Converting the sum : $\sum\limits_{n=1}^{\infty}\frac1n\cot^{-1}(n)$ to an integral. The problem is pretty complicated in my opinion. I wanted to calculate the sum by converting it to integral. Thank you for help.
$$\sum\limits_{n=1}^{\infty}\frac1n\cot^{-1}(n)$$
 A: Note that
$$\cot^{-1}(x)=\int_x^\infty\frac{{\rm d}t}{1+t^2}$$
By letting $t\mapsto x/u$,
$$\cot^{-1}(x)=\int_0^1\frac{x{\rm~d}u}{u^2+x^2}$$
And so,
$$\frac1n\cot^{-1}(n)=\int_0^1\frac{{\rm d}u}{u^2+n^2}$$
Now we may use
$$\sum_{n=1}^\infty\frac1{u^2+n^2}=\frac\pi{2u}\coth(\pi u)-\frac1{2u^2}$$
And so,
$$\sum_{n=1}^\infty\frac1n\cot^{-1}(n)=\frac12\int_0^1\left(\frac\pi u\coth(\pi u)-\frac1{u^2}\right){\rm~d}u$$
which is hopefully what you wanted.
A: As an alternative approach,
$$\sum_{n\geq 1}\frac{1}{n}\arctan\frac{1}{n} = \sum_{n\geq 1}\sum_{k\geq 0}\frac{(-1)^k}{(2k+1) n^{2k+2}}=\sum_{k\geq 0}\frac{(-1)^k \zeta(2k+2)}{(2k+1)} $$
can be written as
$$ \sum_{k\geq 0}\frac{(-1)^k}{(2k+1)\cdot(2k+1)!}\int_{0}^{+\infty}\frac{x^{2k+1}}{e^x-1}\,dx = \color{red}{\int_{0}^{+\infty}\frac{\text{Si}(x)}{e^x-1}\,dx} $$
where $\text{Si}(x)=\int_{0}^{x}\frac{\sin(t)}{t}\,dt$. This is just a consequence of $\mathcal{L}\left(\text{Si } x\right)(s)=\frac{1}{s}\arctan\frac{1}{s}$.
A: Note that
$$\int_0^{\infty } \frac{\exp (-n x) \sin (x)}{x} \, dx=\cot ^{-1}(n)$$
And so,
$$\color{red}{\sum _{n=1}^{\infty } \frac{\cot ^{-1}(n)}{n}}=\sum _{n=1}^{\infty } \frac{\int_0^{\infty } \frac{\exp (-n x) \sin
   (x)}{x} \, dx}{n}=\int_0^{\infty } \frac{\sin (x) \sum _{n=1}^{\infty } \frac{\exp (-n x)}{n}}{x} \,
   dx=\int_0^{\infty } \frac{\sin (x) \left(-\log \left(1-e^{-x}\right)\right)}{x} \, dx=\color{red}{\int_0^{\infty } -\frac{\sin
   (x) \log \left(1-e^{-x}\right)}{x} \, dx}$$
