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.

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    $\begingroup$ Looks good to me. $\endgroup$ – Ali Caglayan Mar 1 '16 at 20:07
  • $\begingroup$ well done (+1) ! $\endgroup$ – tired Mar 1 '16 at 21:17
  • $\begingroup$ 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. $\endgroup$ – tired Mar 2 '16 at 8:26

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

  • $\begingroup$ A rose by any other name... Thank you, I haven't heard much of this transform at the time I asked this $\endgroup$ – Yuriy S Jun 2 '17 at 12:06
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    $\begingroup$ $ \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 $\endgroup$ – reuns Jun 2 '17 at 14:21

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