On the integral $\int_1^\infty\big(\{x\}^n-\frac1{n+1}\big)\frac{dx}x$ According to Dirichlet's test (integral version),
$$
I_n=\int_1^\infty\big(\{x\}^n-\frac1{n+1}\big)\frac{dx}x
$$
converges, where $n$ is a positive integer and $\{x\}$ denotes the fractional part of $x$.  Using series, I found out the values of $I_1$ and $I_2$. $I_1=\frac12\ln(2\pi)-1$ and $I_2=\frac12\ln(2\pi)-\frac12-2\ln A$, where $A$ denotes Glaisher's constant.  
My Attempt to Generalize $I_n$ 
$$I_n=\sum_{m=1}^\infty\int_0^1\big(t^n-\frac1{n+1}\big)\frac{dt}{t+m}\\
=\sum_{m=1}^\infty P_n(m)-\frac1{n+1}\ln\big(1+\frac1m\big)+m^n(-1)^n\ln\big(1+\frac1m\big)\\
=\sum_{m=1}^\infty-\frac1{n+1}\ln\big(1+\frac1m\big)+\frac1{(n+1)m}-\frac1{(n+2)m^2}+\cdots\\
=\frac\gamma{n+1}+\sum_{k=2}^\infty\frac{(-1)^{k-1}\zeta(k)}{n+k}$$where $P_n$ is a polynomial with $\deg P_n=n-1$ and $\gamma$ denotes Euler's constant. 
My questions are:  

(i) Is my answer right? 
(ii) If my answer is right, can I make it a little bit more
  simplified? 
(iii) How to find the value of $I_3$?

Edit: the convergence test of $I_n$
Denote $F(x)=\int_1^x\{t\}^n-\frac1{n+1}dt$, we have
$$F(x+1)-F(x)=\int_x^{x+1}\{t\}^ndt-\frac1{n+1}=0.$$
Obviously, $F(x)$ is bounded in $[0,1]$. So $F(x)$ is bounded in $\mathbb{R}$. Also, $1/x$ is a decreasing function in $[1,+\infty)$ and $\lim_{x\to\infty}1/x=0$ Hence $I_n$ converges. 
 A: Your result in (i) is correct. In fact $I_n=-\int_0^1 x^n\psi(1+x)\,dx$, where $$\psi(1+x)=\frac{\Gamma'(1+x)}{\Gamma(1+x)}=-\gamma+\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+x}\right)=-\gamma-\sum_{k=2}^{\infty}\zeta(k)(-x)^{k-1}$$ is the digamma function (and the last series is valid for $|x|<1$).
Numerical experiments (and the values you have already) suggest $$-I_n=R_n+\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\zeta'(-k),$$ where $R_n$ are rational numbers. The sequence $(R_1,R_2,R_3,\ldots)$ begins with $$1,\ \frac{2}{3},\ \frac{7}{12},\ \frac{47}{90},\ \frac{167}{360},\ \frac{349}{840},\ \frac{481}{1260},\ \frac{6749}{18900},\ \frac{8329}{25200},\ \frac{7157}{23760},\ \ldots$$
What I have at the moment is a (rather tedious) proof of $$\color{blue}{-I_n=\frac{1}{n}+\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\left[\zeta'(-k)+\zeta(-k)\sum_{j=1}^{k}\frac{1}{j}\right]}$$ (thus $R_n=\frac{1}{n}-\sum_{k=0}^{n-1}\frac{(-1)^k}{k+1}\binom{n}{k}$$B_{k+1}$$H_k$; is there a simpler form?..).

My initial proof considered $\int_0^1 e^{-xt}\psi(1+x)\,dx$ as the generating function of the integrals. The Laplace transform of $\psi(1+x)$ (i.e. $\int_0^\infty$, which reduces to this integral using $\psi(1+x)=\psi(x)+1/x$) appears to be closely related to $\sum_{k=0}^{\infty}\xi'(-k)t^k$ where $\xi(s)=\zeta(s)/\Gamma(1-s)$ (I used $\xi(s)=\frac{1}{2\pi i}\int_\lambda\frac{z^{s-1}\,dz}{e^{-z}-1}$, with $\lambda$ encircling the negative real axis, to evaluate the sum); observe that $k!\xi'(-k)$ is "almost" the quantity in square brackets above.
Compared to all of this, the suggestions given by @reuns achieve the result almost instantly: $$I_n=\lim_{s\to0^+}\left(f_n(s)-\frac{1}{(n+1)s}\right),\quad f_n(s)=\int_1^\infty\{x\}^n x^{-s-1}\,dx\quad(\Re s>0)$$ and then, for $n>0$ and $\Re s>1$, using $s\int_1^\infty\lfloor x\rfloor x^{-s-1}\,dx=\zeta(s)$ and integration by parts, $$nf_{n-1}(s-1)=\zeta(s)-1+sf_n(s),\qquad\color{gray}{[\text{put }f_n(s)=(-1)^n n!\Gamma(-s)g_n(s)\ldots]}$$ which then holds analytically continued, and [by induction] we obtain $$\frac{f_n(s)}{\Gamma(-s)}=\frac{(-1)^{n-1}n!}{\Gamma(n+1-s)}+\sum_{k=0}^{n-1}\frac{(-1)^k n!}{(n-k)!}\frac{\zeta(s-k)-1}{\Gamma(k+1-s)}.$$ Now computing $I_n$ amounts to taking derivative at $s=0$.
