If $a_0 = -1$ and $\sum_{k=0}^n\frac{a_{n-k}}{k+1}=0$, then show $a_n = \int_{1}^{\infty}{\frac{dt}{t^n({\pi}^2+\log^2(t-1))}}$ Bonjour. The following is from IMO 2006:
The sequence $(a_n)$ is defined recursively by $a_0=-1$ and $$\sum_{k=0}^{n}{\frac{a_{n-k}}{k+1}}=0, n\geq 1. $$
In the shortlist, the commentator said that this relation holds: 
$$a_n=\int_{1}^{\infty}{\frac{dt}{t^n({\pi}^2+\log^2(t-1))}} \qquad\forall n\geq 1 \tag{1}$$

Question: How to show the relation $(1)$?

 A: The claim is equivalent to
$$
\left(a*\frac1{i+1}\right)_n = -1_{n=0}
$$ holds for every $n\ge 0$, where $$
\left( a*\frac1{i+1}\right)_n =\sum_{k=0}^n a_{n-i}\frac{1}{i+1}
$$ is the convolution of $(a_i)_{i\ge 0}$ and $(\frac1{i+1})_{i\ge 0}$ and
$$
a_n =\int_{1}^{\infty}\frac1{t^n}\frac{\mathrm d t}{{\pi}^2+\log^2(t-1)},\quad n\ge 1,
$$$$
a_0=-1.
$$
In view of the generating function of $(\frac1{i+1})_{i\ge 0}$, which is $-\frac{\ln(1-x)}x$, it suffices to prove that for $0<x<1$,
$$
\sum_{n\ge 0} a_n x^n=-1+\sum_{n\ge 1}a_n x^n = \frac{x}{\ln(1-x)}.
$$  We find that
$$\begin{eqnarray}
\sum_{n=1}^\infty a_n x^n &=& \int_{1}^{\infty}\sum_{n=1}^\infty \left(\frac x t\right)^n\frac{\mathrm d t}{{\pi}^2+\log^2(t-1)}\\ &=&\int_{-\infty}^{\infty}\sum_{n=1}^\infty \left(\frac x {1+e^u}\right)^n\frac{e^u\mathrm d u}{{\pi}^2+u^2}\ \  \quad(t=1+e^u)\\
&=& \int_{-\infty}^{\infty}\frac{x}{(1-x)+e^u}\frac{e^u\mathrm d u}{{\pi}^2+u^2}\\
&=&\int_{\Im z =0}\frac{x}{(1-x)+e^z}\frac{e^z\mathrm d z}{{\pi}^2+z^2}.
\end{eqnarray}$$ We also observe that
$$
\lim_{c\to \infty}\int_{\Im z =c}\frac{x}{(1-x)+e^z}\frac{e^z\mathrm d z}{{\pi}^2+z^2}=0, 
$$ which implies by residue theorem:
$$\begin{eqnarray}
\int_{\Im z =0}\frac{x}{(1-x)+e^z}\frac{e^z\mathrm d z}{{\pi}^2+z^2}&=&2\pi i \sum_{z:\Im(z)>0} \text{res}_{z}\frac{x}{(1-x)+e^z}\frac{e^z}{{\pi}^2+z^2}\\
&=&1+2\pi i x\sum_{j\ge 0} \frac{1}{\left((2j+1)\pi i + \ln(1-x)\right)^2 +\pi^2}\\
&=&1+x\sum_{j\ge 0} \left(\frac{1}{(2j+1)\pi i + \ln(1-x)-\pi i} \\ -\frac{1}{(2j+1)\pi i + \ln(1-x)+\pi i}\right)\\
&=&1+ \frac{x}{\ln(1-x)}.
\end{eqnarray}$$ (Here, residues are calculated at $z=\pi i$ and $(2j+1)\pi i +\ln(1-x)$ where $j\ge 0$.)
This shows $$
\sum_{n\ge 0}a_n x^n = \frac{x}{\ln(1-x)}
$$ and the claim follows.
