Solving the sequential equation $\sum_{k=0}^{n-1}\frac{a_{n-k}}{k+1}=\frac{1}{n+1}$ $a_1,a_2,a_3,...$ is a sequence of real numbers such that for $n\in \mathbb{N}$:
$$\sum_{k=0}^{n-1}\frac{a_{n-k}}{k+1}=\frac{1}{n+1}$$
How can I prove:
$$a_n=\int_1^\infty{\frac{dx}{x^n(\pi^2+\ln^2(x-1))}}$$
 A: As already noted by Greg Martin, if you set
$$
\phi(z):=\sum_{n\ge 1} a_n z^n,
$$
then the recurrence immediately gives
$$
\phi(z) (-\frac{\log (1-z)}{z})=-\frac{\log (1-z)}{z}-1.
$$
Therefore,
$$
\phi(z)=1+\frac{z}{\log (1-z)},
$$
so, for $n\ge 1$,
$$
a_n=[z^n] \frac{z}{\log (1-z)}.
$$
This can be rewritten as
$$
a_n=\frac{1}{2\pi i} \int \frac{1}{z^{n} \log (1-z)} \, dz \qquad (1)
$$
where the integral is taken over a small circle $C$ around the origin, traversed in the counterclockwise direction.  
Let the function $z\mapsto \log (1-z)$ have a branch cut along the positive real axis starting at $z=1$.  Let $C'$ be the integration path $L\cup D\cup L'\cup E$, where $L$ is the line segment from $1+\epsilon+\epsilon i$ to $M+\epsilon i$ (on one side of the branch cut), $D$ is the arc of the circle $|z|=\sqrt{M^2+\epsilon^2}$ from $M+\epsilon i$ to $M-\epsilon i$, $L'$ is the line segment from $M-\epsilon i$ to $1+\epsilon-\epsilon i$  (on the other side of the branch cut), and $E$ is the arc of the circle $|z-1|=\sqrt{2}\epsilon $ from $1+\epsilon-\epsilon i$ to $1+\epsilon+\epsilon i$.  The path is traversed in an overall counterclockwise direction around the origin and never crosses the branch cut.
Since $C$ can be deformed into $C'$ within the region where the integrand is analytic, the integration path in $(1)$ can be changed from $C$ to $C'$ without affecting the value of the integral.  Let $\epsilon\to 0$.  Then the integral on $E$ is $O(\epsilon |\log \epsilon|^{-1})$ and so approaches $0$, the integral over $L$ approaches
$$
\int_1^M \frac{1}{z^n (\log(z-1)-\pi i)} \, dz,\qquad (2)
$$
and the integral over $L'$ approaches
$$
-\int_1^M \frac{1}{z^n (\log(z-1) + \pi i)} \, dz.\qquad (3)
$$
Since the integral on $D$ is $O(M^{-(n-1)} (\log M)^{-1})$, combining $(1)$, $(2)$ and $(3)$ gives
$$a_n=\int_1^M \frac{1}{z^n (\log^2 (z-1) + \pi^2)} \, dz+O(M^{-(n-1)} (\log M)^{-1}).$$
Now let $M\to\infty$ to give the desired result.
