-1} \log(x)^n \;dx = \pi\csc(\pi s)(-1)^{1+n+s} s^{-n-1}\Gamma(1+n)$$ and $$\mathcal{M}^{-1}_s\left[(-1)^{1+n+s} \pi s^{-n-1}\csc(\pi s)\Gamma(1+n)\right](a)= (-1)^n n! \text{Li}_{n+1}\left(\frac{1}{a}\right)$$ letting $a=1$ and using that $\text{Li}_n(1)=\zeta(n)$ then $$\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$$

• A rose by any other name... Thank you, I haven't heard much of this transform at the time I asked this Jun 2, 2017 at 12:06
• $\int_0^\infty x^{s-1} e^{-nx}dx =\int_0^\infty (y/n)^{s-1} e^{-y}d(y/n)= n^{-s} \Gamma(s)$ therefore for $\Re(s) > 1$ : $$\zeta(s) \Gamma(s) = \lim_{N \to \infty} \sum_{n=1}^N \int_0^\infty x^{s-1} e^{-nx}dx = \lim_{N \to \infty} \int_0^\infty x^{s-1}\frac{e^{-x}-e^{-(N+1)x}}{1-e^{-x}}dx= \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$ and $$\zeta(n+1) n! =\int_0^\infty \frac{x^{n}}{e^x-1}dx= \int_0^1\frac{(-\log y)^n}{1/y-1}d(-\log y) =(-1)^n \int_0^1 \frac{\log^ny}{1-y}dy$$ @YuriyS Jun 2, 2017 at 14:21

We have 2 nice solutions. I want to add one more elementary method using a reduction formula.

First of all, let’s expand the integrand into a power series. $$\int_{0}^{1} \frac{(\ln x)^{n}}{1-x} d x =\int_{0}^{n}(\ln x)^{n} \sum_{k=0}^{\infty} x^{k} d x =\sum_{k=0}^{\infty} \int_{0}^{1} x^{k}(\ln x)^{n} d x$$

Then we are going to find a reduction formula for the last integral

$$I_{n}=\int_{0}^{1} x^{k}(\ln x)^{n} d x$$ Applying integration by parts yields \begin{aligned} I_n&=\int_{0}^{1}(\ln x)^{n} d\left(\frac{x^{k+1}}{k+1}\right) \\ &=\left[\frac{x^{k+1}(\ln x)^{n}}{k+1}\right]_{0}^{1}-\frac{n}{k+1} \int_{0}^{1} (\ln x)^{n-1} x^{k} d x \\ &=-\frac{n}{k+1} I_{n-1} \\ &\qquad\qquad \vdots \\ &=-\frac{n}{k+1}\left(-\frac{n-1}{k+1}\right) \cdots\left(-\frac{1}{k+1}\right) \int_{0}^{1} x^{k} d x \\ &=\frac{(-1)^{n} n !}{(k+1)^{n+1}} \end{aligned} We can now conclude that $$\boxed{I =\sum_{k=0}^{\infty} \frac{(-1)^{n} n !}{(k+1)^{n+1}} =(-1)^{n} n ! \zeta (n+1)}$$