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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.

Ben Sheller
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Yuriy S
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    Looks good to me. – Ali Caglayan Mar 01 '16 at 20:07
  • well done (+1) ! – tired Mar 01 '16 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. – tired Mar 02 '16 at 08:26

3 Answers3

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

Lai
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Seems to work with a Mellin Transform under the integral sign. We can introduce a parameter $a$ \begin{equation} \mathcal{M}_a\left[\frac{\log(x)^n}{a-x}\right](s) = \pi\csc(\pi s) (-x)^{s-1}\log(x)^n \end{equation} 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 $$ \int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1) $$

Benedict W. J. Irwin
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  • A rose by any other name... Thank you, I haven't heard much of this transform at the time I asked this – Yuriy S Jun 02 '17 at 12:06
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    $ \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 – reuns Jun 02 '17 at 14:21
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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 $$ \boxed{I =\sum_{k=0}^{\infty} \frac{(-1)^{n} n !}{(k+1)^{n+1}} =(-1)^{n} n ! \zeta (n+1)} $$

Lai
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